100 results found for "variable-numerator" in Class 10.
एक भिन्न का हर अंश से (4) अधिक है। यदि अंश और हर दोनों में (2) जोड़ने पर भिन्न \(\frac{3}{5}\) हो जाती है तो मूल अंश क्या है?
The denominator of a fraction is (4) more than its numerator. If (2) is added to both numerator and denominator, the fraction becomes \(\frac{3}{5}\). What is the original numerator?
#quadratic equations
#fraction word problem
#linear reduction
A (4)
B (5)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
The numerator is (x) and denominator is (x+4). From \(\frac{x+2}{x+6}=\frac{3}{5}\), (x=4).
Step 2
Why this answer is correct
The correct answer is A. (4). The numerator is (x) and denominator is (x+4). From \(\frac{x+2}{x+6}=\frac{3}{5}\), (x=4).
Step 3
Exam Tip
अंश (x) और हर (x+4) है। \(\frac{x+2}{x+6}=\frac{3}{5}\) से (x=4) मिलता है।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश में (3) और हर में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (3) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#word-problem-fraction-substitution
A \(\frac{7}{12}\)
B \(\frac{8}{13}\)
C \(\frac{9}{14}\)
D \(\frac{10}{15}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{12}\)
Step 1
Concept
Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{12}\). Let the numerator be (x) and denominator be (x+5). From \(\frac{x+3}{x+6}=\frac{2}{3}\), solve carefully and verify the original fraction.
Step 3
Exam Tip
अंश (x) और हर (x+5) लें। \(\frac{x+3}{x+6}=\frac{2}{3}\) से (x=3), इसलिए मूल भिन्न \(\frac{3}{8}\) नहीं; विकल्प जांचें।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश और हर दोनों में (2) जोड़ने पर भिन्न \(\frac{4}{5}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to both numerator and denominator, the fraction becomes \(\frac{4}{5}\). What is the original fraction?
#word-problem
#fraction
#substitution
A \(\frac{9}{12}\)
B \(\frac{10}{13}\)
C \(\frac{11}{14}\)
D \(\frac{12}{15}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{10}{13}\)
Step 1
Concept
Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10}{13}\). Let the denominator be (y), so the numerator is (y-3). From \(\frac{y-1}{y+2}=\frac{4}{5}\), (y=13), so the fraction is \(\frac{10}{13}\).
Step 3
Exam Tip
मान लें हर (y) है तो अंश (y-3)। \(\frac{y-1}{y+2}=\frac{4}{5}\) से (y=13), इसलिए भिन्न \(\frac{10}{13}\) है।
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एक भिन्न में अंश हर से (3) कम है। यदि अंश में (2) और हर में (1) जोड़ने पर भिन्न \(\frac{3}{4}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the numerator is (3) less than the denominator. If (2) is added to the numerator and (1) to the denominator, the fraction becomes \(\frac{3}{4}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{5}{8}\)
B \(\frac{6}{9}\)
C \(\frac{7}{10}\)
D \(\frac{8}{11}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{7}{10}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7}{10}\). Let the numerator be (x) and denominator be (y), giving (y-x=3) and \(\frac{x+2}{y+1}=\frac{3}{4}\). In exams, solve the simple linear equations after cross multiplication.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y-x=3) और \(\frac{x+2}{y+1}=\frac{3}{4}\) बनता है। परीक्षा में क्रॉस गुणा के बाद सरल रैखिक समीकरण हल करें।
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एक भिन्न में हर अंश से (5) अधिक है। यदि अंश और हर दोनों में (1) जोड़ने पर भिन्न \(\frac{2}{3}\) हो जाती है, तो मूल भिन्न क्या है?
In a fraction, the denominator is (5) more than the numerator. If (1) is added to both numerator and denominator, the fraction becomes \(\frac{2}{3}\). What is the original fraction?
#linear equations
#fraction
#substitution
#class 10
A \(\frac{9}{14}\)
B \(\frac{8}{13}\)
C \(\frac{7}{12}\)
D \(\frac{6}{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{14}\)
Step 1
Concept
Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{14}\). Let the numerator be (x) and denominator be (y), so (y=x+5) and \(\frac{x+1}{y+1}=\frac{2}{3}\). In exams, cross multiply when converting a fraction into an equation.
Step 3
Exam Tip
अंश (x) और हर (y) मानकर (y=x+5) और \(\frac{x+1}{y+1}=\frac{2}{3}\) बनता है। परीक्षा में भिन्न को समीकरण में बदलते समय क्रॉस गुणा करें।
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एक भिन्न का हर अंश से (3) अधिक है। यदि भिन्न और उसके व्युत्क्रम का योग \(\frac{29}{10}\) है तो अंश क्या है?
The denominator of a fraction is (3) more than its numerator. If the sum of the fraction and its reciprocal is \(\frac{29}{10}\), what is the numerator?
#quadratic equations
#fraction
#reciprocal
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 2
Why this answer is correct
The correct answer is A. (2). The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 3
Exam Tip
भिन्न \(\frac{x}{x+3}\) है। \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\) से (x=2) या (x=15) आता है और विकल्पों में (2) सही है।
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एक चर के बहुपद में चर की घातें कैसी होनी चाहिए?
What should the powers of the variable be in a polynomial in one variable?
#powers
#definition
#polynomial
A ऋणात्मक पूर्णांक / Negative integers
B भिन्न संख्याएं / Fractions
C अऋणात्मक पूर्णांक / Non-negative integers
D केवल अभाज्य संख्याएं / Only prime numbers
Explanation opens after your attempt
Correct Answer
C. अऋणात्मक पूर्णांक / Non-negative integers
Step 1
Concept
In a polynomial, powers of the variable are non-negative integers like \(0,1,2,\ldots\). This is the basic identification rule.
Step 2
Why this answer is correct
The correct answer is C. अऋणात्मक पूर्णांक / Non-negative integers. In a polynomial, powers of the variable are non-negative integers like \(0,1,2,\ldots\). This is the basic identification rule.
Step 3
Exam Tip
बहुपद में चर की घातें \(0,1,2,\ldots\) जैसी अऋणात्मक पूर्णांक होती हैं। यही बहुपद की मूल पहचान है।
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बहुपद (p(u)=13u-3 -2u+9) में चर कौन सा है?
What is the variable in the polynomial (p(u)=13u-3 -2u+9)?
#variable
#one variable
#polynomial
A (p)
B (u)
C (13)
D (9)
Explanation opens after your attempt
Step 1
Concept
The letter whose values can change is the variable. Here the variable is (u).
Step 2
Why this answer is correct
The correct answer is B. (u). The letter whose values can change is the variable. Here the variable is (u).
Step 3
Exam Tip
जिस अक्षर के मान बदल सकते हैं वह चर होता है। यहां चर (u) है।
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बहुपद (p(t)=6t-2 -7t+3) में चर कौन सा है?
What is the variable in the polynomial (p(t)=6t-2 -7t+3)?
#variable
#one variable
#polynomial
A (p)
B (t)
C (6)
D (3)
Explanation opens after your attempt
Step 1
Concept
The letter whose value can vary is the variable. Here the variable is (t).
Step 2
Why this answer is correct
The correct answer is B. (t). The letter whose value can vary is the variable. Here the variable is (t).
Step 3
Exam Tip
जिस अक्षर का मान बदल सकता है वह चर है। यहां चर (t) है।
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एक धनात्मक भिन्न का हर अंश से (4) अधिक है। भिन्न और उसके व्युत्क्रम का योग \(\frac{41}{20}\) है। भिन्न क्या है?
In a positive fraction, the denominator is (4) more than the numerator. The sum of the fraction and its reciprocal is \(\frac{41}{20}\). What is the fraction?
#quadratic equations
#fraction
#reciprocal
A \(\frac{4}{8}\)
B \(\frac{5}{9}\)
C \(\frac{6}{10}\)
D \(\frac{8}{12}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{5}{9}\)
Step 1
Concept
Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5}{9}\). Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 3
Exam Tip
भिन्न \(\frac{x}{x+4}\) हो, तो \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\)। इससे (x=5), इसलिए भिन्न \(\frac{5}{9}\) है।
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यदि सरलतम हर \(q=2^5\cdot 5^5\) है और अंश (10) से विभाज्य नहीं है, तो दशमलव प्रसार के बारे में क्या निश्चित है?
If the reduced denominator is \(q=2^5\cdot 5^5\) and the numerator is not divisible by (10), what is certain about the decimal expansion?
#powers-of-10
#terminating-decimal
#lowest-form
#concept
A ठीक (5) स्थानों पर समाप्त / Terminates exactly after (5) places
B ठीक (10) स्थानों पर समाप्त / Terminates exactly after (10) places
C असांत आवर्ती / Non-terminating recurring
D अंश पर निर्भर करके कभी भी / Depends only on the numerator
Explanation opens after your attempt
Correct Answer
A. ठीक (5) स्थानों पर समाप्त / Terminates exactly after (5) places
Step 1
Concept
The reduced denominator is \(10^5\), so the decimal terminates exactly after (5) places. The numerator condition indicates no further cancellation.
Step 2
Why this answer is correct
The correct answer is A. ठीक (5) स्थानों पर समाप्त / Terminates exactly after (5) places. The reduced denominator is \(10^5\), so the decimal terminates exactly after (5) places. The numerator condition indicates no further cancellation.
Step 3
Exam Tip
सरलतम हर \(10^5\) है, इसलिए दशमलव ठीक (5) स्थानों पर समाप्त होगा। अंश की दी गई बात अतिरिक्त कटौती न होने का संकेत देती है।
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\(\frac{1}{2^4\cdot 5^6}\) के हर को \(10^6\) बनाने के लिए अंश और हर को किससे गुणा करना चाहिए?
By what should numerator and denominator of \(\frac{1}{2^4\cdot 5^6}\) be multiplied to make the denominator \(10^6\)?
#denominator-conversion
#powers-of-10
#terminating-decimal
#real-numbers
A \(2^2\)
B \(5^2\)
C \(10^2\)
D \(2^4\)
Explanation opens after your attempt
Correct Answer
A. \(2^2\)
Step 1
Concept
\(10^6=2^6\cdot 5^6\).
Step 2
Why this answer is correct
The denominator \(2^4\cdot 5^6\) lacks \(2^2\). So multiply numerator and denominator by \(2^2\).
Step 3
Exam Tip
Complete the deficiency of the prime with the smaller exponent. चरण 1: \(10^6=2^6\cdot 5^6\) होता है। चरण 2: हर \(2^4\cdot 5^6\) में \(2^2\) की कमी है। इसलिए अंश और हर को \(2^2\) से गुणा करेंगे। चरण 3: जिसकी घात कम हो, उसी की कमी पूरी करें।
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\(\frac{1}{2^r5^s}\) को किसी पूर्णांक अंश के साथ \(10^8\) हर वाली भिन्न के रूप में लिखना हो, तो कौन-सी शर्त आवश्यक है?
To write \(\frac{1}{2^r5^s}\) as a fraction with denominator \(10^8\) and an integer numerator, which condition is necessary?
#powers-of-10
#terminating-decimal
#exponents
#advanced
A \(r\leq 8\) और \(s\leq 8\) / \(r\leq 8\) and \(s\leq 8\)
B (r+s=8)
C (r=s=8)
D (r>8) या (s>8) / (r>8) or (s>8)
Explanation opens after your attempt
Correct Answer
A. \(r\leq 8\) और \(s\leq 8\) / \(r\leq 8\) and \(s\leq 8\)
Step 1
Concept
\(10^8=2^8\cdot 5^8\).
Step 2
Why this answer is correct
The denominator \(2^r5^s\) must divide \(10^8\), so \(r\leq 8\) and \(s\leq 8\).
Step 3
Exam Tip
When converting to denominator \(10^k\), remember the divisor condition. चरण 1: \(10^8=2^8\cdot 5^8\) है। चरण 2: हर \(2^r5^s\) को \(10^8\) का भाजक होना चाहिए, इसलिए \(r\leq 8\) और \(s\leq 8\)। चरण 3: हर को \(10^k\) में बदलते समय भाजक की शर्त याद रखें।
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यदि \(x=\frac{3}{2^m5^n}\) और (m<n), तो (x) को सांत दशमलव में लिखने के लिए अंश और हर को किससे गुणा करना चाहिए?
If \(x=\frac{3}{2^m5^n}\) and (m<n), by what should numerator and denominator be multiplied to write (x) as a terminating decimal?
#decimal-conversion
#exponents
#terminating-decimal
#class-10
A \(2^{n-m}\)
B \(5^{n-m}\)
C \(10^{n-m}\)
D \(2^m5^n\)
Explanation opens after your attempt
Correct Answer
A. \(2^{n-m}\)
Step 1
Concept
The denominator is \(2^m5^n\), and the power of (5) is larger.
Step 2
Why this answer is correct
To make \(10^n=2^n5^n\), the power of (2) must be increased to (n). So multiply by \(2^{n-m}\).
Step 3
Exam Tip
First identify which prime power is short. चरण 1: हर \(2^m5^n\) है और (5) की घात अधिक है। चरण 2: \(10^n=2^n5^n\) बनाने के लिए (2) की घात (n) तक बढ़ानी होगी। इसलिए \(2^{n-m}\) से गुणा करेंगे। चरण 3: कमी किस अभाज्य घात में है, पहले वही पहचानें।
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किसी परिमेय संख्या का सरलतम हर \(q=2^4\cdot 5^4\) है। यदि उसका अंश (10) से विभाज्य नहीं है, तो दशमलव प्रसार के बारे में सबसे उचित निष्कर्ष क्या है?
A rational number has reduced denominator \(q=2^4\cdot 5^4\). If its numerator is not divisible by (10), what is the most suitable conclusion about its decimal expansion?
#terminating-decimal
#powers-of-10
#real-numbers
#exam-tip
A ठीक (4) दशमलव स्थानों पर समाप्त होगा / It terminates exactly after (4) decimal places
B ठीक (8) दशमलव स्थानों पर समाप्त होगा / It terminates exactly after (8) decimal places
C असांत आवर्ती होगा / It will be non-terminating recurring
D दशमलव स्थान अंश पर निर्भर नहीं करते / Decimal places do not depend on the numerator at all
Explanation opens after your attempt
Correct Answer
A. ठीक (4) दशमलव स्थानों पर समाप्त होगा / It terminates exactly after (4) decimal places
Step 1
Concept
\(2^4\cdot 5^4=10^4\).
Step 2
Why this answer is correct
A reduced denominator of \(10^4\) gives a decimal terminating after (4) places. The numerator condition assures no hidden further reduction.
Step 3
Exam Tip
If the reduced denominator is \(10^k\), think of (k) decimal places. चरण 1: \(2^4\cdot 5^4=10^4\) है। चरण 2: सरलतम हर \(10^4\) होने से दशमलव (4) स्थानों पर समाप्त होगा। अंश (10) से विभाज्य नहीं होने की बात यह भरोसा देती है कि आगे और सरलता नहीं छिपी है। चरण 3: सरलतम हर \(10^k\) हो तो (k) दशमलव स्थान सोचें।
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यदि कोई भिन्न सरलतम रूप में है, तो उसके अंश और हर के बारे में क्या सही है?
If a fraction is in lowest form, what is true about its numerator and denominator?
#lowest form
#coprime
#fraction
#class 10
A वे दोनों बराबर होते हैं / They are equal
B वे सहअभाज्य होते हैं / They are coprime
C वे दोनों शून्य होते हैं / They are both zero
D वे दोनों अपरिमेय होते हैं / They are both irrational
Explanation opens after your attempt
Correct Answer
B. वे सहअभाज्य होते हैं / They are coprime
Step 1
Concept
Lowest form means the fraction cannot be reduced further.
Step 2
Why this answer is correct
So the numerator and denominator have only (1) as a common factor.
Step 3
Exam Tip
This fact is important in irrationality proofs. चरण 1: सरलतम रूप का अर्थ है कि भिन्न को और छोटा नहीं किया जा सकता। चरण 2: इसलिए अंश और हर का साझा गुणनखंड केवल (1) होता है। चरण 3: अपरिमेयता के प्रमाण में यही बात जरूरी होती है।
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निम्न में से एक चर वाला बहुपद कौन सा नहीं है?
Which of the following is not a polynomial in one variable?
#one variable
#two variables
#identification
A \(2x^2+3x+1\)
B \(5t^3-t\)
C \(x^2+y+1\)
D (9z-4)
Explanation opens after your attempt
Correct Answer
C. \(x^2+y+1\)
Step 1
Concept
\(x^2+y+1\) has two variables (x) and (y). Therefore it is not a polynomial in one variable.
Step 2
Why this answer is correct
The correct answer is C. \(x^2+y+1\). \(x^2+y+1\) has two variables (x) and (y). Therefore it is not a polynomial in one variable.
Step 3
Exam Tip
\(x^2+y+1\) में दो चर (x) और (y) हैं। इसलिए यह एक चर वाला बहुपद नहीं है।
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इनमें से कौन सा एक चर में बहुपद है?
Which of the following is a polynomial in one variable?
#polynomial
#one variable
#identification
A (p(x)=3x-2 -5x+7)
B (p(x)=\frac{2}{x}+1)
C (p(x)=\sqrt{x}+4)
D (p(x)=x^{-2}+3)
Explanation opens after your attempt
Correct Answer
A. (p(x)=3x-2 -5x+7)
Step 1
Concept
In a polynomial in one variable, powers of the variable are non-negative integers. In exams, reject negative powers and roots of the variable.
Step 2
Why this answer is correct
The correct answer is A. (p(x)=3x-2 -5x+7). In a polynomial in one variable, powers of the variable are non-negative integers. In exams, reject negative powers and roots of the variable.
Step 3
Exam Tip
एक चर के बहुपद में चर की घात केवल अशून्य पूर्णांक या शून्य होती है। परीक्षा में ऋणात्मक घात और मूल से बचें।
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कौन सा बहुपद एक चर (x) में है?
Which polynomial is in one variable (x)?
#polynomials
#one-variable
#easy
#concept
A \(x^2+2x+1\)
B (x+y+1)
C \(a^2+b^2\)
D (xy+3)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x+1\)
Step 1
Concept
A polynomial in one variable (x) contains only the variable (x). The other options have more than one variable.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x+1\). A polynomial in one variable (x) contains only the variable (x). The other options have more than one variable.
Step 3
Exam Tip
एक चर (x) में बहुपद में केवल (x) चर होता है। अन्य विकल्पों में एक से अधिक चर हैं।
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बहुपद (p(t)=8t-2 -3t+4) में चर कौन-सा है?
Which is the variable in the polynomial (p(t)=8t-2 -3t+4)?
#variable
#polynomial_notation
#basics
A (p)
B (t)
C (8)
D (4)
Explanation opens after your attempt
Step 1
Concept
In (p(t)), the changing symbol is (t). So the variable is (t).
Step 2
Why this answer is correct
The correct answer is B. (t). In (p(t)), the changing symbol is (t). So the variable is (t).
Step 3
Exam Tip
(p(t)) में बदलने वाला चिन्ह (t) है। इसलिए चर (t) है।
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कौन सा व्यंजक बहुपद नहीं है क्योंकि चर की घात भिन्न है?
Which expression is not a polynomial because the variable has a fractional power?
#not polynomial
#fractional power
#definition
A \(x^4-2x+5\)
B \(3x^2+\sqrt{2}x-7\)
C \(x^{\frac{3}{2}}+x+1\)
D \(9x^3-4x^2\)
Explanation opens after your attempt
Correct Answer
C. \(x^{\frac{3}{2}}+x+1\)
Step 1
Concept
In \(x^{\frac{3}{2}}\), the power of the variable is fractional, so it is not a polynomial. In a polynomial, powers are non-negative integers.
Step 2
Why this answer is correct
The correct answer is C. \(x^{\frac{3}{2}}+x+1\). In \(x^{\frac{3}{2}}\), the power of the variable is fractional, so it is not a polynomial. In a polynomial, powers are non-negative integers.
Step 3
Exam Tip
\(x^{\frac{3}{2}}\) में चर की घात भिन्न है, इसलिए यह बहुपद नहीं है। बहुपद में घातें अऋणात्मक पूर्णांक होती हैं।
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कौन सा विकल्प एक चर में बहुपद है लेकिन रैखिक नहीं है?
Which option is a polynomial in one variable but not linear?
#classification
#quadratic
#not linear
A (5x-8)
B \(x^2+2x+3\)
C \(x+\frac{1}{x}\)
D \(\sqrt{x}+2\)
Explanation opens after your attempt
Correct Answer
B. \(x^2+2x+3\)
Step 1
Concept
\(x^2+2x+3\) is a polynomial and its degree is (2). Therefore it is not linear.
Step 2
Why this answer is correct
The correct answer is B. \(x^2+2x+3\). \(x^2+2x+3\) is a polynomial and its degree is (2). Therefore it is not linear.
Step 3
Exam Tip
\(x^2+2x+3\) बहुपद है और इसकी घात (2) है। इसलिए यह रैखिक नहीं है।
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निम्न में से कौन सा बहुपद (x) में है लेकिन (y) में नहीं है?
Which of the following is a polynomial in (x) but not in (y)?
#one-variable
#variable-identification
#polynomial
A \(x^2+3x+1\)
B \(y^2+3y+1\)
C (xy+1)
D (x+y+1)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x+1\)
Step 1
Concept
\(x^2+3x+1\) has only the variable (x). So it is a one-variable polynomial in (x).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x+1\). \(x^2+3x+1\) has only the variable (x). So it is a one-variable polynomial in (x).
Step 3
Exam Tip
\(x^2+3x+1\) में केवल (x) चर है। इसलिए यह (x) में एक चर वाला बहुपद है।
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कौन सा विकल्प एक चर में बहुपद है लेकिन द्विघात नहीं है?
Which option is a polynomial in one variable but not quadratic?
#classification
#cubic
#not quadratic
A \(x^2+2x+1\)
B \(4x^3+x-9\)
C \(\frac{1}{x}+2\)
D \(\sqrt{x}+3\)
Explanation opens after your attempt
Correct Answer
B. \(4x^3+x-9\)
Step 1
Concept
\(4x^3+x-9\) is a polynomial and its degree is (3). Therefore it is not quadratic.
Step 2
Why this answer is correct
The correct answer is B. \(4x^3+x-9\). \(4x^3+x-9\) is a polynomial and its degree is (3). Therefore it is not quadratic.
Step 3
Exam Tip
\(4x^3+x-9\) बहुपद है और इसकी घात (3) है। इसलिए यह द्विघात नहीं है।
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कौन-सा व्यंजक एक चर वाला बहुपद है जिसमें वास्तविक गुणांक हैं?
Which expression is a polynomial in one variable with real coefficients?
#real_coefficients
#polynomial_identification
#irrational_coefficient
A \(\sqrt{2}x^2-3x+1\)
B \(\sqrt{x}+2\)
C \(x^{-2}+1\)
D \(\frac{1}{x}+3\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}x^2-3x+1\)
Step 1
Concept
\(\sqrt{2}\) is a real number and the powers of (x) are whole numbers. So it is a polynomial.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}x^2-3x+1\). \(\sqrt{2}\) is a real number and the powers of (x) are whole numbers. So it is a polynomial.
Step 3
Exam Tip
\(\sqrt{2}\) एक वास्तविक संख्या है और (x) की घातें पूर्ण संख्याएँ हैं। इसलिए यह बहुपद है।
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कौन-सा व्यंजक बहुपद नहीं है क्योंकि चर हर में है?
Which expression is not a polynomial because the variable is in the denominator?
#variable_denominator
#not_polynomial
#negative_power
A \(x^2+2\)
B \(\frac{3}{x}+1\)
C (4x-5)
D \(8x^3\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{3}{x}+1\)
Step 1
Concept
\(\frac{3}{x}=3x^{-1}\), so the variable has a negative power. Therefore it is not a polynomial.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3}{x}+1\). \(\frac{3}{x}=3x^{-1}\), so the variable has a negative power. Therefore it is not a polynomial.
Step 3
Exam Tip
\(\frac{3}{x}=3x^{-1}\) है और चर की घात ऋणात्मक है। इसलिए यह बहुपद नहीं है।
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कौन-सा व्यंजक (z) में एक चर वाला बहुपद है?
Which expression is a polynomial in one variable (z)?
#one_variable
#polynomial_identification
#z_variable
A \(z^4-2z+1\)
B \(z^{-2}+3\)
C \(\frac{1}{z}+5\)
D (z+w)
Explanation opens after your attempt
Correct Answer
A. \(z^4-2z+1\)
Step 1
Concept
The expression \(z^4-2z+1\) has only (z) and no negative powers. So it is a polynomial in (z).
Step 2
Why this answer is correct
The correct answer is A. \(z^4-2z+1\). The expression \(z^4-2z+1\) has only (z) and no negative powers. So it is a polynomial in (z).
Step 3
Exam Tip
\(z^4-2z+1\) में केवल (z) है और घातें ऋणात्मक नहीं हैं। इसलिए यह (z) में बहुपद है।
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\(2x^2+3xy+1\) एक चर वाला बहुपद क्यों नहीं है?
Why is \(2x^2+3xy+1\) not a polynomial in one variable?
#one_variable
#not_polynomial
#concept
A क्योंकि इसमें दो चर हैं / Because it has two variables
B क्योंकि इसमें कोई स्थिर पद नहीं है / Because it has no constant term
C क्योंकि इसकी घात (2) है / Because its degree is (2)
D क्योंकि गुणांक वास्तविक नहीं हैं / Because coefficients are not real
Explanation opens after your attempt
Correct Answer
A. क्योंकि इसमें दो चर हैं / Because it has two variables
Step 1
Concept
It contains both (x) and (y). A polynomial in one variable should contain only one variable.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि इसमें दो चर हैं / Because it has two variables. It contains both (x) and (y). A polynomial in one variable should contain only one variable.
Step 3
Exam Tip
इसमें (x) और (y) दोनों चर हैं। एक चर वाले बहुपद में केवल एक ही चर होना चाहिए।
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कौन-सा व्यंजक (x) में एक चर वाला बहुपद है?
Which expression is a polynomial in one variable (x)?
#polynomials
#one_variable
#identification
A \(3x^2-5x+7\)
B \(\frac{2}{x}+1\)
C \(\sqrt{x}+4\)
D (x+y)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-5x+7\)
Step 1
Concept
The expression \(3x^2-5x+7\) has only (x) and non-negative integer powers. In exams, the variable power must not be negative or fractional.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-5x+7\). The expression \(3x^2-5x+7\) has only (x) and non-negative integer powers. In exams, the variable power must not be negative or fractional.
Step 3
Exam Tip
\(3x^2-5x+7\) में केवल (x) है और घातें पूर्ण संख्याएँ हैं। परीक्षा में चर की घात ऋणात्मक या भिन्न नहीं होनी चाहिए।
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यदि \(\frac{n}{36}\) सबसे सरल रूप में है और (n) का (36) से कोई सामान्य गुणनखंड नहीं है, तो दशमलव प्रसार कैसा होगा?
If \(\frac{n}{36}\) is in lowest form and (n) has no common factor with (36), what type of decimal expansion will it have?
#variable-numerator
#recurring-decimal
#denominator-test
A समाप्त / Terminating
B असमाप्त आवर्ती / Non-terminating recurring
C असमाप्त अनावर्ती / Non-terminating non-recurring
D हमेशा पूर्णांक / Always an integer
Explanation opens after your attempt
Correct Answer
B. असमाप्त आवर्ती / Non-terminating recurring
Step 1
Concept
\(36=2^2\times3^2\).
Step 2
Why this answer is correct
Since the fraction is in lowest form, \(3^2\) remains in the denominator.
Step 3
Exam Tip
Exam tip: If factor (3) remains in the reduced denominator, the decimal does not terminate. चरण 1: \(36=2^2\times3^2\) है। चरण 2: भिन्न सबसे सरल रूप में है, इसलिए हर में \(3^2\) बचा रहेगा। चरण 3: परीक्षा सुझाव: सरल रूप में (3) बचने पर दशमलव समाप्त नहीं होता।
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यदि \(\frac{a}{40}\) सबसे सरल रूप में है, तो दशमलव प्रसार अधिकतम कितने स्थानों पर समाप्त हो सकता है?
If \(\frac{a}{40}\) is in lowest form, after at most how many decimal places can its decimal expansion terminate?
#variable-numerator
#decimal-places
#terminating
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(40=2^3\times5\).
Step 2
Why this answer is correct
The larger exponent is (3), so there can be at most (3) decimal places.
Step 3
Exam Tip
Exam tip: Whatever (a) is, the denominator in lowest form decides the decimal places. चरण 1: \(40=2^3\times5\) है। चरण 2: हर में (2) की बड़ी घात (3) है, इसलिए अधिकतम (3) स्थान होंगे। चरण 3: परीक्षा सुझाव: अंश (a) चाहे जो हो, न्यूनतम रूप में हर ही दशमलव स्थान तय करता है।
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यदि (p(x)=x-4 -5x-2 +4) है, तो (p(x)) का घात कितना है?
If (p(x)=x-4 -5x-2 +4), what is the degree of (p(x))?
#degree
#polynomial
#one-variable
A (4)
B (2)
C (5)
D (0)
Explanation opens after your attempt
Step 1
Concept
The highest power is \(x^4\), so the degree is (4). Terms with zero coefficients do not affect the degree.
Step 2
Why this answer is correct
The correct answer is A. (4). The highest power is \(x^4\), so the degree is (4). Terms with zero coefficients do not affect the degree.
Step 3
Exam Tip
सबसे बड़ी घात \(x^4\) है, इसलिए घात (4) है। शून्य गुणांक वाले पदों को घात तय करने में नहीं गिनते।
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यदि (p(x)=ax-2 +bx+c) में \(a\neq0\) है और (p(1)=p(-1)=0) है, तो (b) का मान क्या होगा?
If (p(x)=ax-2 +bx+c) with \(a\neq0\) and (p(1)=p(-1)=0), what is the value of (b)?
#polynomials
#one-variable
#values
A (0)
B (a)
C (c)
D (-a)
Explanation opens after your attempt
Step 1
Concept
(p(1)=a+b+c) and (p(-1)=a-b+c); subtracting gives (2b=0). In exams, use addition or subtraction for symmetric inputs.
Step 2
Why this answer is correct
The correct answer is A. (0). (p(1)=a+b+c) and (p(-1)=a-b+c); subtracting gives (2b=0). In exams, use addition or subtraction for symmetric inputs.
Step 3
Exam Tip
(p(1)=a+b+c) और (p(-1)=a-b+c) हैं, घटाने पर (2b=0) मिलता है। परीक्षा में सममित मानों पर जोड़-घटाव जल्दी करें।
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निम्न में से (y) में बहुपद कौन सा है?
Which of the following is a polynomial in (y)?
#polynomial-in-y
#one-variable
A \(y^2-3y+2\)
B \(\frac{1}{y}+4\)
C \(\sqrt{y}+5\)
D \(y^{-2}+1\)
Explanation opens after your attempt
Correct Answer
A. \(y^2-3y+2\)
Step 1
Concept
In \(y^2-3y+2\), the powers of (y) are whole numbers. A polynomial can be in one variable such as (y).
Step 2
Why this answer is correct
The correct answer is A. \(y^2-3y+2\). In \(y^2-3y+2\), the powers of (y) are whole numbers. A polynomial can be in one variable such as (y).
Step 3
Exam Tip
\(y^2-3y+2\) में (y) की घातें पूर्ण संख्याएँ हैं। बहुपद किसी एक चर जैसे (y) में भी हो सकता है।
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क्या \(4x^2-3\sqrt{x}+1\) (x) में बहुपद है?
Is \(4x^2-3\sqrt{x}+1\) a polynomial in (x)?
#not-polynomial
#root-variable
A हाँ / Yes
B नहीं / No
C केवल जब (x) धनात्मक हो / Only when (x) is positive
D केवल जब (x=1) हो / Only when (x=1)
Explanation opens after your attempt
Correct Answer
B. नहीं / No
Step 1
Concept
\(\sqrt{x}=x^{\frac{1}{2}}\), and \(\frac{1}{2}\) is not a whole number. So it is not a polynomial in (x).
Step 2
Why this answer is correct
The correct answer is B. नहीं / No. \(\sqrt{x}=x^{\frac{1}{2}}\), and \(\frac{1}{2}\) is not a whole number. So it is not a polynomial in (x).
Step 3
Exam Tip
\(\sqrt{x}=x^{\frac{1}{2}}\) है और \(\frac{1}{2}\) पूर्ण संख्या नहीं है। इसलिए यह (x) में बहुपद नहीं है।
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(p(x)=x-2 +3x+2) के लिए (p(1)) का मान क्या है?
For (p(x)=x-2 +3x+2), what is the value of (p(1))?
#polynomials
#one-variable
#value
#substitution
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
(p(1)=12 +3\cdot1+2=6). While evaluating, replace every (x) with the given number.
Step 2
Why this answer is correct
The correct answer is C. (6). (p(1)=12 +3\cdot1+2=6). While evaluating, replace every (x) with the given number.
Step 3
Exam Tip
(p(1)=12 +3\cdot1+2=6) है। मान निकालते समय हर (x) की जगह दी गई संख्या रखें।
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यदि (p(x)=2x-3 -5x+9) है, तो इसमें \(x^2\) का गुणांक क्या है?
If (p(x)=2x-3 -5x+9), what is the coefficient of \(x^2\)?
#polynomials
#one-variable
#coefficient
#missing-term
A (2)
B (-5)
C (9)
D (0)
Explanation opens after your attempt
Step 1
Concept
There is no \(x^2\)-term in this polynomial, so its coefficient is (0). The coefficient of a missing term is always taken as (0).
Step 2
Why this answer is correct
The correct answer is D. (0). There is no \(x^2\)-term in this polynomial, so its coefficient is (0). The coefficient of a missing term is always taken as (0).
Step 3
Exam Tip
इस बहुपद में \(x^2\) पद नहीं है, इसलिए उसका गुणांक (0) है। अनुपस्थित पद का गुणांक हमेशा (0) माना जाता है।
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निम्न में से कौन सा (x) में बहुपद है?
Which of the following is a polynomial in (x)?
#polynomials
#one-variable
#easy
#definition
A \(x^2+3x+5\)
B \(\frac{1}{x}+2\)
C \(\sqrt{x}+1\)
D \(x^{-2}+4\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x+5\)
Step 1
Concept
A polynomial has only non-negative integer powers of (x). In exams check the power of every term.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x+5\). A polynomial has only non-negative integer powers of (x). In exams check the power of every term.
Step 3
Exam Tip
बहुपद में (x) की घात केवल पूर्णांक और अशून्य नहीं बल्कि ऋणात्मक नहीं होनी चाहिए। परीक्षा में हर पद की घात जांचें।
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\(\frac{23}{2^5\cdot 5^9}\) को \(\frac{N}{10^9}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{23}{2^5\cdot 5^9}\) is written as \(\frac{N}{10^9}\), what is (N)?
#powers-of-10
#numerator-adjustment
#decimal-conversion
#expert
A (184)
B (368)
C (736)
D (1472)
Explanation opens after your attempt
Step 1
Concept
Since \(10^9=2^9\cdot 5^9\), the denominator lacks \(2^4\). Therefore \(N=23\cdot 16=368\).
Step 2
Why this answer is correct
The correct answer is B. (368). Since \(10^9=2^9\cdot 5^9\), the denominator lacks \(2^4\). Therefore \(N=23\cdot 16=368\).
Step 3
Exam Tip
\(10^9=2^9\cdot 5^9\) है इसलिए हर में \(2^4\) की कमी है। \(N=23\cdot 16=368\) होगा।
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\(\frac{11}{2^8\cdot 5^5}\) को \(\frac{N}{10^8}\) में बदलने पर (N) क्या होगा?
When \(\frac{11}{2^8\cdot 5^5}\) is converted into \(\frac{N}{10^8}\), what is (N)?
#denominator-conversion
#powers-of-10
#numerator-adjustment
#real-numbers
A (275)
B (1375)
C (2750)
D (6875)
Explanation opens after your attempt
Step 1
Concept
Since \(10^8=2^8\cdot 5^8\), the denominator lacks \(5^3\). Thus \(N=11\cdot 125=1375\).
Step 2
Why this answer is correct
The correct answer is B. (1375). Since \(10^8=2^8\cdot 5^8\), the denominator lacks \(5^3\). Thus \(N=11\cdot 125=1375\).
Step 3
Exam Tip
\(10^8=2^8\cdot 5^8\) है इसलिए हर में \(5^3\) की कमी है। \(N=11\cdot 125=1375\) होगा।
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\(\frac{37}{2^4\cdot 5^8}\) को \(\frac{N}{10^8}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{37}{2^4\cdot 5^8}\) is written as \(\frac{N}{10^8}\), what is (N)?
#powers-of-10
#numerator-adjustment
#terminating-decimal
#expert
A (296)
B (592)
C (1184)
D (23125)
Explanation opens after your attempt
Step 1
Concept
Since \(10^8=2^8\cdot 5^8\), the denominator lacks \(2^4\). Thus \(N=37\cdot 16=592\).
Step 2
Why this answer is correct
The correct answer is B. (592). Since \(10^8=2^8\cdot 5^8\), the denominator lacks \(2^4\). Thus \(N=37\cdot 16=592\).
Step 3
Exam Tip
\(10^8=2^8\cdot 5^8\) है इसलिए हर में \(2^4\) की कमी है। \(N=37\cdot 16=592\) होगा।
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\(\frac{19}{2^5\cdot 5^2}\) को \(\frac{N}{10^5}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{19}{2^5\cdot 5^2}\) is written as \(\frac{N}{10^5}\), what is (N)?
#powers-of-10
#numerator-adjustment
#terminating-decimal
#expert
A (475)
B (950)
C (2375)
D (4750)
Explanation opens after your attempt
Step 1
Concept
Since \(10^5=2^5\cdot 5^5\), the denominator lacks \(5^3\). Therefore \(N=19\cdot 125=2375\); multiply by the missing factor when making a power of (10).
Step 2
Why this answer is correct
The correct answer is C. (2375). Since \(10^5=2^5\cdot 5^5\), the denominator lacks \(5^3\). Therefore \(N=19\cdot 125=2375\); multiply by the missing factor when making a power of (10).
Step 3
Exam Tip
\(10^5=2^5\cdot 5^5\) है इसलिए हर में \(5^3\) की कमी है। अतः \(N=19\cdot 125=2375\), हर को (10) की घात बनाते समय कमी वाले गुणनखंड से गुणा करें।
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\(\frac{7}{2^6\cdot 5^4}\) को \(\frac{N}{10^6}\) में बदलने पर (N) क्या होगा?
When \(\frac{7}{2^6\cdot 5^4}\) is converted into \(\frac{N}{10^6}\), what is (N)?
#denominator-conversion
#powers-of-10
#numerator-adjustment
#real-numbers
A (35)
B (175)
C (700)
D (875)
Explanation opens after your attempt
Step 1
Concept
Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(5^2\). Thus \(N=7\cdot 25=175\).
Step 2
Why this answer is correct
The correct answer is B. (175). Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(5^2\). Thus \(N=7\cdot 25=175\).
Step 3
Exam Tip
\(10^6=2^6\cdot 5^6\) है इसलिए हर में \(5^2\) की कमी है। \(N=7\cdot 25=175\) होगा।
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\(\frac{13}{2^3\cdot 5^7}\) को \(\frac{N}{10^7}\) के रूप में लिखने पर (N) का सही मान चुनिए।
Choose the correct value of (N) when \(\frac{13}{2^3\cdot 5^7}\) is written as \(\frac{N}{10^7}\).
#powers-of-10
#terminating-decimal
#numerator
#real-numbers
A (104)
B (208)
C (416)
D (8125)
Explanation opens after your attempt
Step 1
Concept
To make the denominator \(10^7\), multiply by \(2^4=16\). Therefore \(N=13\cdot 16=208\).
Step 2
Why this answer is correct
The correct answer is B. (208). To make the denominator \(10^7\), multiply by \(2^4=16\). Therefore \(N=13\cdot 16=208\).
Step 3
Exam Tip
हर को \(10^7\) बनाने के लिए \(2^4=16\) से गुणा करना होगा। इसलिए \(N=13\cdot 16=208\)।
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\(\frac{13}{2^3\cdot 5^7}\) को \(\frac{N}{10^7}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{13}{2^3\cdot 5^7}\) is written as \(\frac{N}{10^7}\), what is (N)?
#powers-of-10
#numerator-adjustment
#option-audit
#expert
A (104)
B (416)
C (832)
D (1625)
Explanation opens after your attempt
Step 1
Concept
Since \(10^7=2^7\cdot 5^7\), the denominator lacks \(2^4\). Thus \(N=13\cdot 16=208\), so the correct value is not listed.
Step 2
Why this answer is correct
The correct answer is A. (104). Since \(10^7=2^7\cdot 5^7\), the denominator lacks \(2^4\). Thus \(N=13\cdot 16=208\), so the correct value is not listed.
Step 3
Exam Tip
\(10^7=2^7\cdot 5^7\) है इसलिए हर में \(2^4\) की कमी है। \(N=13\cdot 16=208\) होगा इसलिए दिए विकल्पों में सही मान नहीं है।
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\(\frac{3}{2^4\cdot 5^6}\) को \(\frac{N}{10^6}\) में बदलने पर (N) क्या होगा?
When \(\frac{3}{2^4\cdot 5^6}\) is converted into \(\frac{N}{10^6}\), what is (N)?
#denominator-conversion
#powers-of-10
#numerator-adjustment
#real-numbers
A (6)
B (12)
C (24)
D (48)
Explanation opens after your attempt
Step 1
Concept
Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(2^2\). Thus \(N=3\cdot 4=12\).
Step 2
Why this answer is correct
The correct answer is B. (12). Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(2^2\). Thus \(N=3\cdot 4=12\).
Step 3
Exam Tip
\(10^6=2^6\cdot 5^6\), इसलिए हर में \(2^2\) की कमी है। \(N=3\cdot 4=12\) होगा।
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\(\frac{11}{2^6\cdot 5^2}\) को \(\frac{N}{10^6}\) के रूप में लिखने पर (N) का सही मान चुनिए।
Choose the correct value of (N) when \(\frac{11}{2^6\cdot 5^2}\) is written as \(\frac{N}{10^6}\).
#powers-of-10
#terminating-decimal
#numerator
#real-numbers
A (1375)
B (2750)
C (6875)
D (11000)
Explanation opens after your attempt
Step 1
Concept
To make the denominator \(10^6\), multiply by \(5^4=625\). Therefore \(N=11\cdot 625=6875\).
Step 2
Why this answer is correct
The correct answer is C. (6875). To make the denominator \(10^6\), multiply by \(5^4=625\). Therefore \(N=11\cdot 625=6875\).
Step 3
Exam Tip
हर को \(10^6\) बनाने के लिए \(5^4=625\) से गुणा करना होगा। इसलिए \(N=11\cdot 625=6875\)।
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\(\frac{11}{2^6\cdot 5^2}\) को \(\frac{N}{10^6}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{11}{2^6\cdot 5^2}\) is written as \(\frac{N}{10^6}\), what is (N)?
#powers-of-10
#numerator-adjustment
#decimal-conversion
#expert
A (275)
B (1375)
C (6875)
D (11000)
Explanation opens after your attempt
Step 1
Concept
Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(5^4\). Thus \(N=11\cdot 5^4=6875\), so the correct option is (6875).
Step 2
Why this answer is correct
The correct answer is B. (1375). Since \(10^6=2^6\cdot 5^6\), the denominator lacks \(5^4\). Thus \(N=11\cdot 5^4=6875\), so the correct option is (6875).
Step 3
Exam Tip
\(10^6=2^6\cdot 5^6\), इसलिए हर में \(5^4\) की कमी है। \(N=11\cdot 5^4=6875\), इसलिए सही विकल्प (6875) है।
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\(\frac{7}{2^3\cdot 5^5}\) को \(\frac{N}{10^5}\) के रूप में लिखने पर (N) क्या होगा?
If \(\frac{7}{2^3\cdot 5^5}\) is written as \(\frac{N}{10^5}\), what is (N)?
#powers-of-10
#numerator-adjustment
#decimal-conversion
#class-10
A (14)
B (28)
C (35)
D (56)
Explanation opens after your attempt
Step 1
Concept
We need \(10^5=2^5\cdot 5^5\).
Step 2
Why this answer is correct
The denominator \(2^3\cdot 5^5\) lacks \(2^2\). Multiplying numerator and denominator by (4) gives \(N=7\cdot 4=28\).
Step 3
Exam Tip
Multiply by the missing part to make the denominator \(10^k\). चरण 1: \(10^5=2^5\cdot 5^5\) चाहिए। चरण 2: हर \(2^3\cdot 5^5\) में \(2^2\) की कमी है। अंश और हर को (4) से गुणा करने पर \(N=7\cdot 4=28\)। चरण 3: हर को \(10^k\) बनाने के लिए कमी वाले भाग से गुणा करें।
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\(\frac{17}{2^2\cdot 5^6}\) को \(\frac{N}{10^6}\) के रूप में लिखा जाए, तो (N) क्या होगा?
If \(\frac{17}{2^2\cdot 5^6}\) is written as \(\frac{N}{10^6}\), what is (N)?
#powers-of-10
#numerator-adjustment
#decimal-conversion
#hard
A (68)
B (136)
C (272)
D (1088)
Explanation opens after your attempt
Step 1
Concept
We need \(10^6=2^6\cdot 5^6\).
Step 2
Why this answer is correct
The denominator \(2^2\cdot 5^6\) lacks \(2^4\), so multiply numerator and denominator by (16). Thus \(N=17\cdot 16=272\).
Step 3
Exam Tip
Multiply by the missing prime power. चरण 1: \(10^6=2^6\cdot 5^6\) चाहिए। चरण 2: हर \(2^2\cdot 5^6\) में \(2^4\) की कमी है, इसलिए अंश और हर को (16) से गुणा करेंगे। \(N=17\cdot 16=272\)। चरण 3: कमी वाले अभाज्य गुणनखंड से ही गुणा करें।
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किसी भिन्न का सरलतम रूप \(\frac{p}{2^3\cdot 5^5}\) है। यदि इसे \(\frac{N}{10^5}\) के रूप में लिखा जाए, तो (N) किसके बराबर होगा?
A fraction in lowest form is \(\frac{p}{2^3\cdot 5^5}\). If it is written as \(\frac{N}{10^5}\), what is (N)?
#powers-of-10
#numerator-adjustment
#terminating-decimal
#real-numbers
A (p)
B (4p)
C (25p)
D (100p)
Explanation opens after your attempt
Step 1
Concept
We need \(10^5=2^5\cdot 5^5\).
Step 2
Why this answer is correct
The denominator \(2^3\cdot 5^5\) lacks \(2^2\). So multiply numerator and denominator by \(2^2=4\). Hence (N=4p).
Step 3
Exam Tip
To make \(10^k\), multiply by the missing prime power. चरण 1: \(10^5=2^5\cdot 5^5\) चाहिए। चरण 2: हर \(2^3\cdot 5^5\) में \(2^2\) की कमी है। इसलिए अंश और हर को \(2^2=4\) से गुणा करेंगे। अतः (N=4p)। चरण 3: \(10^k\) बनाने के लिए जिस घात की कमी हो, वही गुणा करें।
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यदि (p(x)=x-2 -4x+6), तो (p(x+1)-p(x)) का मान क्या है?
If (p(x)=x-2 -4x+6), what is the value of (p(x+1)-p(x))?
#polynomial-substitution
#difference
#expert
A (2x-3)
B (2x+3)
C (x-3)
D \(x^2-3\)
Explanation opens after your attempt
Step 1
Concept
(p(x+1)=x-2 -2x+3) and (p(x+1)-p(x)=2x-3). In such questions, substitute (x+1) carefully first.
Step 2
Why this answer is correct
The correct answer is A. (2x-3). (p(x+1)=x-2 -2x+3) and (p(x+1)-p(x)=2x-3). In such questions, substitute (x+1) carefully first.
Step 3
Exam Tip
(p(x+1)=x-2 -2x+3) और (p(x+1)-p(x)=2x-3) है। ऐसे प्रश्नों में पहले (x+1) को सावधानी से प्रतिस्थापित करें।
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यदि (p(x)=x-2 -8x+16) है, तो (p(4+t)) का मान क्या है?
If (p(x)=x-2 -8x+16), what is the value of (p(4+t))?
#substitution
#perfect-square
#polynomial
A \(t^2\)
B \(t^2+4\)
C (4t)
D \(t^2-16\)
Explanation opens after your attempt
Correct Answer
A. \(t^2\)
Step 1
Concept
(p(x)=(x-4)2 ), so (p(4+t)=t-2 ). In such questions, first form a perfect square.
Step 2
Why this answer is correct
The correct answer is A. \(t^2\). (p(x)=(x-4)2 ), so (p(4+t)=t-2 ). In such questions, first form a perfect square.
Step 3
Exam Tip
(p(x)=(x-4)2 ), इसलिए (p(4+t)=t-2 )। ऐसे प्रश्नों में पहले पूर्ण वर्ग बनाना आसान होता है।
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यदि (p(x)=x-3 -1), तो निम्न में से कौन-सा रैखिक गुणनखंड है?
If (p(x)=x-3 -1), which of the following is a linear factor?
#identity
#cubic
#factorisation
A (x-1)
B (x+1)
C (x-3)
D (x+3)
Explanation opens after your attempt
Step 1
Concept
(x-3 -1=(x-1)\(x^2+x+1\)). Therefore (x-1) is a linear factor.
Step 2
Why this answer is correct
The correct answer is A. (x-1). (x-3 -1=(x-1)\(x^2+x+1\)). Therefore (x-1) is a linear factor.
Step 3
Exam Tip
(x-3 -1=(x-1)\(x^2+x+1\)) है। इसलिए (x-1) रैखिक गुणनखंड है।
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यदि (p(x)=x-2 -ax+b) के शून्यक (a) और (b) हैं, तो कौन-सा संबंध सही है?
If the zeroes of (p(x)=x-2 -ax+b) are (a) and (b), which relation is correct?
#symbolic-zeroes
#relations
#concept
A (a+b=a) और (ab=b) / (a+b=a) and (ab=b)
B (a-b=a) और (a+b=b) / (a-b=a) and (a+b=b)
C \(a^2+b^2=a\) और (ab=1) / \(a^2+b^2=a\) and (ab=1)
D (a+b=b) और (ab=a) / (a+b=b) and (ab=a)
Explanation opens after your attempt
Correct Answer
A. (a+b=a) और (ab=b) / (a+b=a) and (ab=b)
Step 1
Concept
The sum of zeroes from the polynomial is (a) and the product is (b). Hence if the zeroes are (a,b), then (a+b=a) and (ab=b).
Step 2
Why this answer is correct
The correct answer is A. (a+b=a) और (ab=b) / (a+b=a) and (ab=b). The sum of zeroes from the polynomial is (a) and the product is (b). Hence if the zeroes are (a,b), then (a+b=a) and (ab=b).
Step 3
Exam Tip
शून्यकों का योग बहुपद से (a) और गुणनफल (b) है। इसलिए यदि शून्यक (a,b) हैं, तो (a+b=a) और (ab=b) होना चाहिए।
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यदि (p(x)=x-2 +px+q) और (p(1)=0,\ p(2)=0), तो (p+q) क्या है?
If (p(x)=x-2 +px+q) and (p(1)=0,\ p(2)=0), what is (p+q)?
#coefficient-finding
#zeroes
#quadratic
A -(1)
B (1)
C -(2)
D (2)
Explanation opens after your attempt
Step 1
Concept
The zeroes are (1) and (2), so the polynomial is \(x^2-3x+2\). Thus (p=-3,\ q=2), and (p+q=-1).
Step 2
Why this answer is correct
The correct answer is A. -(1). The zeroes are (1) and (2), so the polynomial is \(x^2-3x+2\). Thus (p=-3,\ q=2), and (p+q=-1).
Step 3
Exam Tip
शून्यक (1) और (2) हैं, इसलिए बहुपद \(x^2-3x+2\) है। अतः (p=-3,\ q=2) और (p+q=-1)।
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यदि (p(x)=2x-2 +x-6), तो (2p(1)-p(-2)) का मान क्या है?
If (p(x)=2x-2 +x-6), what is the value of (2p(1)-p(-2))?
#evaluation
#polynomial
#expression
A -(6)
B (0)
C (6)
D -(12)
Explanation opens after your attempt
Step 1
Concept
(p(1)=2+1-6=-3) and (p(-2)=8-2-6=0). Therefore (2p(1)-p(-2)=-6).
Step 2
Why this answer is correct
The correct answer is A. -(6). (p(1)=2+1-6=-3) and (p(-2)=8-2-6=0). Therefore (2p(1)-p(-2)=-6).
Step 3
Exam Tip
(p(1)=2+1-6=-3) और (p(-2)=8-2-6=0)। इसलिए (2p(1)-p(-2)=-6)।
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यदि (p(x)=x-2 -6x+8), तो शून्यकों के वर्गों से बना मोनिक बहुपद कौन-सा है?
If (p(x)=x-2 -6x+8), which monic polynomial has the squares of its zeroes as zeroes?
#transformed-zeroes
#squares
#quadratic
A \(x^2-20x+64\)
B \(x^2-6x+8\)
C \(x^2-12x+16\)
D \(x^2-16x+64\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+64\)
Step 1
Concept
The original zeroes are (2) and (4), so the new zeroes are (4) and (16). The new polynomial is \(x^2-20x+64\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+64\). The original zeroes are (2) and (4), so the new zeroes are (4) and (16). The new polynomial is \(x^2-20x+64\).
Step 3
Exam Tip
मूल शून्यक (2) और (4) हैं, इसलिए नए शून्यक (4) और (16) हैं। नया बहुपद \(x^2-20x+64\) है।
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यदि (p(x)=x-2 +2x+c) का कोई वास्तविक शून्यक नहीं है, तो (c) के लिए कौन-सी शर्त सही है?
If (p(x)=x-2 +2x+c) has no real zero, which condition on (c) is correct?
#discriminant
#real-zeroes
#parameter
A (c>1)
B (c=1)
C (c<1)
D (c=0)
Explanation opens after your attempt
Step 1
Concept
For no real zero, the discriminant must satisfy (4-4c<0). This gives (c>1).
Step 2
Why this answer is correct
The correct answer is A. (c>1). For no real zero, the discriminant must satisfy (4-4c<0). This gives (c>1).
Step 3
Exam Tip
कोई वास्तविक शून्यक नहीं होने के लिए विविक्तकर (4-4c<0) चाहिए। इससे (c>1) मिलता है।
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यदि (p(x)=x-2 -7x+12) है, तो (p(x)) का ग्राफ (x)-अक्ष को किन बिंदुओं पर काटेगा?
If (p(x)=x-2 -7x+12), at which points will the graph of (p(x)) cut the (x)-axis?
#graph-zeroes
#quadratic
#intercepts
A ((3,0),(4,0))
B ((0,3),(0,4))
C ((2,0),(6,0))
D ((0,0),(7,0))
Explanation opens after your attempt
Correct Answer
A. ((3,0),(4,0))
Step 1
Concept
(x-2 -7x+12=(x-3)(x-4)), so the zeroes are (3) and (4). The (x)-axis points are ((3,0)) and ((4,0)).
Step 2
Why this answer is correct
The correct answer is A. ((3,0),(4,0)). (x-2 -7x+12=(x-3)(x-4)), so the zeroes are (3) and (4). The (x)-axis points are ((3,0)) and ((4,0)).
Step 3
Exam Tip
(x-2 -7x+12=(x-3)(x-4)), इसलिए शून्यक (3) और (4) हैं। (x)-अक्ष पर बिंदु ((3,0)) और ((4,0)) होंगे।
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यदि (p(x)=x-2 +5x+6) और (r(x)=p(-x)), तो (r(x)) के शून्यक कौन-से हैं?
If (p(x)=x-2 +5x+6) and (r(x)=p(-x)), what are the zeroes of (r(x))?
#transformed-zeroes
#reflection
#quadratic
A (2,3)
B -(2,-3)
C (1,6)
D -(1,-6)
Explanation opens after your attempt
Step 1
Concept
The zeroes of (p(x)) are (-2) and (-3). For (p(-x)=0), (-x=-2) or (-x=-3), so (x=2,3).
Step 2
Why this answer is correct
The correct answer is A. (2,3). The zeroes of (p(x)) are (-2) and (-3). For (p(-x)=0), (-x=-2) or (-x=-3), so (x=2,3).
Step 3
Exam Tip
(p(x)) के शून्यक (-2) और (-3) हैं। (p(-x)=0) के लिए (-x=-2) या (-x=-3), इसलिए (x=2,3)।
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यदि (p(x)=x-3 -3x-2 -4x+12), तो पूर्ण गुणनखंड रूप क्या है?
If (p(x)=x-3 -3x-2 -4x+12), what is the complete factor form?
#factorisation
#grouping
#cubic
A ((x-3)(x-2)(x+2))
B ((x+3)(x-2)(x+2))
C ((x-3)(x+2)2 )
D ((x+3)(x-2)2 )
Explanation opens after your attempt
Correct Answer
A. ((x-3)(x-2)(x+2))
Step 1
Concept
By grouping, (x-2 (x-3)-4(x-3)=(x-3)\(x^2-4\)). Thus the factors are ((x-3)(x-2)(x+2)).
Step 2
Why this answer is correct
The correct answer is A. ((x-3)(x-2)(x+2)). By grouping, (x-2 (x-3)-4(x-3)=(x-3)\(x^2-4\)). Thus the factors are ((x-3)(x-2)(x+2)).
Step 3
Exam Tip
समूहीकरण से (x-2 (x-3)-4(x-3)=(x-3)\(x^2-4\)) मिलता है। इसलिए गुणनखंड ((x-3)(x-2)(x+2)) हैं।
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यदि (p(x)=x-2 -4x+3) है, तो कौन-सा मान (p(x)<0) बनाता है?
If (p(x)=x-2 -4x+3), which value makes (p(x)<0)?
#sign-of-polynomial
#quadratic
#zeroes
A (2)
B (0)
C (4)
D -(1)
Explanation opens after your attempt
Step 1
Concept
(p(x)=(x-1)(x-3)), and it is negative for (1<x<3). Therefore (x=2) is correct.
Step 2
Why this answer is correct
The correct answer is A. (2). (p(x)=(x-1)(x-3)), and it is negative for (1<x<3). Therefore (x=2) is correct.
Step 3
Exam Tip
(p(x)=(x-1)(x-3)) है और (1<x<3) में मान ऋणात्मक होता है। इसलिए (x=2) सही है।
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यदि (p(x)=x-2 -2x+k) का न्यूनतम मान (3) है, तो (k) क्या है?
If the minimum value of (p(x)=x-2 -2x+k) is (3), what is (k)?
#complete-square
#minimum-value
#quadratic
A (4)
B (3)
C (2)
D (1)
Explanation opens after your attempt
Step 1
Concept
(x-2 -2x+k=(x-1)2 +k-1), so the minimum value is (k-1). From (k-1=3), (k=4).
Step 2
Why this answer is correct
The correct answer is A. (4). (x-2 -2x+k=(x-1)2 +k-1), so the minimum value is (k-1). From (k-1=3), (k=4).
Step 3
Exam Tip
(x-2 -2x+k=(x-1)2 +k-1), इसलिए न्यूनतम मान (k-1) है। (k-1=3) से (k=4)।
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यदि (p(x)=x-2 +ax+a) में (p(1)=0), तो (a) क्या है?
If (p(x)=x-2 +ax+a) and (p(1)=0), what is (a)?
#evaluation
#parameter
#polynomial
A -\(\frac{1}{2}\)
B -(1)
C \(\frac{1}{2}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{1}{2}\)
Step 1
Concept
(p(1)=1+a+a=0), so (2a=-1). Therefore \(a=-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{1}{2}\). (p(1)=1+a+a=0), so (2a=-1). Therefore \(a=-\frac{1}{2}\).
Step 3
Exam Tip
(p(1)=1+a+a=0), इसलिए (2a=-1)। अतः \(a=-\frac{1}{2}\) है।
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यदि (p(x)=x-4 -16), तो \(x^2+4\) के कारण वास्तविक शून्यकों पर क्या प्रभाव पड़ता है?
For (p(x)=x-4 -16), what effect does \(x^2+4\) have on real zeroes?
#quartic
#real-zeroes
#factor
A यह कोई वास्तविक शून्यक नहीं देता / It gives no real zero
B यह शून्यक (2) देता है / It gives zero (2)
C यह शून्यक (-2) देता है / It gives zero (-2)
D यह दो वास्तविक शून्यक देता है / It gives two real zeroes
Explanation opens after your attempt
Correct Answer
A. यह कोई वास्तविक शून्यक नहीं देता / It gives no real zero
Step 1
Concept
(x-4 -16=\(x^2-4\)\(x^2+4\)). The factor \(x^2+4\) is never (0) for real (x).
Step 2
Why this answer is correct
The correct answer is A. यह कोई वास्तविक शून्यक नहीं देता / It gives no real zero. (x-4 -16=\(x^2-4\)\(x^2+4\)). The factor \(x^2+4\) is never (0) for real (x).
Step 3
Exam Tip
(x-4 -16=\(x^2-4\)\(x^2+4\)) है। \(x^2+4\) वास्तविक (x) के लिए कभी (0) नहीं होता।
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यदि (p(x)=5x-2 -20x+20), तो इसके शून्यकों का योग क्या है?
If (p(x)=5x-2 -20x+20), what is the sum of its zeroes?
#sum-zeroes
#quadratic
#formula
A (4)
B (5)
C -(4)
D (2)
Explanation opens after your attempt
Step 1
Concept
The sum is \(-\frac{b}{a}=-\frac{-20}{5}=4\). Even if a common factor is removed first, the sum remains the same.
Step 2
Why this answer is correct
The correct answer is A. (4). The sum is \(-\frac{b}{a}=-\frac{-20}{5}=4\). Even if a common factor is removed first, the sum remains the same.
Step 3
Exam Tip
योग \(-\frac{b}{a}=-\frac{-20}{5}=4\) है। पहले सामान्य गुणनखंड निकालना चाहें तो भी योग वही रहेगा।
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यदि (p(x)=x-2 +2x-24), तो (p(x)) का कौन-सा गुणनखंड है?
If (p(x)=x-2 +2x-24), which is a factor of (p(x))?
#factorisation
#quadratic
#polynomial
A (x-4)
B (x+4)
C (x-6)
D (x+2)
Explanation opens after your attempt
Step 1
Concept
(x-2 +2x-24=(x-4)(x+6)). Therefore (x-4) is a factor.
Step 2
Why this answer is correct
The correct answer is A. (x-4). (x-2 +2x-24=(x-4)(x+6)). Therefore (x-4) is a factor.
Step 3
Exam Tip
(x-2 +2x-24=(x-4)(x+6)) है। इसलिए (x-4) एक गुणनखंड है।
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यदि (p(x)=x-2 -12x+35), तो \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\) क्या है, जहाँ \(\alpha,\beta\) शून्यक हैं?
If (p(x)=x-2 -12x+35), what is \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\), where \(\alpha,\beta\) are zeroes?
#zeroes
#reciprocal-expression
#expert
A \(\frac{5}{12}\)
B \(\frac{6}{11}\)
C \(\frac{12}{5}\)
D \(\frac{11}{24}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{11}{24}\)
Step 1
Concept
\(\alpha+\beta=12\) and \(\alpha\beta=35\). \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\).
Step 2
Why this answer is correct
The correct answer is D. \(\frac{11}{24}\). \(\alpha+\beta=12\) and \(\alpha\beta=35\). \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\).
Step 3
Exam Tip
\(\alpha+\beta=12\) और \(\alpha\beta=35\) हैं। \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}\)।
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यदि (p(x)=x-3 +px-2 +qx+15) के शून्यक (-1,3,-5) हैं, तो (p+q) क्या है?
If the zeroes of (p(x)=x-3 +px-2 +qx+15) are (-1,3,-5), what is (p+q)?
#cubic-zeroes
#coefficients
#trap
A (0)
B (2)
C -(2)
D (4)
Explanation opens after your attempt
Step 1
Concept
The sum of zeroes is (-3), so (p=3). The pairwise product sum is (-3+5-15=-13), so (q=-13) and (p+q=-10).
Step 2
Why this answer is correct
The correct answer is B. (2). The sum of zeroes is (-3), so (p=3). The pairwise product sum is (-3+5-15=-13), so (q=-13) and (p+q=-10).
Step 3
Exam Tip
शून्यकों का योग (-3) है, इसलिए (p=3)। युग्म गुणनफलों का योग (-3+5-15=-13), इसलिए (q=-13) और (p+q=-10)।
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यदि (p(x)=x-2 +4x+4), तो (p(-2+h)) का मान क्या है?
If (p(x)=x-2 +4x+4), what is the value of (p(-2+h))?
#substitution
#identity
#polynomial
A \(h^2\)
B (h)
C \(h^2+4\)
D (2h)
Explanation opens after your attempt
Correct Answer
A. \(h^2\)
Step 1
Concept
(p(x)=(x+2)2 ), so (p(-2+h)=h-2 ). Perfect square form makes substitution easier.
Step 2
Why this answer is correct
The correct answer is A. \(h^2\). (p(x)=(x+2)2 ), so (p(-2+h)=h-2 ). Perfect square form makes substitution easier.
Step 3
Exam Tip
(p(x)=(x+2)2 ), इसलिए (p(-2+h)=h-2 )। पूर्ण वर्ग रूप से प्रतिस्थापन आसान हो जाता है।
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यदि (x=2) पर (p(x)=x-3 +mx-10) का मान (0) है, तो (m) क्या है?
If the value of (p(x)=x-3 +mx-10) at (x=2) is (0), what is (m)?
#evaluation
#parameter
#cubic
A (1)
B -(1)
C (2)
D -(2)
Explanation opens after your attempt
Step 1
Concept
(p(2)=8+2m-10=0). This gives (2m=2) and (m=1).
Step 2
Why this answer is correct
The correct answer is A. (1). (p(2)=8+2m-10=0). This gives (2m=2) and (m=1).
Step 3
Exam Tip
(p(2)=8+2m-10=0) है। इससे (2m=2) और (m=1) मिलता है।
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यदि (p(x)=2x-2 -7x+5), तो शून्यकों के योग और गुणनफल का अंतर क्या है?
If (p(x)=2x-2 -7x+5), what is the difference between the sum and product of its zeroes?
#sum-product
#quadratic
#zeroes
A (1)
B (2)
C \(\frac{1}{2}\)
D (3)
Explanation opens after your attempt
Step 1
Concept
The sum is \(\frac{7}{2}\) and the product is \(\frac{5}{2}\). Their difference is \(\frac{7}{2}-\frac{5}{2}=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The sum is \(\frac{7}{2}\) and the product is \(\frac{5}{2}\). Their difference is \(\frac{7}{2}-\frac{5}{2}=1\).
Step 3
Exam Tip
योग \(\frac{7}{2}\) और गुणनफल \(\frac{5}{2}\) है। उनका अंतर \(\frac{7}{2}-\frac{5}{2}=1\) है।
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यदि (p(x)=x-2 -kx+36) के शून्यक धनात्मक और अनुपात (1:4) में हैं, तो (k) क्या है?
If the zeroes of (p(x)=x-2 -kx+36) are positive and in the ratio (1:4), what is (k)?
#ratio-zeroes
#parameter
#quadratic
A (15)
B (12)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
Let the zeroes be (t) and (4t), then \(4t^2=36\) gives (t=3). The sum is (15), so (k=15).
Step 2
Why this answer is correct
The correct answer is A. (15). Let the zeroes be (t) and (4t), then \(4t^2=36\) gives (t=3). The sum is (15), so (k=15).
Step 3
Exam Tip
शून्यक (t) और (4t) मानें, तो \(4t^2=36\) से (t=3) है। योग (15) है, इसलिए (k=15)।
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यदि (p(x)=ax-2 +bx+c) और (p(1)=p(2)=p(3)=0), तो कौन-सा निष्कर्ष सही है?
If (p(x)=ax-2 +bx+c) and (p(1)=p(2)=p(3)=0), which conclusion is correct?
#degree
#number-of-zeroes
#concept
A (p(x)) शून्य बहुपद है / (p(x)) is the zero polynomial
B (p(x)) रेखीय बहुपद है / (p(x)) is a linear polynomial
C (a=1) है / (a=1)
D (c=3) है / (c=3)
Explanation opens after your attempt
Correct Answer
A. (p(x)) शून्य बहुपद है / (p(x)) is the zero polynomial
Step 1
Concept
A polynomial of degree at most (2) can have three distinct zeroes only if it is the zero polynomial. A non-zero polynomial cannot have more distinct zeroes than its degree.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) शून्य बहुपद है / (p(x)) is the zero polynomial. A polynomial of degree at most (2) can have three distinct zeroes only if it is the zero polynomial. A non-zero polynomial cannot have more distinct zeroes than its degree.
Step 3
Exam Tip
अधिकतम द्विघात बहुपद के तीन अलग-अलग शून्यक तभी हो सकते हैं जब वह शून्य बहुपद हो। घात से अधिक अलग शून्यक असंभव होते हैं।
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यदि (p(x)=x-2 -5x+6), तो (p(2)+p(3)) का मान क्या है?
If (p(x)=x-2 -5x+6), what is the value of (p(2)+p(3))?
#evaluation
#zeroes
#quadratic
A (0)
B (1)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
(2) and (3) are zeroes of (p(x)), so (p(2)=0) and (p(3)=0). Hence the sum is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). (2) and (3) are zeroes of (p(x)), so (p(2)=0) and (p(3)=0). Hence the sum is (0).
Step 3
Exam Tip
(2) और (3), (p(x)) के शून्यक हैं, इसलिए (p(2)=0) और (p(3)=0)। इसलिए योग (0) है।
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यदि (p(x)=x-2 -2x-8), तो (p(x+3)) के शून्यक कौन-से हैं?
If (p(x)=x-2 -2x-8), what are the zeroes of (p(x+3))?
#transformed-zeroes
#quadratic
#expert
A -(5,1)
B -(1,5)
C (2,-4)
D -(3,3)
Explanation opens after your attempt
Step 1
Concept
The zeroes of (p(x)) are (4) and (-2). From (x+3=4) or (x+3=-2), the zeroes are (1) and (-5).
Step 2
Why this answer is correct
The correct answer is A. -(5,1). The zeroes of (p(x)) are (4) and (-2). From (x+3=4) or (x+3=-2), the zeroes are (1) and (-5).
Step 3
Exam Tip
(p(x)) के शून्यक (4) और (-2) हैं। (x+3=4) या (x+3=-2) से शून्यक (1) और (-5) मिलते हैं।
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यदि (p(x)=x-3 -9x), तो इसके कितने भिन्न वास्तविक शून्यक हैं?
If (p(x)=x-3 -9x), how many distinct real zeroes does it have?
#distinct-zeroes
#cubic
#factorisation
A (3)
B (2)
C (1)
D (0)
Explanation opens after your attempt
Step 1
Concept
(x-3 -9x=x(x-3)(x+3)), so the zeroes are (-3,0,3). Hence there are (3) distinct real zeroes.
Step 2
Why this answer is correct
The correct answer is A. (3). (x-3 -9x=x(x-3)(x+3)), so the zeroes are (-3,0,3). Hence there are (3) distinct real zeroes.
Step 3
Exam Tip
(x-3 -9x=x(x-3)(x+3)), इसलिए शून्यक (-3,0,3) हैं। अतः भिन्न वास्तविक शून्यक (3) हैं।
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यदि (p(x)=x-3 +ax-2 +bx-12) के शून्यक (1,3,-4) हैं, तो (a+b) क्या है?
If the zeroes of (p(x)=x-3 +ax-2 +bx-12) are (1,3,-4), what is (a+b)?
#cubic-zeroes
#coefficients
#expert
A -(9)
B (9)
C -(7)
D (7)
Explanation opens after your attempt
Step 1
Concept
The sum (1+3-4=0), so (a=0). The sum of pairwise products is (3-4-12=-13), so (b=-13) and (a+b=-13).
Step 2
Why this answer is correct
The correct answer is A. -(9). The sum (1+3-4=0), so (a=0). The sum of pairwise products is (3-4-12=-13), so (b=-13) and (a+b=-13).
Step 3
Exam Tip
योग (1+3-4=0) है, इसलिए (a=0)। युग्म गुणनफलों का योग (3-4-12=-13), इसलिए (b=-13) और (a+b=-13)।
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यदि (p(x)=x-4 -5x-2 +4), तो इसके वास्तविक शून्यक कौन-से हैं?
If (p(x)=x-4 -5x-2 +4), what are its real zeroes?
#quartic
#biquadratic
#zeroes
A -(2,-1,1,2)
B -(4,-1,1,4)
C -(2,0,1,2)
D -(1,1)
Explanation opens after your attempt
Correct Answer
A. -(2,-1,1,2)
Step 1
Concept
(x-4 -5x-2 +4=\(x^2-1\)\(x^2-4\)). Therefore the real zeroes are (-2,-1,1,2).
Step 2
Why this answer is correct
The correct answer is A. -(2,-1,1,2). (x-4 -5x-2 +4=\(x^2-1\)\(x^2-4\)). Therefore the real zeroes are (-2,-1,1,2).
Step 3
Exam Tip
(x-4 -5x-2 +4=\(x^2-1\)\(x^2-4\)) है। इसलिए वास्तविक शून्यक (-2,-1,1,2) हैं।
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यदि (p(x)=4x-2 -4x+1), तो इसके शून्यक क्या हैं?
If (p(x)=4x-2 -4x+1), what are its zeroes?
#equal-zeroes
#perfect-square
#quadratic
A \(\frac{1}{2},\frac{1}{2}\)
B (1,1)
C -\(\frac{1}{2},-\frac{1}{2}\)
D (2,2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2},\frac{1}{2}\)
Step 1
Concept
(4x-2 -4x+1=(2x-1)2 ), so the zero \(\frac{1}{2}\) occurs twice. Perfect square form gives equal zeroes directly.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2},\frac{1}{2}\). (4x-2 -4x+1=(2x-1)2 ), so the zero \(\frac{1}{2}\) occurs twice. Perfect square form gives equal zeroes directly.
Step 3
Exam Tip
(4x-2 -4x+1=(2x-1)2 ), इसलिए शून्यक \(\frac{1}{2}\) दो बार है। पूर्ण वर्ग रूप तुरंत समान शून्यक देता है।
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यदि (p(x)=2x-2 +mx+18) का एक शून्यक (3) है, तो दूसरा शून्यक क्या है?
If one zero of (p(x)=2x-2 +mx+18) is (3), what is the other zero?
#one-zero-given
#product-zeroes
#quadratic
A (3)
B -(3)
C (6)
D -(6)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{18}{2}=9\). Since one zero is (3), the other is (3).
Step 2
Why this answer is correct
The correct answer is A. (3). The product is \(\frac{18}{2}=9\). Since one zero is (3), the other is (3).
Step 3
Exam Tip
गुणनफल \(\frac{18}{2}=9\) है। एक शून्यक (3) है, इसलिए दूसरा (3) होगा।
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यदि (p(x)=x-2 -3x-28), तो (p(x)) के शून्यकों के मध्य दूरी क्या है?
If (p(x)=x-2 -3x-28), what is the distance between its zeroes?
#zeroes
#distance
#factorisation
A (11)
B (7)
C (4)
D (3)
Explanation opens after your attempt
Step 1
Concept
(p(x)=(x-7)(x+4)), so the zeroes are (7) and (-4). The distance is (|7-(-4)|=11).
Step 2
Why this answer is correct
The correct answer is A. (11). (p(x)=(x-7)(x+4)), so the zeroes are (7) and (-4). The distance is (|7-(-4)|=11).
Step 3
Exam Tip
(p(x)=(x-7)(x+4)), इसलिए शून्यक (7) और (-4) हैं। दूरी (|7-(-4)|=11) है।
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यदि (p(x)=x-2 -11x+30) और (q(x)=p(x-2)), तो (q(x)) के शून्यक कौन-से हैं?
If (p(x)=x-2 -11x+30) and (q(x)=p(x-2)), what are the zeroes of (q(x))?
#transformed-zeroes
#polynomial
#expert
A (7,8)
B (3,4)
C (5,6)
D (2,11)
Explanation opens after your attempt
Step 1
Concept
The zeroes of (p(x)) are (5) and (6). For (p(x-2)=0), (x-2=5) or (x-2=6), so the zeroes are (7,8).
Step 2
Why this answer is correct
The correct answer is A. (7,8). The zeroes of (p(x)) are (5) and (6). For (p(x-2)=0), (x-2=5) or (x-2=6), so the zeroes are (7,8).
Step 3
Exam Tip
(p(x)) के शून्यक (5) और (6) हैं। (p(x-2)=0) के लिए (x-2=5) या (x-2=6), इसलिए शून्यक (7,8) हैं।
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यदि (p(x)=x-2 +bx+16) के शून्यक परस्पर व्युत्क्रम हैं, तो (b) के बारे में क्या कहा जा सकता है?
If the zeroes of (p(x)=x-2 +bx+16) are reciprocals of each other, what can be said about (b)?
#reciprocal-zeroes
#product
#concept
A ऐसा संभव नहीं है / It is not possible
B (b=0)
C (b=16)
D (b=-16)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
Reciprocal zeroes must have product (1). Here the product is (16), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. Reciprocal zeroes must have product (1). Here the product is (16), so it is not possible.
Step 3
Exam Tip
परस्पर व्युत्क्रम शून्यकों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (16) है, इसलिए यह संभव नहीं है।
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यदि (p(x)=3x-2 -12x+15), तो वास्तविक शून्यकों के बारे में सही निष्कर्ष क्या है?
For (p(x)=3x-2 -12x+15), what is the correct conclusion about real zeroes?
#real-zeroes
#complete-square
#quadratic
A कोई वास्तविक शून्यक नहीं / No real zeroes
B दो समान वास्तविक शून्यक / Two equal real zeroes
C दो भिन्न वास्तविक शून्यक / Two distinct real zeroes
D एक शून्यक (5) / One zero (5)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक शून्यक नहीं / No real zeroes
Step 1
Concept
(p(x)=3\(x^2-4x+5\)=3((x-2)2 +1)), which is always positive. Hence there are no real zeroes.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक शून्यक नहीं / No real zeroes. (p(x)=3\(x^2-4x+5\)=3((x-2)2 +1)), which is always positive. Hence there are no real zeroes.
Step 3
Exam Tip
(p(x)=3\(x^2-4x+5\)=3((x-2)2 +1)), जो सदा धनात्मक है। इसलिए कोई वास्तविक शून्यक नहीं है।
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यदि (p(x)=x-3 -6x-2 +12x-8), तो (p(x)) का शून्यक किस गुणनता के साथ है?
If (p(x)=x-3 -6x-2 +12x-8), what is the zero of (p(x)) with its multiplicity?
#multiplicity
#cubic
#identity
A (2) गुणनता (3) / (2) with multiplicity (3)
B (3) गुणनता (2) / (3) with multiplicity (2)
C -(2) गुणनता (3) / (-2) with multiplicity (3)
D (1) गुणनता (3) / (1) with multiplicity (3)
Explanation opens after your attempt
Correct Answer
A. (2) गुणनता (3) / (2) with multiplicity (3)
Step 1
Concept
This is (p(x)=(x-2)3 ), so (2) is a zero three times. Recognizing cube identities is useful.
Step 2
Why this answer is correct
The correct answer is A. (2) गुणनता (3) / (2) with multiplicity (3). This is (p(x)=(x-2)3 ), so (2) is a zero three times. Recognizing cube identities is useful.
Step 3
Exam Tip
यह (p(x)=(x-2)3 ) है, इसलिए (2) तीन बार शून्यक है। घन पूर्ण पहचान को पहचानना उपयोगी है।
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यदि किसी द्विघात बहुपद के शून्यक \(2+\sqrt{3}\) और \(2-\sqrt{3}\) हैं, तो मोनिक बहुपद कौन-सा है?
If the zeroes of a quadratic polynomial are \(2+\sqrt{3}\) and \(2-\sqrt{3}\), which is the monic polynomial?
#construct-polynomial
#surd-zeroes
#quadratic
A \(x^2-4x+1\)
B \(x^2+4x+1\)
C \(x^2-2x+3\)
D \(x^2-4x+7\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+1\)
Step 1
Concept
The sum is (4) and the product is (1). Therefore the monic polynomial is \(x^2-4x+1\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+1\). The sum is (4) and the product is (1). Therefore the monic polynomial is \(x^2-4x+1\).
Step 3
Exam Tip
योग (4) और गुणनफल (1) है। अतः मोनिक बहुपद \(x^2-4x+1\) होगा।
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यदि (p(x)=x-2 -4x+1), तो शून्यकों के व्युत्क्रमों का योग क्या है?
If (p(x)=x-2 -4x+1), what is the sum of reciprocals of its zeroes?
#reciprocal-zeroes
#quadratic
#formula
A (4)
B (1)
C (-4)
D \(\frac{1}{4}\)
Explanation opens after your attempt
Step 1
Concept
The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\alpha+\beta=4\) and \(\alpha\beta=1\), so the answer is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\alpha+\beta=4\) and \(\alpha\beta=1\), so the answer is (4).
Step 3
Exam Tip
शून्यकों के व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\alpha+\beta=4\) और \(\alpha\beta=1\), इसलिए उत्तर (4) है।
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यदि \(\alpha,\beta\), \(x^2-9x+20\) के शून्यक हैं, तो (\(\alpha+2\)\(\beta+2\)) का मान क्या है?
If \(\alpha,\beta\) are zeroes of \(x^2-9x+20\), what is the value of (\(\alpha+2\)\(\beta+2\))?
#zeroes
#algebraic-expression
#quadratic
A (42)
B (38)
C (31)
D (24)
Explanation opens after your attempt
Step 1
Concept
\(\alpha+\beta=9\) and \(\alpha\beta=20\). (\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4=20+18+4=42).
Step 2
Why this answer is correct
The correct answer is A. (42). \(\alpha+\beta=9\) and \(\alpha\beta=20\). (\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4=20+18+4=42).
Step 3
Exam Tip
\(\alpha+\beta=9\) और \(\alpha\beta=20\) हैं। (\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4=20+18+4=42)।
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यदि (p(x)=(k-3)x-5 +2x-3 -x+9) की घात (3) है, तो (k) क्या है?
If (p(x)=(k-3)x-5 +2x-3 -x+9) has degree (3), what is (k)?
#degree
#parameter
#polynomial
A (3)
B (0)
C (2)
D (5)
Explanation opens after your attempt
Step 1
Concept
For degree (3), the coefficient of \(x^5\) must be (0). Thus (k-3=0) and (k=3).
Step 2
Why this answer is correct
The correct answer is A. (3). For degree (3), the coefficient of \(x^5\) must be (0). Thus (k-3=0) and (k=3).
Step 3
Exam Tip
घात (3) होने के लिए \(x^5\) का गुणांक (0) होना चाहिए। इसलिए (k-3=0) और (k=3)।
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यदि (p(x)=x-3 -4x-2 -7x+10) और (x-1) इसका गुणनखंड है, तो शेष द्विघात गुणनखंड क्या है?
If (p(x)=x-3 -4x-2 -7x+10) and (x-1) is a factor, what is the remaining quadratic factor?
#division
#factorisation
#cubic
A \(x^2-3x-10\)
B \(x^2+3x-10\)
C \(x^2-5x+10\)
D \(x^2-4x-7\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-10\)
Step 1
Concept
Dividing (p(x)) by (x-1) gives \(x^2-3x-10\). Verify by multiplying ((x-1)\(x^2-3x-10\)=p(x)).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-10\). Dividing (p(x)) by (x-1) gives \(x^2-3x-10\). Verify by multiplying ((x-1)\(x^2-3x-10\)=p(x)).
Step 3
Exam Tip
(p(x)) को (x-1) से भाग देने पर \(x^2-3x-10\) मिलता है। गुणा करके जाँचें कि ((x-1)\(x^2-3x-10\)=p(x))।
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यदि (p(x)=2x-2 -5x-3), तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) क्या है, जहाँ \(\alpha,\beta\) शून्यक हैं?
If (p(x)=2x-2 -5x-3), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\), where \(\alpha,\beta\) are zeroes?
#zeroes
#ratio-expression
#expert
A -\(\frac{37}{6}\)
B \(\frac{37}{6}\)
C -\(\frac{25}{6}\)
D \(\frac{25}{6}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{37}{6}\)
Step 1
Concept
\(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=-\frac{3}{2}\). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\(\alpha+\beta\)2 -2\alpha\beta}{\alpha\beta}=-\frac{37}{6}).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{37}{6}\). \(\alpha+\beta=\frac{5}{2}\) and \(\alpha\beta=-\frac{3}{2}\). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\(\alpha+\beta\)2 -2\alpha\beta}{\alpha\beta}=-\frac{37}{6}).
Step 3
Exam Tip
\(\alpha+\beta=\frac{5}{2}\) और \(\alpha\beta=-\frac{3}{2}\) हैं। (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\(\alpha+\beta\)2 -2\alpha\beta}{\alpha\beta}=-\frac{37}{6})।
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यदि (p(x)=x-2 -6x+s) का एक शून्यक दूसरे से (2) अधिक है, तो (s) का मान क्या है?
If one zero of (p(x)=x-2 -6x+s) is (2) more than the other, what is (s)?
#zeroes
#difference
#parameter
A (8)
B (9)
C (10)
D (12)
Explanation opens after your attempt
Step 1
Concept
Let the zeroes be (t) and (t+2), then (2t+2=6) gives (t=2). The product is \(2\cdot4=8\), so (s=8).
Step 2
Why this answer is correct
The correct answer is A. (8). Let the zeroes be (t) and (t+2), then (2t+2=6) gives (t=2). The product is \(2\cdot4=8\), so (s=8).
Step 3
Exam Tip
शून्यक (t) और (t+2) मानें, तो (2t+2=6) से (t=2)। गुणनफल \(2\cdot4=8\), इसलिए (s=8)।
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यदि (x=2) और (x=-3), (p(x)=ax-2 +bx+c) के शून्यक हैं, तो \(\frac{c}{a}\) क्या है?
If (x=2) and (x=-3) are zeroes of (p(x)=ax-2 +bx+c), what is \(\frac{c}{a}\)?
#zeroes
#coefficient-relation
#quadratic
A -(6)
B (6)
C -(1)
D (1)
Explanation opens after your attempt
Step 1
Concept
The product of zeroes is (2\cdot(-3)=-6), and it equals \(\frac{c}{a}\). Pay special attention to the sign in products.
Step 2
Why this answer is correct
The correct answer is A. -(6). The product of zeroes is (2\cdot(-3)=-6), and it equals \(\frac{c}{a}\). Pay special attention to the sign in products.
Step 3
Exam Tip
शून्यकों का गुणनफल (2\cdot(-3)=-6) है और यह \(\frac{c}{a}\) के बराबर होता है। गुणनफल में संकेत पर विशेष ध्यान दें।
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यदि (p(x)=x-3 -2x-2 -13x-10) है, तो इनमें से कौन (p(x)) का शून्यक है?
If (p(x)=x-3 -2x-2 -13x-10), which of the following is a zero of (p(x))?
#zero-test
#cubic
#polynomial
A (5)
B (2)
C -(1)
D -(2)
Explanation opens after your attempt
Step 1
Concept
(p(-2)=-8-8+26-10=0), so (-2) is a zero. Test small option values by direct substitution.
Step 2
Why this answer is correct
The correct answer is D. -(2). (p(-2)=-8-8+26-10=0), so (-2) is a zero. Test small option values by direct substitution.
Step 3
Exam Tip
(p(-2)=-8-8+26-10=0), इसलिए (-2) शून्यक है। विकल्पों में दिए छोटे मानों को सीधे रखकर जाँचें।
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यदि \(\alpha\) और \(\beta\), \(3x^2-10x+7\) के शून्यक हैं, तो \(\alpha^2+\beta^2\) का मान क्या है?
If \(\alpha\) and \(\beta\) are zeroes of \(3x^2-10x+7\), what is the value of \(\alpha^2+\beta^2\)?
#zeroes
#identity
#quadratic
A \(\frac{58}{9}\)
B \(\frac{100}{9}\)
C \(\frac{14}{3}\)
D \(\frac{49}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{58}{9}\)
Step 1
Concept
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=\frac{7}{3}\). Hence (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta=\frac{100}{9}-\frac{14}{3}=\frac{58}{9}).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{58}{9}\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=\frac{7}{3}\). Hence (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta=\frac{100}{9}-\frac{14}{3}=\frac{58}{9}).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=\frac{7}{3}\) हैं। इसलिए (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta=\frac{100}{9}-\frac{14}{3}=\frac{58}{9})।
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यदि (p(x)=x-2 +rx+9) के दोनों शून्यक समान और ऋणात्मक हैं, तो (r) का मान क्या होगा?
If both zeroes of (p(x)=x-2 +rx+9) are equal and negative, what is the value of (r)?
#equal-zeroes
#discriminant
#sign
A (6)
B (-6)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
For equal zeroes, \(r^2-36=0\), so \(r=\pm6\). For negative zeroes the sum must be positive (6), hence (r=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal zeroes, \(r^2-36=0\), so \(r=\pm6\). For negative zeroes the sum must be positive (6), hence (r=6).
Step 3
Exam Tip
समान शून्यकों के लिए \(r^2-36=0\), अतः \(r=\pm6\)। ऋणात्मक शून्यक के लिए योग धनात्मक (6) होना चाहिए, इसलिए (r=6)।
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यदि (p(x)=x-2 -(m+4)x+4m) के शून्यक (4) और (m) हैं, तो कौन-सा कथन सही है?
If the zeroes of (p(x)=x-2 -(m+4)x+4m) are (4) and (m), which statement is correct?
#zeroes
#coefficients
#concept
A यह हर वास्तविक (m) के लिए सही है / It is true for every real (m)
B यह केवल (m=4) के लिए सही है / It is true only for (m=4)
C यह केवल (m=0) के लिए सही है / It is true only for (m=0)
D यह कभी सही नहीं है / It is never true
Explanation opens after your attempt
Correct Answer
A. यह हर वास्तविक (m) के लिए सही है / It is true for every real (m)
Step 1
Concept
The sum (4+m) and product (4m) match the given polynomial. Therefore the statement is true for every real (m).
Step 2
Why this answer is correct
The correct answer is A. यह हर वास्तविक (m) के लिए सही है / It is true for every real (m). The sum (4+m) and product (4m) match the given polynomial. Therefore the statement is true for every real (m).
Step 3
Exam Tip
योग (4+m) और गुणनफल (4m) हैं, जो दिए बहुपद से मिलते हैं। इसलिए कथन हर वास्तविक (m) के लिए सही है।
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यदि (p(x)=2x-3 +kx-2 -11x+5) को (x+1) से भाग देने पर शेष (7) है, तो (k) क्या है?
If (p(x)=2x-3 +kx-2 -11x+5) leaves remainder (7) when divided by (x+1), what is (k)?
#remainder-theorem
#parameter
#trap
A (5)
B (-9)
C (-11)
D (3)
Explanation opens after your attempt
Step 1
Concept
By the remainder theorem (p(-1)=7), so (-2+k+11+5=7). This gives (k=-7), so none of the listed options is correct.
Step 2
Why this answer is correct
The correct answer is B. (-9). By the remainder theorem (p(-1)=7), so (-2+k+11+5=7). This gives (k=-7), so none of the listed options is correct.
Step 3
Exam Tip
शेष प्रमेय से (p(-1)=7), इसलिए (-2+k+11+5=7)। इससे (k=-7) नहीं बल्कि (k=-7) आता है, इसलिए विकल्पों में सही मान नहीं है।
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