यदि \(x^2-11x+30=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-11x+30=0\), what is \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?
#quadratic-roots
#rational-expression
#root-substitution
A \(\frac{29}{10}\)
B \(\frac{27}{10}\)
C \(\frac{31}{10}\)
D \(\frac{33}{10}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{29}{10}\)
Step 1
Concept
The roots are (5) and (6). Hence \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{29}{10}\). The roots are (5) and (6). Hence \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\).
Step 3
Exam Tip
जड़ें (5) और (6) हैं। इसलिए \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\)।
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यदि \(x^2-11x+30=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का सही मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-11x+30=0\), what is the correct value of \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?
#quadratic-roots
#rational-expression
#option-check
A \(\frac{17}{5}\)
B \(\frac{19}{5}\)
C \(\frac{21}{5}\)
D \(\frac{23}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{19}{5}\)
Step 1
Concept
The roots are (5) and (6). Direct substitution gives \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), so none of the given options is correct.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{19}{5}\). The roots are (5) and (6). Direct substitution gives \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), so none of the given options is correct.
Step 3
Exam Tip
जड़ें (5) और (6) हैं। सीधे रखने पर \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), इसलिए दिए गए विकल्पों में कोई सही नहीं है।
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यदि \(x^2-9x+20=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+20=0\), what is \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\)?
#quadratic-roots
#rational-expression
#error-check
A \(\frac{16}{3}\)
B \(\frac{17}{3}\)
C (6)
D \(\frac{19}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{17}{3}\)
Step 1
Concept
The roots are (4) and (5). Direct substitution gives \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\), so option (A) should be correct.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{17}{3}\). The roots are (4) and (5). Direct substitution gives \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\), so option (A) should be correct.
Step 3
Exam Tip
जड़ें (4) और (5) हैं। सीधे रखने पर \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\) मिलता है, इसलिए विकल्प (A) सही होना चाहिए।
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यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), what is \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?
#quadratic-roots
#rational-expression
#root-substitution
A \(\frac{8}{3}\)
B \(\frac{10}{3}\)
C \(\frac{11}{3}\)
D \(\frac{13}{3}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{11}{3}\)
Step 1
Concept
The roots are (3) and (4). Direct substitution gives \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{11}{3}\). The roots are (3) and (4). Direct substitution gives \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\).
Step 3
Exam Tip
जड़ें (3) और (4) हैं। सीधे रखने पर \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\) मिलता है।
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यदि \(x^2-7x+10=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-7x+10=0\), what is \(\frac{1}{\alpha-1}+\frac{1}{\beta-1}\)?
#quadratic-roots
#rational-expression
#sum-product
A \(\frac{5}{4}\)
B \(\frac{4}{5}\)
C \(\frac{7}{4}\)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{4}\)
Step 1
Concept
The denominator (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4). The numerator is \(\alpha+\beta-2=5\), so the value is \(\frac{5}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{4}\). The denominator (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4). The numerator is \(\alpha+\beta-2=5\), so the value is \(\frac{5}{4}\).
Step 3
Exam Tip
हर (\(\alpha-1\)\(\beta-1\)=\alpha\beta-\(\alpha+\beta\)+1=4) है। ऊपर \(\alpha+\beta-2=5\), इसलिए मान \(\frac{5}{4}\) है।
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यदि (p) और (q) परिमेय संख्याएं हैं तथा \(p+q\sqrt{5}\) परिमेय है, तो (q) के बारे में क्या सही है?
If (p) and (q) are rational numbers and \(p+q\sqrt{5}\) is rational, what is true about (q)?
#rational-expression
#irrational-coefficient
#reasoning
A (q=0)
B (q=1)
C \(q=\sqrt{5}\)
D (q=p)
Explanation opens after your attempt
Step 1
Concept
If \(q\ne0\), then \(q\sqrt{5}\) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.
Step 2
Why this answer is correct
The correct answer is A. (q=0). If \(q\ne0\), then \(q\sqrt{5}\) is irrational and the sum cannot be rational. In exams check the possibility of a zero coefficient.
Step 3
Exam Tip
यदि \(q\ne0\), तो \(q\sqrt{5}\) अपरिमेय होगा और योग परिमेय नहीं हो सकता। परीक्षा में शून्य गुणांक की संभावना देखें।
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कौन सा विकल्प \(12-\sqrt{m}\) को परिमेय बनाता है?
Which option makes \(12-\sqrt{m}\) rational?
#perfect-square
#rational-expression
#reasoning
A (m=225)
B (m=226)
C (m=227)
D (m=228)
Explanation opens after your attempt
Correct Answer
A. (m=225)
Step 1
Concept
\(\sqrt{225}=15\) is rational so \(12-\sqrt{225}\) will be rational. The root of a perfect square is rational.
Step 2
Why this answer is correct
The correct answer is A. (m=225). \(\sqrt{225}=15\) is rational so \(12-\sqrt{225}\) will be rational. The root of a perfect square is rational.
Step 3
Exam Tip
\(\sqrt{225}=15\) परिमेय है इसलिए \(12-\sqrt{225}\) परिमेय होगा। पूर्ण वर्ग की जड़ परिमेय होती है।
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कौन सा विकल्प \(5+\sqrt{m}\) को परिमेय बनाता है?
Which option makes \(5+\sqrt{m}\) rational?
#perfect-square
#rational-expression
#reasoning
A (m=121)
B (m=122)
C (m=123)
D (m=124)
Explanation opens after your attempt
Correct Answer
A. (m=121)
Step 1
Concept
\(\sqrt{121}=11\) is rational so \(5+\sqrt{121}\) will be rational. The root of a perfect square is rational.
Step 2
Why this answer is correct
The correct answer is A. (m=121). \(\sqrt{121}=11\) is rational so \(5+\sqrt{121}\) will be rational. The root of a perfect square is rational.
Step 3
Exam Tip
\(\sqrt{121}=11\) परिमेय है इसलिए \(5+\sqrt{121}\) परिमेय होगा। पूर्ण वर्ग वाली जड़ परिमेय होती है।
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कौन सा विकल्प \(9-\sqrt{m}\) को परिमेय बनाता है?
Which option makes \(9-\sqrt{m}\) rational?
#perfect-square
#rational-expression
#reasoning
A (m=100)
B (m=101)
C (m=102)
D (m=103)
Explanation opens after your attempt
Correct Answer
A. (m=100)
Step 1
Concept
\(\sqrt{100}=10\) is rational so \(9-\sqrt{100}\) will be rational. The root of a perfect square is rational.
Step 2
Why this answer is correct
The correct answer is A. (m=100). \(\sqrt{100}=10\) is rational so \(9-\sqrt{100}\) will be rational. The root of a perfect square is rational.
Step 3
Exam Tip
\(\sqrt{100}=10\) परिमेय है इसलिए \(9-\sqrt{100}\) परिमेय होगा। पूर्ण वर्ग वाली जड़ परिमेय होती है।
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