यदि (f(x)=2x-3) और (g(x)=x-2 +1) हों, तो (h(x)=2f(x)-g(x)) के लिए (h(2)) का मान क्या होगा?
If (f(x)=2x-3) and (g(x)=x-2 +1), what is the value of (h(2)) for (h(x)=2f(x)-g(x))?
#relations-functions
#algebra-real-functions
#function-evaluation
#medium
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A (-3)
B (3)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
Here (f(2)=1) and (g(2)=5), so (h(2)=2\cdot1-5=-3). In exams, first find individual function values, then apply the algebraic operation.
Step 2
Why this answer is correct
The correct answer is A. (-3). Here (f(2)=1) and (g(2)=5), so (h(2)=2\cdot1-5=-3). In exams, first find individual function values, then apply the algebraic operation.
Step 3
Exam Tip
यहाँ (f(2)=1) और (g(2)=5), इसलिए (h(2)=2\cdot1-5=-3)। परीक्षा में पहले अलग-अलग फलनों का मान निकालें, फिर बीजगणितीय क्रिया करें।
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यदि (f(x)=4x-7) और (g(x)=x+2) हों, तो ((f-g)(x)) का मान क्या है?
If (f(x)=4x-7) and (g(x)=x+2), what is ((f-g)(x))?
#relations functions
#subtraction of functions
#medium
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A (3x-9)
B (5x-5)
C (3x-5)
D (5x-9)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=f(x)-g(x)), so (4x-7-(x+2)=3x-9). Apply the minus sign to the whole (g(x)).
Step 2
Why this answer is correct
The correct answer is A. (3x-9). ((f-g)(x)=f(x)-g(x)), so (4x-7-(x+2)=3x-9). Apply the minus sign to the whole (g(x)).
Step 3
Exam Tip
((f-g)(x)=f(x)-g(x)), इसलिए (4x-7-(x+2)=3x-9)। ऋण चिह्न को पूरे (g(x)) पर लगाएं।
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वास्तविक फलनों (f(x)=\sqrt{x-1}) और (g(x)=\frac{1}{x-3}) के लिए गुणनफल (fg) का प्रांत क्या होगा?
For real functions (f(x)=\sqrt{x-1}) and (g(x)=\frac{1}{x-3}), what is the domain of the product (fg)?
#relations-functions
#algebra-real-functions
#domain-product-functions
#medium
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A \([1,\infty\)-{3})
B (\(1,\infty\))
C ([1,3))
D \(\mathbb{R}-{3}\)
Explanation opens after your attempt
Correct Answer
A. \([1,\infty\)-{3})
Step 1
Concept
For (f(x)), \(x\geq1\), and for (g(x)), \(x\neq3\), so the domain of (fg) is \([1,\infty\)-{3}). For product domains, take the intersection of the domains of both functions.
Step 2
Why this answer is correct
The correct answer is A. \([1,\infty\)-{3}). For (f(x)), \(x\geq1\), and for (g(x)), \(x\neq3\), so the domain of (fg) is \([1,\infty\)-{3}). For product domains, take the intersection of the domains of both functions.
Step 3
Exam Tip
(f(x)) के लिए \(x\geq1\) और (g(x)) के लिए \(x\neq3\), इसलिए (fg) का प्रांत \([1,\infty\)-{3}) है। उत्पाद के प्रांत में दोनों फलनों के प्रांतों का प्रतिच्छेद लें।
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यदि (f(x)=x-2 -9) और (g(x)=x-3) हों, तो \(x \ne 3\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या होगा?
If (f(x)=x-2 -9) and (g(x)=x-3), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \ne 3\)?
#relations functions
#quotient of functions
#domain
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A (x+3)
B (x-3)
C \(x^2+9\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), but (x=3) is excluded. In division, the denominator must not be zero.
Step 2
Why this answer is correct
The correct answer is A. (x+3). (\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), but (x=3) is excluded. In division, the denominator must not be zero.
Step 3
Exam Tip
(\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), लेकिन (x=3) हटेगा। भाग में हर शून्य न हो।
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यदि (f(x)=3x-2 ) और (g(x)=2x+1) हों, तो ((f+2g)(x)) क्या है?
If (f(x)=3x-2 ) and (g(x)=2x+1), what is ((f+2g)(x))?
#relations functions
#linear combination
#algebra
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A \(3x^2+4x+2\)
B \(6x^2+2x+1\)
C \(3x^2+2x+1\)
D \(5x^2+1\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+4x+2\)
Step 1
Concept
((f+2g)(x)=f(x)+2g(x)=3x-2 +2(2x+1)). First multiply, then add.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+4x+2\). ((f+2g)(x)=f(x)+2g(x)=3x-2 +2(2x+1)). First multiply, then add.
Step 3
Exam Tip
((f+2g)(x)=f(x)+2g(x)=3x-2 +2(2x+1))। पहले गुणा करें, फिर जोड़ें।
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यदि (f(x)=\sqrt{x-2}) और (g(x)=\sqrt{5-x}) हों, तो ((f+g)(x)) का डोमेन क्या है?
If (f(x)=\sqrt{x-2}) and (g(x)=\sqrt{5-x}), what is the domain of ((f+g)(x))?
#relations functions
#domain
#square root
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A ([2,5])
B ((2,5))
C (\(-\infty,2]\)
D \([5,\infty\))
Explanation opens after your attempt
Correct Answer
A. ([2,5])
Step 1
Concept
For both roots, \(x-2\ge 0\) and \(5-x\ge 0\), so \(2\le x\le 5\). For a sum, take the intersection of domains.
Step 2
Why this answer is correct
The correct answer is A. ([2,5]). For both roots, \(x-2\ge 0\) and \(5-x\ge 0\), so \(2\le x\le 5\). For a sum, take the intersection of domains.
Step 3
Exam Tip
दोनों मूलों के लिए \(x-2\ge 0\) और \(5-x\ge 0\), इसलिए \(2\le x\le 5\)। जोड़ में डोमेन प्रतिच्छेद होता है।
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यदि (f(x)=\frac{1}{x-1}) और (g(x)=\frac{1}{x+2}) हों, तो ((f+g)(x)) का डोमेन क्या है?
If (f(x)=\frac{1}{x-1}) and (g(x)=\frac{1}{x+2}), what is the domain of ((f+g)(x))?
#relations functions
#rational functions
#domain
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A \(\mathbb{R}-{-2,1}\)
B \(\mathbb{R}-{1}\)
C \(\mathbb{R}-{-2}\)
D ({-2,1})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-2,1}\)
Step 1
Concept
The denominators require \(x-1\ne 0\) and \(x+2\ne 0\), so \(x\ne 1,-2\). For addition, use the common domain of both functions.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-2,1}\). The denominators require \(x-1\ne 0\) and \(x+2\ne 0\), so \(x\ne 1,-2\). For addition, use the common domain of both functions.
Step 3
Exam Tip
हर में \(x-1\ne 0\) और \(x+2\ne 0\), इसलिए \(x\ne 1,-2\)। जोड़ के लिए दोनों फलनों का साझा डोमेन लें।
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यदि (f(x)=x-2 +1) और (g(x)=x-2) हों, तो ((fg)(3)) का मान क्या है?
If (f(x)=x-2 +1) and (g(x)=x-2), what is the value of ((fg)(3))?
#relations functions
#function value
#product
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A (10)
B (12)
C (8)
D (6)
Explanation opens after your attempt
Step 1
Concept
(f(3)=10) and (g(3)=1), so ((fg)(3)=10). It is safer to find each value first.
Step 2
Why this answer is correct
The correct answer is A. (10). (f(3)=10) and (g(3)=1), so ((fg)(3)=10). It is safer to find each value first.
Step 3
Exam Tip
(f(3)=10) और (g(3)=1), इसलिए ((fg)(3)=10)। पहले अलग-अलग मान निकालना सुरक्षित रहता है।
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यदि (f(x)=2x-1) और (g(x)=x-2 ) हों, तो ((f+g)(-2)) क्या होगा?
If (f(x)=2x-1) and (g(x)=x-2 ), what is ((f+g)(-2))?
#relations functions
#evaluation
#addition
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A (-1)
B (1)
C (9)
D (-9)
Explanation opens after your attempt
Step 1
Concept
(f(-2)=-5) and (g(-2)=4), so the sum is (-1). The square of a negative number is positive.
Step 2
Why this answer is correct
The correct answer is A. (-1). (f(-2)=-5) and (g(-2)=4), so the sum is (-1). The square of a negative number is positive.
Step 3
Exam Tip
(f(-2)=-5) और (g(-2)=4), इसलिए योग (-1) है। ऋणात्मक संख्या का वर्ग धनात्मक होता है।
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यदि (f(x)=5x+4) और (g(x)=2x-3) हों, तो ((f-2g)(x)) क्या होगा?
If (f(x)=5x+4) and (g(x)=2x-3), what is ((f-2g)(x))?
#relations functions
#linear combination
#subtraction
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A (x+10)
B (9x-2)
C (x-2)
D (3x+1)
Explanation opens after your attempt
Step 1
Concept
((f-2g)(x)=5x+4-2(2x-3)=x+10). Be careful with signs while opening brackets.
Step 2
Why this answer is correct
The correct answer is A. (x+10). ((f-2g)(x)=5x+4-2(2x-3)=x+10). Be careful with signs while opening brackets.
Step 3
Exam Tip
((f-2g)(x)=5x+4-2(2x-3)=x+10)। कोष्ठक खोलते समय चिह्न न बदलें।
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यदि (f(x)=x-2 ) और (g(x)=3x) हों, तो वह (x) कौन-सा है जिसके लिए ((f-g)(x)=0)?
If (f(x)=x-2 ) and (g(x)=3x), which (x) satisfies ((f-g)(x)=0)?
#relations functions
#equation
#zero product
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A (x=0) या (x=3) / (x=0) or (x=3)
B (x=1) या (x=3) / (x=1) or (x=3)
C (x=-3) या (x=0) / (x=-3) or (x=0)
D केवल (x=3) / only (x=3)
Explanation opens after your attempt
Correct Answer
A. (x=0) या (x=3) / (x=0) or (x=3)
Step 1
Concept
((f-g)(x)=x-2 -3x=x(x-3)), so (x=0) or (x=3). Remember the zero product rule.
Step 2
Why this answer is correct
The correct answer is A. (x=0) या (x=3) / (x=0) or (x=3). ((f-g)(x)=x-2 -3x=x(x-3)), so (x=0) or (x=3). Remember the zero product rule.
Step 3
Exam Tip
((f-g)(x)=x-2 -3x=x(x-3)), इसलिए (x=0) या (x=3)। शून्य गुणनफल नियम याद रखें।
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यदि (f(x)=x+4) और (g(x)=2x-1) हों, तो ((f+g)(x)=(fg)(x)) के लिए कौन-सा समीकरण बनेगा?
If (f(x)=x+4) and (g(x)=2x-1), which equation is formed for ((f+g)(x)=(fg)(x))?
#relations functions
#equation setup
#product
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A \(2x^2+6x-4=3x+3\)
B \(2x^2+7x-4=3x+3\)
C \(2x^2+7x-4=x+3\)
D \(2x^2+8x-4=3x+3\)
Explanation opens after your attempt
Correct Answer
B. \(2x^2+7x-4=3x+3\)
Step 1
Concept
((f+g)(x)=3x+3) and ((fg)(x)=(x+4)(2x-1)=2x-2 +7x-4). Write both sides separately first.
Step 2
Why this answer is correct
The correct answer is B. \(2x^2+7x-4=3x+3\). ((f+g)(x)=3x+3) and ((fg)(x)=(x+4)(2x-1)=2x-2 +7x-4). Write both sides separately first.
Step 3
Exam Tip
((f+g)(x)=3x+3) और ((fg)(x)=(x+4)(2x-1)=2x-2 +7x-4)। पहले दोनों पक्ष अलग लिखें।
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यदि (f(x)=\frac{x+1}{x-2}) और (g(x)=x-2 -4) हों, तो ((fg)(x)) का डोमेन क्या है?
If (f(x)=\frac{x+1}{x-2}) and (g(x)=x-2 -4), what is the domain of ((fg)(x))?
#relations functions
#product domain
#rational
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A \(\mathbb{R}-{2}\)
B \(\mathbb{R}-{-2,2}\)
C \(\mathbb{R}\)
D \(\mathbb{R}-{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{2}\)
Step 1
Concept
(g(x)) is defined for all real (x), but (f(x)) needs \(x-2\ne 0\). For product also, take the common domain.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{2}\). (g(x)) is defined for all real (x), but (f(x)) needs \(x-2\ne 0\). For product also, take the common domain.
Step 3
Exam Tip
(g(x)) सभी वास्तविक (x) पर परिभाषित है, पर (f(x)) में \(x-2\ne 0\) चाहिए। गुणन में भी साझा डोमेन लेते हैं।
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यदि (f(x)=x-1) और (g(x)=x+1) हों, तो (\left\(\frac{f}{g}\right\)(x)) का डोमेन क्या है?
If (f(x)=x-1) and (g(x)=x+1), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations functions
#quotient domain
#real functions
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A \(\mathbb{R}-{-1}\)
B \(\mathbb{R}-{1}\)
C \(\mathbb{R}-{-1,1}\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-1}\)
Step 1
Concept
In division, (g(x)\ne 0), so \(x+1\ne 0\) and \(x\ne -1\). The numerator may be zero.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-1}\). In division, (g(x)\ne 0), so \(x+1\ne 0\) and \(x\ne -1\). The numerator may be zero.
Step 3
Exam Tip
भाग में (g(x)\ne 0) चाहिए, इसलिए \(x+1\ne 0\) और \(x\ne -1\)। अंश शून्य हो सकता है।
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यदि (f(x)=2x-2 -3) और (g(x)=x-2 +5) हों, तो ((f-g)(x)) क्या होगा?
If (f(x)=2x-2 -3) and (g(x)=x-2 +5), what is ((f-g)(x))?
#relations functions
#polynomial functions
#subtraction
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A \(x^2-8\)
B \(3x^2+2\)
C \(x^2+2\)
D \(3x^2-8\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8\)
Step 1
Concept
((f-g)(x)=2x-2 -3-\(x^2+5\)=x-2 -8). The whole (g(x)) is subtracted.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8\). ((f-g)(x)=2x-2 -3-\(x^2+5\)=x-2 -8). The whole (g(x)) is subtracted.
Step 3
Exam Tip
((f-g)(x)=2x-2 -3-\(x^2+5\)=x-2 -8)। पूरा (g(x)) घटाया जाता है।
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यदि (f(x)=3) और (g(x)=x-2 -2x) हों, तो ((fg)(x)) क्या होगा?
If (f(x)=3) and (g(x)=x-2 -2x), what is ((fg)(x))?
#relations functions
#constant function
#product
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A \(3x^2-6x\)
B \(x^2-2x+3\)
C \(3x^2-2x\)
D \(x^2-6x\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-6x\)
Step 1
Concept
The constant function (3) multiplies the whole (g(x)). Thus ((fg)(x)=3\(x^2-2x\)=3x-2 -6x).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-6x\). The constant function (3) multiplies the whole (g(x)). Thus ((fg)(x)=3\(x^2-2x\)=3x-2 -6x).
Step 3
Exam Tip
स्थिर फलन (3) पूरे (g(x)) से गुणा होगा। इसलिए ((fg)(x)=3\(x^2-2x\)=3x-2 -6x)।
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यदि (f(x)=|x|) और (g(x)=x) हों, तो ((f+g)(-4)) क्या होगा?
If (f(x)=|x|) and (g(x)=x), what is ((f+g)(-4))?
#relations functions
#modulus function
#value
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A (0)
B (-8)
C (8)
D (-4)
Explanation opens after your attempt
Step 1
Concept
(f(-4)=|-4|=4) and (g(-4)=-4), so the sum is (0). Modulus is always non-negative.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(-4)=|-4|=4) and (g(-4)=-4), so the sum is (0). Modulus is always non-negative.
Step 3
Exam Tip
(f(-4)=|-4|=4) और (g(-4)=-4), इसलिए योग (0) है। मापांक हमेशा अऋणात्मक होता है।
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यदि (f(x)=\frac{1}{x}) और (g(x)=x-2 ) हों, तो ((fg)(x)) को सरल करें।
If (f(x)=\frac{1}{x}) and (g(x)=x-2 ), simplify ((fg)(x)).
#relations functions
#domain restriction
#product
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A (x), \(x\ne 0\)
B \(x^2\), \(x\ne 0\)
C \(\frac{1}{x^2}\), \(x\ne 0\)
D (x), \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (x), \(x\ne 0\)
Step 1
Concept
((fg)(x)=\frac{1}{x}\cdot x-2 =x), but the original domain keeps \(x\ne 0\). Restrictions do not disappear after simplification.
Step 2
Why this answer is correct
The correct answer is A. (x), \(x\ne 0\). ((fg)(x)=\frac{1}{x}\cdot x-2 =x), but the original domain keeps \(x\ne 0\). Restrictions do not disappear after simplification.
Step 3
Exam Tip
((fg)(x)=\frac{1}{x}\cdot x-2 =x), पर मूल डोमेन में \(x\ne 0\) रहेगा। सरल करने से हटे प्रतिबंध वापस नहीं आते।
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यदि (f(x)=x-2 +2x) और (g(x)=x-2 -2x) हों, तो ((f-g)(x)) क्या है?
If (f(x)=x-2 +2x) and (g(x)=x-2 -2x), what is ((f-g)(x))?
#relations functions
#like terms
#subtraction
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A (4x)
B \(2x^2\)
C \(x^2+4x\)
D (0)
Explanation opens after your attempt
Step 1
Concept
The \(x^2\) terms cancel and (2x-(-2x)=4x). The minus sign changes the second term.
Step 2
Why this answer is correct
The correct answer is A. (4x). The \(x^2\) terms cancel and (2x-(-2x)=4x). The minus sign changes the second term.
Step 3
Exam Tip
समान \(x^2\) पद कट जाते हैं और (2x-(-2x)=4x)। ऋण चिह्न के कारण दूसरा पद बदलता है।
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यदि (f(x)=2x+1) और (g(x)=x-2 -1) हों, तो ((f+g)(2)) क्या होगा?
If (f(x)=2x+1) and (g(x)=x-2 -1), what is ((f+g)(2))?
#relations functions
#evaluation
#addition
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A (8)
B (6)
C (10)
D (4)
Explanation opens after your attempt
Step 1
Concept
(f(2)=5) and (g(2)=3), so ((f+g)(2)=8). Substitute (2) for (x) carefully.
Step 2
Why this answer is correct
The correct answer is A. (8). (f(2)=5) and (g(2)=3), so ((f+g)(2)=8). Substitute (2) for (x) carefully.
Step 3
Exam Tip
(f(2)=5) और (g(2)=3), इसलिए ((f+g)(2)=8)। मान निकालते समय (x) की जगह (2) रखें।
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यदि (f(x)=\sqrt{x}) और (g(x)=\frac{1}{x-4}) हों, तो ((fg)(x)) का डोमेन क्या है?
If (f(x)=\sqrt{x}) and (g(x)=\frac{1}{x-4}), what is the domain of ((fg)(x))?
#relations functions
#domain
#product
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A \([0,\infty\)-{4})
B (\(0,\infty\))
C ([0,4))
D \(\mathbb{R}-{4}\)
Explanation opens after your attempt
Correct Answer
A. \([0,\infty\)-{4})
Step 1
Concept
\(\sqrt{x}\) needs \(x\ge 0\), and the denominator needs \(x\ne 4\). Take the intersection of both conditions.
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)-{4}). \(\sqrt{x}\) needs \(x\ge 0\), and the denominator needs \(x\ne 4\). Take the intersection of both conditions.
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर के लिए \(x\ne 4\) चाहिए। दोनों शर्तों का प्रतिच्छेद लें।
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यदि (f(x)=x-2 +4x+4) और (g(x)=x+2) हों, तो \(x\ne -2\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या है?
If (f(x)=x-2 +4x+4) and (g(x)=x+2), what is (\left\(\frac{f}{g}\right\)(x)) for \(x\ne -2\)?
#relations functions
#quotient
#factorisation
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A (x+2)
B (x-2)
C \(x^2+2\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(x-2 +4x+4=(x+2)2 ), so the quotient is (x+2) and (x=-2) is excluded. The square identity is useful.
Step 2
Why this answer is correct
The correct answer is A. (x+2). (x-2 +4x+4=(x+2)2 ), so the quotient is (x+2) and (x=-2) is excluded. The square identity is useful.
Step 3
Exam Tip
(x-2 +4x+4=(x+2)2 ), इसलिए भागफल (x+2) है और (x=-2) हटेगा। वर्ग पहचान उपयोगी है।
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यदि (f(x)=x-3) और (g(x)=x+5) हों, तो ((fg)(0)) का मान क्या है?
If (f(x)=x-3) and (g(x)=x+5), what is the value of ((fg)(0))?
#relations functions
#function value
#product
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A (-15)
B (15)
C (2)
D (-8)
Explanation opens after your attempt
Step 1
Concept
(f(0)=-3) and (g(0)=5), so the product is (-15). When (x=0), constant terms remain.
Step 2
Why this answer is correct
The correct answer is A. (-15). (f(0)=-3) and (g(0)=5), so the product is (-15). When (x=0), constant terms remain.
Step 3
Exam Tip
(f(0)=-3) और (g(0)=5), इसलिए गुणनफल (-15) है। (x=0) रखने पर स्थिर पद बचते हैं।
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यदि (f(x)=\frac{x}{x-1}) और (g(x)=\frac{x-1}{x+1}) हों, तो ((fg)(x)) का सरल रूप और प्रतिबंध क्या है?
If (f(x)=\frac{x}{x-1}) and (g(x)=\frac{x-1}{x+1}), what is the simplified form and restriction of ((fg)(x))?
#relations functions
#product
#simplification
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A \(\frac{x}{x+1}\), \(x\ne 1,-1\)
B \(\frac{x}{x+1}\), \(x\ne -1\)
C (x), \(x\ne 1,-1\)
D \(\frac{x-1}{x}\), \(x\ne 0\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{x}{x+1}\), \(x\ne 1,-1\)
Step 1
Concept
The simplified form is \(\frac{x}{x+1}\), but original denominators keep \(x\ne 1,-1\). Remember restrictions from cancelled factors too.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x}{x+1}\), \(x\ne 1,-1\). The simplified form is \(\frac{x}{x+1}\), but original denominators keep \(x\ne 1,-1\). Remember restrictions from cancelled factors too.
Step 3
Exam Tip
सरल रूप \(\frac{x}{x+1}\) है, पर मूल हरों से \(x\ne 1,-1\) रहेगा। कटे हुए गुणनखंड का प्रतिबंध भी याद रखें।
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यदि (f(x)=2x), (g(x)=x+3) और (h(x)=x-1) हों, तो ((f+g-h)(x)) क्या होगा?
If (f(x)=2x), (g(x)=x+3), and (h(x)=x-1), what is ((f+g-h)(x))?
#relations functions
#three functions
#algebra
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A (2x+4)
B (4x+2)
C (2x+2)
D (x+4)
Explanation opens after your attempt
Step 1
Concept
(2x+(x+3)-(x-1)=2x+4). Even with three functions, combining like terms is the key step.
Step 2
Why this answer is correct
The correct answer is A. (2x+4). (2x+(x+3)-(x-1)=2x+4). Even with three functions, combining like terms is the key step.
Step 3
Exam Tip
(2x+(x+3)-(x-1)=2x+4)। तीन फलनों में भी समान पद जोड़ना ही मुख्य काम है।
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यदि (f(x)=x-2 ) और (g(x)=2) हों, तो ((f+g)(x)) और ((fg)(x)) बराबर कब होंगे?
If (f(x)=x-2 ) and (g(x)=2), when will ((f+g)(x)) and ((fg)(x)) be equal?
#relations functions
#equality
#equation
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A \(x=\pm \sqrt{2}\)
B \(x=\pm 2\)
C (x=0)
D कोई वास्तविक मान नहीं / no real value
Explanation opens after your attempt
Correct Answer
A. \(x=\pm \sqrt{2}\)
Step 1
Concept
The equation \(x^2+2=2x^2\) gives \(x^2=2\), so \(x=\pm\sqrt{2}\). Equate both expressions and solve.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm \sqrt{2}\). The equation \(x^2+2=2x^2\) gives \(x^2=2\), so \(x=\pm\sqrt{2}\). Equate both expressions and solve.
Step 3
Exam Tip
समीकरण \(x^2+2=2x^2\) से \(x^2=2\), इसलिए \(x=\pm\sqrt{2}\)। दोनों पक्षों को बराबर रखकर हल करें।
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यदि (f(x)=x-2 -1) और (g(x)=x+1) हों, तो \(x\ne -1\) के लिए (\left\(\frac{f}{g}\right\)(2)) क्या होगा?
If (f(x)=x-2 -1) and (g(x)=x+1), what is (\left\(\frac{f}{g}\right\)(2)) for \(x\ne -1\)?
#relations functions
#quotient
#evaluation
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A (1)
B (3)
C (2)
D (5)
Explanation opens after your attempt
Step 1
Concept
\(\frac{x^2-1}{x+1}=x-1\), so at (x=2) the value is (1). Directly, (f(2)=3) and (g(2)=3), so the value is (1).
Step 2
Why this answer is correct
The correct answer is B. (3). \(\frac{x^2-1}{x+1}=x-1\), so at (x=2) the value is (1). Directly, (f(2)=3) and (g(2)=3), so the value is (1).
Step 3
Exam Tip
\(\frac{x^2-1}{x+1}=x-1\), अतः (x=2) पर मान (1) नहीं बल्कि \(\frac{3}{3}\) से (1) होता है। सही गणना में (f(2)=3) और (g(2)=3), इसलिए मान (1) है।
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यदि (f(x)=\sqrt{x+1}) और (g(x)=\sqrt{x-3}) हों, तो ((f-g)(x)) का डोमेन क्या होगा?
If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{x-3}), what is the domain of ((f-g)(x))?
#relations functions
#domain
#square root
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A \([3,\infty\))
B \([-1,\infty\))
C ([-1,3])
D (\(3,\infty\))
Explanation opens after your attempt
Correct Answer
A. \([3,\infty\))
Step 1
Concept
Both square roots must be defined, so \(x\ge -1\) and \(x\ge 3\). The common domain is \([3,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([3,\infty\)). Both square roots must be defined, so \(x\ge -1\) and \(x\ge 3\). The common domain is \([3,\infty\)).
Step 3
Exam Tip
दोनों वर्गमूल परिभाषित होने चाहिए, इसलिए \(x\ge -1\) और \(x\ge 3\)। साझा डोमेन \([3,\infty\)) है।
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यदि (f(x)=x-2 +3x+2) और (g(x)=x+1) हों, तो \(x\ne -1\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या है?
If (f(x)=x-2 +3x+2) and (g(x)=x+1), what is (\left\(\frac{f}{g}\right\)(x)) for \(x\ne -1\)?
#relations functions
#quotient
#factorisation
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A (x+2)
B (x+1)
C (x-2)
D \(x^2+2\)
Explanation opens after your attempt
Step 1
Concept
(x-2 +3x+2=(x+1)(x+2)), so the quotient is (x+2). Exclude the zero of the denominator first.
Step 2
Why this answer is correct
The correct answer is A. (x+2). (x-2 +3x+2=(x+1)(x+2)), so the quotient is (x+2). Exclude the zero of the denominator first.
Step 3
Exam Tip
(x-2 +3x+2=(x+1)(x+2)), इसलिए भागफल (x+2) है। हर का शून्य मान पहले हटाएं।
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यदि (f(x)=2x-2 +x) और (g(x)=x-2 -x) हों, तो ((2f-g)(x)) क्या होगा?
If (f(x)=2x-2 +x) and (g(x)=x-2 -x), what is ((2f-g)(x))?
#relations functions
#linear combination
#polynomial
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A \(3x^2+3x\)
B \(5x^2+x\)
C \(3x^2+x\)
D \(4x^2+2x\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+3x\)
Step 1
Concept
(2f(x)=4x-2 +2x), then (4x-2 +2x-\(x^2-x\)=3x-2 +3x). Find (2f(x)) first.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+3x\). (2f(x)=4x-2 +2x), then (4x-2 +2x-\(x^2-x\)=3x-2 +3x). Find (2f(x)) first.
Step 3
Exam Tip
(2f(x)=4x-2 +2x), फिर (4x-2 +2x-\(x^2-x\)=3x-2 +3x)। पहले (2f(x)) निकालें।
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यदि (f(x)=x+2) और (g(x)=x-2) हों, तो ((f+g)(x)\cdot(f-g)(x)) क्या होगा?
If (f(x)=x+2) and (g(x)=x-2), what is ((f+g)(x)\cdot(f-g)(x))?
#relations functions
#combined operations
#product
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A (8x)
B (4x)
C \(x^2-4\)
D (2x+4)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=2x) and ((f-g)(x)=4), so the product is (8x). Solve combined expressions step by step.
Step 2
Why this answer is correct
The correct answer is A. (8x). ((f+g)(x)=2x) and ((f-g)(x)=4), so the product is (8x). Solve combined expressions step by step.
Step 3
Exam Tip
((f+g)(x)=2x) और ((f-g)(x)=4), इसलिए गुणनफल (8x) है। संयुक्त अभिव्यक्ति को चरणों में हल करें।
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यदि (f(x)=\frac{1}{x-2 -4}) और (g(x)=x+2) हों, तो ((fg)(x)) का डोमेन क्या है?
If (f(x)=\frac{1}{x-2 -4}) and (g(x)=x+2), what is the domain of ((fg)(x))?
#relations functions
#domain
#rational product
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A \(\mathbb{R}-{-2,2}\)
B \(\mathbb{R}-{2}\)
C \(\mathbb{R}-{-2}\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-2,2}\)
Step 1
Concept
(x-2 -4=(x-2)(x+2)), so \(x\ne 2,-2\). Even if cancellation appears after multiplication, do not forget the original domain.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-2,2}\). (x-2 -4=(x-2)(x+2)), so \(x\ne 2,-2\). Even if cancellation appears after multiplication, do not forget the original domain.
Step 3
Exam Tip
(x-2 -4=(x-2)(x+2)), इसलिए \(x\ne 2,-2\)। गुणा के बाद कटाव दिखे तो भी मूल डोमेन न भूलें।
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यदि (f(x)=x-2 +1) और (g(x)=2x) हों, तो ((f+g)(x)) किसके बराबर है?
If (f(x)=x-2 +1) and (g(x)=2x), which expression equals ((f+g)(x))?
#relations functions
#addition
#polynomial
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A \(x^2+2x+1\)
B \(x^2-2x+1\)
C \(2x^3+2x\)
D \(x^2+1\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x+1\)
Step 1
Concept
((f+g)(x)=x-2 +1+2x=x-2 +2x+1). Changing the order of terms does not change the value.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x+1\). ((f+g)(x)=x-2 +1+2x=x-2 +2x+1). Changing the order of terms does not change the value.
Step 3
Exam Tip
((f+g)(x)=x-2 +1+2x=x-2 +2x+1)। पदों का क्रम बदलने से मान नहीं बदलता।
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यदि (f(x)=x-2 -4x+4) और (g(x)=x-2) हों, तो \(x\ne 2\) के लिए (\left\(\frac{f}{g}\right\)(5)) क्या है?
If (f(x)=x-2 -4x+4) and (g(x)=x-2), what is (\left\(\frac{f}{g}\right\)(5)) for \(x\ne 2\)?
#relations functions
#quotient
#evaluation
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A (3)
B (9)
C (7)
D (1)
Explanation opens after your attempt
Step 1
Concept
(\frac{(x-2)2 }{x-2}=x-2), so at (x=5) the value is (3). Simplify first, then substitute.
Step 2
Why this answer is correct
The correct answer is A. (3). (\frac{(x-2)2 }{x-2}=x-2), so at (x=5) the value is (3). Simplify first, then substitute.
Step 3
Exam Tip
(\frac{(x-2)2 }{x-2}=x-2), इसलिए (x=5) पर मान (3) है। पहले सरल करें, फिर मान रखें।
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यदि (f(x)=3x-2) और (g(x)=x+4) हों, तो ((fg)(1)) क्या होगा?
If (f(x)=3x-2) and (g(x)=x+4), what is ((fg)(1))?
#relations functions
#product
#evaluation
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A (5)
B (7)
C (3)
D (9)
Explanation opens after your attempt
Step 1
Concept
(f(1)=1) and (g(1)=5), so ((fg)(1)=5). In a product function, multiply the values.
Step 2
Why this answer is correct
The correct answer is A. (5). (f(1)=1) and (g(1)=5), so ((fg)(1)=5). In a product function, multiply the values.
Step 3
Exam Tip
(f(1)=1) और (g(1)=5), इसलिए ((fg)(1)=5)। गुणनफल फलन में मानों को गुणा करें।
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यदि (f(x)=\frac{x+2}{x-3}) और (g(x)=x-3) हों, तो ((fg)(x)) का सरल रूप और डोमेन क्या है?
If (f(x)=\frac{x+2}{x-3}) and (g(x)=x-3), what is the simplified form and domain of ((fg)(x))?
#relations functions
#product
#domain restriction
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A (x+2), \(x\ne 3\)
B (x+2), \(x\in\mathbb{R}\)
C (x-3), \(x\ne 3\)
D \(\frac{x+2}{x-3}\), \(x\ne 3\)
Explanation opens after your attempt
Correct Answer
A. (x+2), \(x\ne 3\)
Step 1
Concept
Multiplication gives (x+2), but \(x\ne 3\) remains because of (f(x)). Always check the original function conditions.
Step 2
Why this answer is correct
The correct answer is A. (x+2), \(x\ne 3\). Multiplication gives (x+2), but \(x\ne 3\) remains because of (f(x)). Always check the original function conditions.
Step 3
Exam Tip
गुणन से (x+2) मिलता है, लेकिन (f(x)) के कारण \(x\ne 3\) रहेगा। मूल फलन की शर्तें हमेशा जांचें।
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यदि (f(x)=x-3 ) और (g(x)=x) हों, तो ((f-g)(x)) किस रूप में लिखा जा सकता है?
If (f(x)=x-3 ) and (g(x)=x), how can ((f-g)(x)) be written?
#relations functions
#factorisation
#difference
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A (x(x-1)(x+1))
B (x(x+1)2 )
C (x-2 (x-1))
D (x\(x^2+1\))
Explanation opens after your attempt
Correct Answer
A. (x(x-1)(x+1))
Step 1
Concept
((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1)). Use the difference of squares identity.
Step 2
Why this answer is correct
The correct answer is A. (x(x-1)(x+1)). ((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1)). Use the difference of squares identity.
Step 3
Exam Tip
((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1))। अंतर वर्ग पहचान का उपयोग करें।
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यदि (f(x)=2x+5) और (g(x)=5-2x) हों, तो ((f+g)(x)) क्या है?
If (f(x)=2x+5) and (g(x)=5-2x), what is ((f+g)(x))?
#relations functions
#addition
#constant result
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A (10)
B (4x)
C (10-4x)
D (0)
Explanation opens after your attempt
Step 1
Concept
(2x) and (-2x) cancel, so the sum is (10). Identifying opposite terms saves time.
Step 2
Why this answer is correct
The correct answer is A. (10). (2x) and (-2x) cancel, so the sum is (10). Identifying opposite terms saves time.
Step 3
Exam Tip
(2x) और (-2x) कट जाते हैं, इसलिए योग (10) है। विपरीत पदों को पहचानना समय बचाता है।
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यदि (f(x)=x-2 +6x+9) और (g(x)=x+3) हों, तो (\left\(\frac{f}{g}\right\)(0)) क्या होगा, जहाँ \(x\ne -3\)?
If (f(x)=x-2 +6x+9) and (g(x)=x+3), what is (\left\(\frac{f}{g}\right\)(0)), where \(x\ne -3\)?
#relations functions
#quotient
#evaluation
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A (3)
B (9)
C (0)
D (6)
Explanation opens after your attempt
Step 1
Concept
(\frac{(x+3)2 }{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).
Step 2
Why this answer is correct
The correct answer is A. (3). (\frac{(x+3)2 }{x+3}=x+3), so at (x=0) the value is (3). Keep the restriction \(x\ne -3\).
Step 3
Exam Tip
(\frac{(x+3)2 }{x+3}=x+3), इसलिए (x=0) पर मान (3) है। प्रतिबंध \(x\ne -3\) साथ रखें।
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यदि (f(x)=\frac{1}{\sqrt{x-1}}) और (g(x)=x-2 ) हों, तो ((f+g)(x)) का डोमेन क्या है?
If (f(x)=\frac{1}{\sqrt{x-1}}) and (g(x)=x-2 ), what is the domain of ((f+g)(x))?
#relations functions
#domain
#radical denominator
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A (\(1,\infty\))
B \([1,\infty\))
C \(\mathbb{R}\)
D (\(-\infty,1\))
Explanation opens after your attempt
Correct Answer
A. (\(1,\infty\))
Step 1
Concept
The denominator has \(\sqrt{x-1}\), so (x-1>0) is needed. Hence the domain is (\(1,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(1,\infty\)). The denominator has \(\sqrt{x-1}\), so (x-1>0) is needed. Hence the domain is (\(1,\infty\)).
Step 3
Exam Tip
हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) चाहिए। अतः डोमेन (\(1,\infty\)) है।
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यदि (f(x)=x-2 ) और (g(x)=x+1) हों, तो ((f+g)(a)) क्या होगा?
If (f(x)=x-2 ) and (g(x)=x+1), what is ((f+g)(a))?
#relations functions
#symbolic value
#addition
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A \(a^2+a+1\)
B \(a^2+1\)
C \(a^2+a\)
D (2a+1)
Explanation opens after your attempt
Correct Answer
A. \(a^2+a+1\)
Step 1
Concept
((f+g)(x)=x-2 +x+1), so putting (x=a) gives \(a^2+a+1\). The rule stays the same when the variable changes.
Step 2
Why this answer is correct
The correct answer is A. \(a^2+a+1\). ((f+g)(x)=x-2 +x+1), so putting (x=a) gives \(a^2+a+1\). The rule stays the same when the variable changes.
Step 3
Exam Tip
((f+g)(x)=x-2 +x+1), इसलिए (x=a) रखने पर \(a^2+a+1\)। चर बदलने पर नियम वही रहता है।
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यदि (f(x)=2x-3) और (g(x)=x-2 +1) हों, तो ((3f+g)(x)) क्या है?
If (f(x)=2x-3) and (g(x)=x-2 +1), what is ((3f+g)(x))?
#relations functions
#linear combination
#algebra
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A \(x^2+6x-8\)
B \(3x^2+2x-2\)
C \(x^2+6x+10\)
D \(6x^2+3x-2\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+6x-8\)
Step 1
Concept
(3f(x)=6x-9), then \(6x-9+x^2+1=x^2+6x-8\). Add constant terms correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+6x-8\). (3f(x)=6x-9), then \(6x-9+x^2+1=x^2+6x-8\). Add constant terms correctly.
Step 3
Exam Tip
(3f(x)=6x-9), फिर \(6x-9+x^2+1=x^2+6x-8\)। स्थिर पदों को सही जोड़ें।
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यदि (f(x)=x-2 -5x+6) और (g(x)=x-2) हों, तो \(x\ne 2\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या है?
If (f(x)=x-2 -5x+6) and (g(x)=x-2), what is (\left\(\frac{f}{g}\right\)(x)) for \(x\ne 2\)?
#relations functions
#quotient
#factorisation
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A (x-3)
B (x+3)
C (x-2)
D \(x^2-3\)
Explanation opens after your attempt
Step 1
Concept
(x-2 -5x+6=(x-2)(x-3)), so the quotient is (x-3). Factorisation helps in quotient problems.
Step 2
Why this answer is correct
The correct answer is A. (x-3). (x-2 -5x+6=(x-2)(x-3)), so the quotient is (x-3). Factorisation helps in quotient problems.
Step 3
Exam Tip
(x-2 -5x+6=(x-2)(x-3)), इसलिए भागफल (x-3) है। गुणनखंड बनाना भागफल में मदद करता है।
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यदि (f(x)=x+1) और (g(x)=\frac{1}{x+1}) हों, तो ((fg)(x)) का सही कथन कौन-सा है?
If (f(x)=x+1) and (g(x)=\frac{1}{x+1}), which statement about ((fg)(x)) is correct?
#relations functions
#product
#domain
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A ((fg)(x)=1), \(x\ne -1\)
B ((fg)(x)=1), \(x\in\mathbb{R}\)
C ((fg)(x)=x+1), \(x\ne -1\)
D ((fg)(x)=0), (x=-1)
Explanation opens after your attempt
Correct Answer
A. ((fg)(x)=1), \(x\ne -1\)
Step 1
Concept
The product is (1), but \(x\ne -1\) because of the denominator of (g(x)). It is important to write the domain with the simplified form.
Step 2
Why this answer is correct
The correct answer is A. ((fg)(x)=1), \(x\ne -1\). The product is (1), but \(x\ne -1\) because of the denominator of (g(x)). It is important to write the domain with the simplified form.
Step 3
Exam Tip
गुणनफल (1) है, पर (g(x)) के हर के कारण \(x\ne -1\)। सरल रूप के साथ डोमेन लिखना जरूरी है।
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यदि (f(x)=x-2 +2) और (g(x)=2x-2 -1) हों, तो ((g-f)(x)) क्या होगा?
If (f(x)=x-2 +2) and (g(x)=2x-2 -1), what is ((g-f)(x))?
#relations functions
#subtraction
#order
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A \(x^2-3\)
B \(3x^2+1\)
C \(x^2+3\)
D \(3x^2-3\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3\)
Step 1
Concept
((g-f)(x)=2x-2 -1-\(x^2+2\)=x-2 -3). Changing the order can change the subtraction result.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3\). ((g-f)(x)=2x-2 -1-\(x^2+2\)=x-2 -3). Changing the order can change the subtraction result.
Step 3
Exam Tip
((g-f)(x)=2x-2 -1-\(x^2+2\)=x-2 -3)। क्रम बदलने से घटाव का उत्तर बदल सकता है।
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यदि (f(x)=\frac{x-2}{x+3}) और (g(x)=\frac{x+3}{x-2}) हों, तो ((fg)(x)) का डोमेन क्या है?
If (f(x)=\frac{x-2}{x+3}) and (g(x)=\frac{x+3}{x-2}), what is the domain of ((fg)(x))?
#relations functions
#rational functions
#domain
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A \(\mathbb{R}-{-3,2}\)
B \(\mathbb{R}\)
C \(\mathbb{R}-{2}\)
D \(\mathbb{R}-{-3}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-3,2}\)
Step 1
Concept
The product becomes (1), but original denominators give \(x\ne -3,2\). Always decide the domain from the original functions.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-3,2}\). The product becomes (1), but original denominators give \(x\ne -3,2\). Always decide the domain from the original functions.
Step 3
Exam Tip
गुणनफल (1) बनता है, पर मूल हरों से \(x\ne -3,2\) है। डोमेन हमेशा मूल फलनों से तय करें।
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यदि (f(x)=x-2 +x) और (g(x)=x-2 -x) हों, तो ((f+g)(x)) क्या होगा?
If (f(x)=x-2 +x) and (g(x)=x-2 -x), what is ((f+g)(x))?
#relations functions
#addition
#polynomial
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A \(2x^2\)
B (2x)
C \(x^2\)
D \(2x^2+2x\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2\)
Step 1
Concept
(x) and (-x) cancel, so ((f+g)(x)=2x-2 ). Identify opposite terms.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2\). (x) and (-x) cancel, so ((f+g)(x)=2x-2 ). Identify opposite terms.
Step 3
Exam Tip
(x) और (-x) कट जाते हैं, इसलिए ((f+g)(x)=2x-2 )। विपरीत पद पहचानें।
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यदि (f(x)=\sqrt{4-x}) और (g(x)=\frac{1}{x}) हों, तो ((f+g)(x)) का डोमेन क्या है?
If (f(x)=\sqrt{4-x}) and (g(x)=\frac{1}{x}), what is the domain of ((f+g)(x))?
#relations functions
#domain
#combined functions
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A (\(-\infty,4]-{0}\)
B (\(-\infty,4]\)
C \([4,\infty\))
D \(\mathbb{R}-{0}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,4]-{0}\)
Step 1
Concept
The square root needs \(4-x\ge 0\), i.e. \(x\le 4\), and the denominator needs \(x\ne 0\). Take both conditions together.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,4]-{0}\). The square root needs \(4-x\ge 0\), i.e. \(x\le 4\), and the denominator needs \(x\ne 0\). Take both conditions together.
Step 3
Exam Tip
वर्गमूल के लिए \(4-x\ge 0\), यानी \(x\le 4\), और हर के लिए \(x\ne 0\)। दोनों शर्तें साथ लें।
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यदि (f(x)=2x+1) और (g(x)=3x-4) हों, तो ((fg)(-1)) क्या है?
If (f(x)=2x+1) and (g(x)=3x-4), what is ((fg)(-1))?
#relations functions
#evaluation
#product
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A (7)
B (-7)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
(f(-1)=-1) and (g(-1)=-7), so the product is (7). The product of two negative numbers is positive.
Step 2
Why this answer is correct
The correct answer is A. (7). (f(-1)=-1) and (g(-1)=-7), so the product is (7). The product of two negative numbers is positive.
Step 3
Exam Tip
(f(-1)=-1) और (g(-1)=-7), इसलिए गुणनफल (7) है। दो ऋणात्मक संख्याओं का गुणन धनात्मक होता है।
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यदि (f(x)=x-2 -2x+1) और (g(x)=1-x) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सरल रूप क्या है, जहाँ \(x\ne 1\)?
If (f(x)=x-2 -2x+1) and (g(x)=1-x), what is the simplified form of (\left\(\frac{f}{g}\right\)(x)), where \(x\ne 1\)?
#relations functions
#quotient
#sign error
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A (1-x)
B (x-1)
C (x+1)
D \(x^2-1\)
Explanation opens after your attempt
Step 1
Concept
(f(x)=(x-1)2 ) and (g(x)=-(x-1)), so the quotient is (-(x-1)=1-x). Pay special attention to the sign.
Step 2
Why this answer is correct
The correct answer is A. (1-x). (f(x)=(x-1)2 ) and (g(x)=-(x-1)), so the quotient is (-(x-1)=1-x). Pay special attention to the sign.
Step 3
Exam Tip
(f(x)=(x-1)2 ) और (g(x)=-(x-1)), इसलिए भागफल (-(x-1)=1-x) है। चिह्न पर विशेष ध्यान दें।
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