यदि (f(x)=\frac{1}{x-2}) और (g(x)=\sqrt{x+3}) हों, तो ((fg)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{1}{x-2}) and (g(x)=\sqrt{x+3}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
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A \( [-3,\infty\)\setminus{2} )
B \( [-3,\infty\) )
C \( \mathbb{R}\setminus{2} \)
D ( (-3,2) )
Explanation opens after your attempt
Correct Answer
A. \( [-3,\infty\)\setminus{2} )
Step 1
Concept
The root needs \(x\ge -3\) and the denominator needs \(x\ne 2\). For a product, take the common domain of both functions.
Step 2
Why this answer is correct
The correct answer is A. \( [-3,\infty\)\setminus{2} ). The root needs \(x\ge -3\) and the denominator needs \(x\ne 2\). For a product, take the common domain of both functions.
Step 3
Exam Tip
मूल के लिए \(x\ge -3\) और हर के लिए \(x\ne 2\) चाहिए। गुणन में भी दोनों फलनों का सामान्य प्रांत लिया जाता है।
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यदि (f(x)=x-2 -4) और (g(x)=x-2) हों, तो (\left\(\frac{f}{g}\right\)(x)) किस रूप में सही है?
If (f(x)=x-2 -4) and (g(x)=x-2), which form is correct for (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-functions
#domain
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A \(x+2,\ x\ne 2\)
B \(x+2,\ x\in\mathbb{R}\)
C \(x-2,\ x\ne -2\)
D \(x^2-4,\ x\ne 2\)
Explanation opens after your attempt
Correct Answer
A. \(x+2,\ x\ne 2\)
Step 1
Concept
Even after simplification, the original denominator must satisfy \(x-2\ne 0\). Cancelled denominator values do not return to the domain.
Step 2
Why this answer is correct
The correct answer is A. \(x+2,\ x\ne 2\). Even after simplification, the original denominator must satisfy \(x-2\ne 0\). Cancelled denominator values do not return to the domain.
Step 3
Exam Tip
सरलीकरण के बाद भी मूल हर \(x-2\ne 0\) होना चाहिए। रद्द करने से हटे हुए मान वापस प्रांत में नहीं आते।
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यदि (f(x)=|x-3|) और (g(x)=2x+1) हों, तो ((f-g)(4)) का मान क्या है?
If (f(x)=|x-3|) and (g(x)=2x+1), what is the value of ((f-g)(4))?
#relations-functions
#function-values
#algebra
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A ( -8 )
B ( 8 )
C ( -6 )
D ( 10 )
Explanation opens after your attempt
Step 1
Concept
(f(4)=1) and (g(4)=9), so ((f-g)(4)=1-9=-8). Find individual function values first.
Step 2
Why this answer is correct
The correct answer is A. ( -8 ). (f(4)=1) and (g(4)=9), so ((f-g)(4)=1-9=-8). Find individual function values first.
Step 3
Exam Tip
(f(4)=1) और (g(4)=9), इसलिए ((f-g)(4)=1-9=-8)। मान निकालते समय पहले अलग-अलग फलनों के मान निकालें।
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यदि (f(x)=\frac{x+1}{x-1}) और (g(x)=\frac{x-1}{x+1}) हों, तो ((fg)(x)) का सही प्रांत क्या होगा?
If (f(x)=\frac{x+1}{x-1}) and (g(x)=\frac{x-1}{x+1}), what is the correct domain of ((fg)(x))?
#relations-functions
#product-functions
#domain
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A \( \mathbb{R}\setminus{-1,1} \)
B \( \mathbb{R}\setminus{1} \)
C \( \mathbb{R}\setminus{-1} \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{-1,1} \)
Step 1
Concept
The denominators require \(x\ne 1\) and \(x\ne -1\). The product becomes (1), but the restrictions remain.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-1,1} \). The denominators require \(x\ne 1\) and \(x\ne -1\). The product becomes (1), but the restrictions remain.
Step 3
Exam Tip
दोनों हरों के कारण \(x\ne 1\) और \(x\ne -1\) चाहिए। गुणन (1) बनता है पर प्रतिबंध बने रहते हैं।
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यदि (f(x)=\sqrt{9-x-2 }) और (g(x)=\frac{1}{x}) हों, तो ((f+g)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{9-x-2 }) and (g(x)=\frac{1}{x}), what is the domain of ((f+g)(x))?
#relations-functions
#domain
#radical
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A \( [-3,3]\setminus{0} \)
B ( [-3,3] )
C \( \mathbb{R}\setminus{0} \)
D ( (-3,3) )
Explanation opens after your attempt
Correct Answer
A. \( [-3,3]\setminus{0} \)
Step 1
Concept
The root needs \(9-x^2\ge 0\), so \(x\in[-3,3]\), and \(x\ne 0\). The domain of the sum is the intersection.
Step 2
Why this answer is correct
The correct answer is A. \( [-3,3]\setminus{0} \). The root needs \(9-x^2\ge 0\), so \(x\in[-3,3]\), and \(x\ne 0\). The domain of the sum is the intersection.
Step 3
Exam Tip
मूल के लिए \(9-x^2\ge 0\), इसलिए \(x\in[-3,3]\), और \(x\ne 0\)। योग का प्रांत दोनों शर्तों का प्रतिच्छेद है।
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यदि (f(x)=x-2 +2x) और (g(x)=3x-5) हों, तो ((2f-3g)(x)) क्या होगा?
If (f(x)=x-2 +2x) and (g(x)=3x-5), what is ((2f-3g)(x))?
#relations-functions
#linear-combination
#algebra
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A \(2x^2-5x+15\)
B \(2x^2+13x-15\)
C \(2x^2-13x+15\)
D \(2x^2+x-15\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-5x+15\)
Step 1
Concept
\(2f=2x^2+4x\) and (3g=9x-15), so the difference is \(2x^2-5x+15\). Watch signs carefully.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-5x+15\). \(2f=2x^2+4x\) and (3g=9x-15), so the difference is \(2x^2-5x+15\). Watch signs carefully.
Step 3
Exam Tip
\(2f=2x^2+4x\) और (3g=9x-15), इसलिए अंतर \(2x^2-5x+15\) है। संकेतों पर विशेष ध्यान दें।
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यदि (f(x)=\frac{1}{x+1}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का सरल रूप क्या है?
If (f(x)=\frac{1}{x+1}) and (g(x)=\frac{1}{x-1}), what is the simplified form of ((f+g)(x))?
#relations-functions
#sum-functions
#rational
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A \(\frac{2x}{x^2-1},\ x\ne \pm1\)
B \(\frac{2}{x^2-1},\ x\ne \pm1\)
C \(\frac{x}{x^2-1},\ x\ne \pm1\)
D \(\frac{2x}{x^2+1},\ x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2x}{x^2-1},\ x\ne \pm1\)
Step 1
Concept
With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2x}{x^2-1},\ x\ne \pm1\). With common denominator ((x+1)(x-1)), the numerator is ((x-1)+(x+1)=2x). Do not forget excluded denominator values.
Step 3
Exam Tip
समान हर ((x+1)(x-1)) लेने पर अंश ((x-1)+(x+1)=2x) मिलता है। हर के शून्य मान हटाना न भूलें।
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यदि (f(x)=x-2 ) और (g(x)=\frac{1}{x}) हों, तो ((f+g)(x)=2) के लिए कौन-सा समीकरण मिलेगा?
If (f(x)=x-2 ) and (g(x)=\frac{1}{x}), which equation is obtained from ((f+g)(x)=2)?
#relations-functions
#equations
#domain
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A \(x^3-2x+1=0,\ x\ne 0\)
B \(x^2-2x+1=0,\ x\ne 0\)
C \(x^3+2x+1=0,\ x\ne 0\)
D \(x^3-2x-1=0,\ x\ne 0\)
Explanation opens after your attempt
Correct Answer
A. \(x^3-2x+1=0,\ x\ne 0\)
Step 1
Concept
Multiplying \(x^2+\frac{1}{x}=2\) by (x) gives \(x^3+1=2x\). The condition \(x\ne 0\) is necessary.
Step 2
Why this answer is correct
The correct answer is A. \(x^3-2x+1=0,\ x\ne 0\). Multiplying \(x^2+\frac{1}{x}=2\) by (x) gives \(x^3+1=2x\). The condition \(x\ne 0\) is necessary.
Step 3
Exam Tip
\(x^2+\frac{1}{x}=2\) को (x) से गुणा करने पर \(x^3+1=2x\) मिलता है। \(x\ne 0\) शर्त जरूरी है।
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यदि (f(x)=\sqrt{x}) और (g(x)=x-4) हों, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{x}) and (g(x)=x-4), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-domain
#radical
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A \( [0,\infty\)\setminus{4} )
B ( \(0,\infty\) )
C ( [0,4) )
D \( \mathbb{R}\setminus{4} \)
Explanation opens after your attempt
Correct Answer
A. \( [0,\infty\)\setminus{4} )
Step 1
Concept
\(\sqrt{x}\) needs \(x\ge 0\) and the denominator needs \(x\ne 4\). In a quotient, the denominator function must not be zero.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\)\setminus{4} ). \(\sqrt{x}\) needs \(x\ge 0\) and the denominator needs \(x\ne 4\). In a quotient, the denominator function must not be zero.
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर के लिए \(x\ne 4\) चाहिए। भागफल में भाजक फलन शून्य नहीं होना चाहिए।
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यदि (f(x)=x+2) और (g(x)=x-2 -1) हों, तो ((fg)(-2)) का मान क्या होगा?
If (f(x)=x+2) and (g(x)=x-2 -1), what is the value of ((fg)(-2))?
#relations-functions
#function-values
#product
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A (0)
B (3)
C (6)
D (-3)
Explanation opens after your attempt
Step 1
Concept
(f(-2)=0) and (g(-2)=3), so the product is (0). If one factor value is (0), the product is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). (f(-2)=0) and (g(-2)=3), so the product is (0). If one factor value is (0), the product is (0).
Step 3
Exam Tip
(f(-2)=0) और (g(-2)=3), इसलिए गुणन (0) है। यदि किसी एक गुणक का मान (0) हो, तो गुणनफल (0) होता है।
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यदि (f(x)=\frac{x}{x-3}) और (g(x)=\frac{x-3}{x}) हों, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{x}{x-3}) and (g(x)=\frac{x-3}{x}), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-functions
#domain
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A \( \mathbb{R}\setminus{0,3} \)
B \( \mathbb{R}\setminus{3} \)
C \( \mathbb{R}\setminus{0} \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{0,3} \)
Step 1
Concept
(f) needs \(x\ne 3\), (g) needs \(x\ne 0\), and divisor (g(x)\ne 0) gives \(x\ne 3\). Together, \(x\ne 0,3\).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{0,3} \). (f) needs \(x\ne 3\), (g) needs \(x\ne 0\), and divisor (g(x)\ne 0) gives \(x\ne 3\). Together, \(x\ne 0,3\).
Step 3
Exam Tip
(f) के लिए \(x\ne 3\), (g) के लिए \(x\ne 0\), और भाजक (g(x)\ne 0) से \(x\ne 3\)। सभी प्रतिबंध मिलाकर \(x\ne 0,3\) मिलते हैं।
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यदि (f(x)=\sqrt{x-2}) और (g(x)=\sqrt{x+2}) हों, तो ((f-g)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{x-2}) and (g(x)=\sqrt{x+2}), what is the domain of ((f-g)(x))?
#relations-functions
#difference-domain
#radical
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A \( [2,\infty\) )
B \( [-2,\infty\) )
C ( [-2,2] )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [2,\infty\) )
Step 1
Concept
The first root needs \(x\ge 2\) and the second needs \(x\ge -2\). Their intersection is \( [2,\infty\) ).
Step 2
Why this answer is correct
The correct answer is A. \( [2,\infty\) ). The first root needs \(x\ge 2\) and the second needs \(x\ge -2\). Their intersection is \( [2,\infty\) ).
Step 3
Exam Tip
पहले मूल के लिए \(x\ge 2\) और दूसरे के लिए \(x\ge -2\) चाहिए। प्रतिच्छेद \( [2,\infty\) ) है।
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यदि (f(x)=x-2 -3x+2) और (g(x)=x-1) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सही रूप क्या होगा?
If (f(x)=x-2 -3x+2) and (g(x)=x-1), what is the correct form of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-simplification
#domain
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A \(x-2,\ x\ne 1\)
B \(x-2,\ x\in\mathbb{R}\)
C \(x+2,\ x\ne 1\)
D \(x-1,\ x\ne 2\)
Explanation opens after your attempt
Correct Answer
A. \(x-2,\ x\ne 1\)
Step 1
Concept
(x-2 -3x+2=(x-1)(x-2)), but (x=1) is excluded. Always write the domain restriction with a simplified form.
Step 2
Why this answer is correct
The correct answer is A. \(x-2,\ x\ne 1\). (x-2 -3x+2=(x-1)(x-2)), but (x=1) is excluded. Always write the domain restriction with a simplified form.
Step 3
Exam Tip
(x-2 -3x+2=(x-1)(x-2)), पर (x=1) हटेगा। सरलीकृत रूप के साथ प्रांत प्रतिबंध लिखना जरूरी है।
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यदि (f(x)=2x-1) और (g(x)=x-2 +1) हों, तो ((f+g)(x)) का न्यूनतम मान क्या है?
If (f(x)=2x-1) and (g(x)=x-2 +1), what is the minimum value of ((f+g)(x))?
#relations-functions
#minimum-value
#quadratic
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A (0)
B (1)
C (-1)
D (2)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=x-2 +2x=(x+1)2 -1), so the minimum is (-1). Be careful because \(2x-1+x^2+1=x^2+2x\).
Step 2
Why this answer is correct
The correct answer is A. (0). ((f+g)(x)=x-2 +2x=(x+1)2 -1), so the minimum is (-1). Be careful because \(2x-1+x^2+1=x^2+2x\).
Step 3
Exam Tip
((f+g)(x)=x-2 +2x=(x+1)2 -1) नहीं, सही योग (x-2 +2x=x(x+2)) का न्यूनतम (-1) होता है। ध्यान दें \(2x-1+x^2+1=x^2+2x\)।
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यदि (f(x)=x-2 -1) और (g(x)=x+1) हों, तो ((f+g)(x)) का (x=2) पर मान क्या है?
If (f(x)=x-2 -1) and (g(x)=x+1), what is the value of ((f+g)(x)) at (x=2)?
#relations-functions
#function-values
#sum
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A (6)
B (4)
C (8)
D (10)
Explanation opens after your attempt
Step 1
Concept
(f(2)=3) and (g(2)=3), so the sum is (6). Finding function values first reduces mistakes.
Step 2
Why this answer is correct
The correct answer is A. (6). (f(2)=3) and (g(2)=3), so the sum is (6). Finding function values first reduces mistakes.
Step 3
Exam Tip
(f(2)=3) और (g(2)=3), इसलिए योग (6) है। पहले फलनों के मान निकालना गलती घटाता है।
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यदि (f(x)=\frac{2x+1}{x-1}) और (g(x)=\frac{x-1}{2x+1}) हों, तो ((fg)(0)) का मान क्या है?
If (f(x)=\frac{2x+1}{x-1}) and (g(x)=\frac{x-1}{2x+1}), what is the value of ((fg)(0))?
#relations-functions
#product-values
#rational
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A (1)
B (-1)
C (0)
D (परिभाषित नहीं) / (Not defined)
Explanation opens after your attempt
Step 1
Concept
(x=0) is in both domains and (f(0)g(0)=(-1)(-1)=1). First check whether the given value is in the domain.
Step 2
Why this answer is correct
The correct answer is A. (1). (x=0) is in both domains and (f(0)g(0)=(-1)(-1)=1). First check whether the given value is in the domain.
Step 3
Exam Tip
(x=0) दोनों फलनों के प्रांत में है और (f(0)g(0)=(-1)(-1)=1)। पहले जांचें कि दिया मान प्रांत में है या नहीं।
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यदि (f(x)=\sqrt{4-x}) और (g(x)=\frac{1}{\sqrt{x}}) हों, तो ((fg)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{4-x}) and (g(x)=\frac{1}{\sqrt{x}}), what is the domain of ((fg)(x))?
#relations-functions
#domain
#radical-denominator
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A ( (0,4] )
B ( [0,4] )
C ( (0,4) )
D \( [4,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( (0,4] )
Step 1
Concept
\(\sqrt{4-x}\) needs \(x\le 4\), and \(\sqrt{x}\) in the denominator needs (x>0). The intersection is ( (0,4] ).
Step 2
Why this answer is correct
The correct answer is A. ( (0,4] ). \(\sqrt{4-x}\) needs \(x\le 4\), and \(\sqrt{x}\) in the denominator needs (x>0). The intersection is ( (0,4] ).
Step 3
Exam Tip
\(\sqrt{4-x}\) के लिए \(x\le 4\), और \(\sqrt{x}\) हर में होने से (x>0)। प्रतिच्छेद ( (0,4] ) है।
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यदि (f(x)=x-2 +1) और (g(x)=2x) हों, तो (\left\(\frac{f}{g}\right\)(1)+\left\(\frac{g}{f}\right\)(1)) क्या है?
If (f(x)=x-2 +1) and (g(x)=2x), what is (\left\(\frac{f}{g}\right\)(1)+\left\(\frac{g}{f}\right\)(1))?
#relations-functions
#quotient-values
#algebra
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A (2)
B (1)
C \(\frac{5}{2}\)
D \(\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
(f(1)=2) and (g(1)=2), so both ratios are (1). In quotient questions, the denominator value must be non-zero.
Step 2
Why this answer is correct
The correct answer is A. (2). (f(1)=2) and (g(1)=2), so both ratios are (1). In quotient questions, the denominator value must be non-zero.
Step 3
Exam Tip
(f(1)=2) और (g(1)=2), इसलिए दोनों अनुपात (1) हैं। भागफल प्रश्न में भाजक का मान शून्य नहीं होना चाहिए।
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यदि (f(x)=x-2 -4x+5) और (g(x)=1-x) हों, तो ((f+g)(x)) का न्यूनतम मान क्या है?
If (f(x)=x-2 -4x+5) and (g(x)=1-x), what is the minimum value of ((f+g)(x))?
#relations-functions
#minimum-value
#quadratic
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A \(-\frac{1}{4}\)
B (0)
C \(\frac{1}{4}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{4}\)
Step 1
Concept
((f+g)(x)=x-2 -5x+6=\(x-\frac{5}{2}\)2 -\frac{1}{4}). Hence the minimum value is \(-\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{4}\). ((f+g)(x)=x-2 -5x+6=\(x-\frac{5}{2}\)2 -\frac{1}{4}). Hence the minimum value is \(-\frac{1}{4}\).
Step 3
Exam Tip
((f+g)(x)=x-2 -5x+6=\(x-\frac{5}{2}\)2 -\frac{1}{4})। अतः न्यूनतम मान \(-\frac{1}{4}\) है।
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यदि (f(x)=\frac{x-2 -9}{x-3}) और (g(x)=x+3) हों, तो (f) और (g) के बारे में सही कथन कौन-सा है?
If (f(x)=\frac{x-2 -9}{x-3}) and (g(x)=x+3), which statement about (f) and (g) is correct?
#relations-functions
#equality-functions
#domain
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A (f(x)=g(x)) पर \(x\ne 3\) / (f(x)=g(x)) but \(x\ne 3\)
B (f=g) सभी \(x\in\mathbb{R}\) पर / (f=g) for all \(x\in\mathbb{R}\)
C (f(3)=g(3))
D (f(x)\ne g(x)) हर (x) पर / (f(x)\ne g(x)) for every (x)
Explanation opens after your attempt
Correct Answer
A. (f(x)=g(x)) पर \(x\ne 3\) / (f(x)=g(x)) but \(x\ne 3\)
Step 1
Concept
(f(x)) simplifies to (x+3), but the original function is not defined at (x=3). Same formula and same function require matching domains.
Step 2
Why this answer is correct
The correct answer is A. (f(x)=g(x)) पर \(x\ne 3\) / (f(x)=g(x)) but \(x\ne 3\). (f(x)) simplifies to (x+3), but the original function is not defined at (x=3). Same formula and same function require matching domains.
Step 3
Exam Tip
(f(x)) सरल होकर (x+3) बनता है, पर (x=3) पर मूल फलन परिभाषित नहीं है। समान सूत्र और समान फलन में प्रांत भी मिलना चाहिए।
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यदि (f(x)=|x|+x) और (g(x)=|x|-x) हों, तो ((fg)(x)) का मान क्या होगा?
If (f(x)=|x|+x) and (g(x)=|x|-x), what is ((fg)(x))?
#relations-functions
#modulus
#product
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A (0)
B \(2x^2\)
C \(|x|^2+x^2\)
D (2|x|)
Explanation opens after your attempt
Step 1
Concept
((|x|+x)(|x|-x)=|x|2 -x-2 =0). The identity ((a+b)(a-b)=a-2 -b-2 ) is useful.
Step 2
Why this answer is correct
The correct answer is A. (0). ((|x|+x)(|x|-x)=|x|2 -x-2 =0). The identity ((a+b)(a-b)=a-2 -b-2 ) is useful.
Step 3
Exam Tip
((|x|+x)(|x|-x)=|x|2 -x-2 =0)। पहचान ((a+b)(a-b)=a-2 -b-2 ) उपयोगी रहती है।
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यदि (f(x)=\frac{1}{x-2 -4}) और (g(x)=\frac{1}{x-2}) हों, तो ((f-g)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{1}{x-2 -4}) and (g(x)=\frac{1}{x-2}), what is the domain of ((f-g)(x))?
#relations-functions
#difference-domain
#rational
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A \( \mathbb{R}\setminus{-2,2} \)
B \( \mathbb{R}\setminus{2} \)
C \( \mathbb{R}\setminus{-2} \)
D ( (-2,2) )
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{-2,2} \)
Step 1
Concept
The first function gives \(x^2-4\ne 0\), so \(x\ne \pm2\), and the second gives \(x\ne 2\). The common domain is \( \mathbb{R}\setminus{-2,2} \).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-2,2} \). The first function gives \(x^2-4\ne 0\), so \(x\ne \pm2\), and the second gives \(x\ne 2\). The common domain is \( \mathbb{R}\setminus{-2,2} \).
Step 3
Exam Tip
पहले फलन में \(x^2-4\ne 0\) से \(x\ne \pm2\), और दूसरे में \(x\ne 2\)। संयुक्त प्रांत \( \mathbb{R}\setminus{-2,2} \) है।
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यदि (f(x)=x-3 ) और (g(x)=x-2 ) हों, तो ((f-g)(x)=0) के वास्तविक हल कौन-से हैं?
If (f(x)=x-3 ) and (g(x)=x-2 ), what are the real solutions of ((f-g)(x)=0)?
#relations-functions
#equation-solving
#difference
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A (x=0,1)
B (x=1)
C (x=0,-1)
D (x=-1,1)
Explanation opens after your attempt
Correct Answer
A. (x=0,1)
Step 1
Concept
(x-3 -x-2 =x-2 (x-1)=0), so (x=0) or (x=1). Factoring gives solutions quickly.
Step 2
Why this answer is correct
The correct answer is A. (x=0,1). (x-3 -x-2 =x-2 (x-1)=0), so (x=0) or (x=1). Factoring gives solutions quickly.
Step 3
Exam Tip
(x-3 -x-2 =x-2 (x-1)=0), इसलिए (x=0) या (x=1)। गुणनखंड करने से हल तेजी से मिलते हैं।
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यदि (f(x)=\frac{x+2}{x-2}) और (g(x)=\frac{x-2}{x+2}) हों, तो ((f+g)(x)) का सरल रूप क्या है?
If (f(x)=\frac{x+2}{x-2}) and (g(x)=\frac{x-2}{x+2}), what is the simplified form of ((f+g)(x))?
#relations-functions
#rational-sum
#algebra
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A \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\)
B \(\frac{2x^2-8}{x^2-4},\ x\ne \pm2\)
C \(\frac{2x}{x^2-4},\ x\ne \pm2\)
D \(\frac{x^2+4}{x^2-4},\ x\ne \pm2\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\)
Step 1
Concept
The numerator is ((x+2)2 +(x-2)2 =2x-2 +8). Expand both squares carefully while taking a common denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2x^2+8}{x^2-4},\ x\ne \pm2\). The numerator is ((x+2)2 +(x-2)2 =2x-2 +8). Expand both squares carefully while taking a common denominator.
Step 3
Exam Tip
अंश ((x+2)2 +(x-2)2 =2x-2 +8) है। समान हर बनाते समय पूरे वर्ग ध्यान से खोलें।
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यदि (f(x)=x-2 +ax) और (g(x)=bx+1) हों तथा ((f-g)(x)=x-2 -5x-1), तो ((a,b)) क्या है?
If (f(x)=x-2 +ax) and (g(x)=bx+1), and ((f-g)(x)=x-2 -5x-1), what is ((a,b))?
#relations-functions
#parameters
#difference
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A ((a,b)=(2,7))
B ((a,b)=(7,2))
C ((a,b)=(-2,3))
D ((a,b)=(3,8))
Explanation opens after your attempt
Correct Answer
A. ((a,b)=(2,7))
Step 1
Concept
((f-g)(x)=x-2 +(a-b)x-1), so (a-b=-5). Among the options, only ((2,7)) works.
Step 2
Why this answer is correct
The correct answer is A. ((a,b)=(2,7)). ((f-g)(x)=x-2 +(a-b)x-1), so (a-b=-5). Among the options, only ((2,7)) works.
Step 3
Exam Tip
((f-g)(x)=x-2 +(a-b)x-1), इसलिए (a-b=-5)। दिए विकल्पों में केवल ((2,7)) सही है।
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यदि (f(x)=\frac{1}{x}) और (g(x)=\frac{1}{x+1}) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सही रूप और प्रांत क्या होगा?
If (f(x)=\frac{1}{x}) and (g(x)=\frac{1}{x+1}), what are the correct form and domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-functions
#rational
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A \(\frac{x+1}{x},\ x\ne 0,-1\)
B \(\frac{x}{x+1},\ x\ne 0,-1\)
C \(\frac{x+1}{x},\ x\ne 0\)
D \(\frac{x+1}{x},\ x\ne -1\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{x+1}{x},\ x\ne 0,-1\)
Step 1
Concept
Dividing gives \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\). The original (g) also requires \(x\ne -1\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x+1}{x},\ x\ne 0,-1\). Dividing gives \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\). The original (g) also requires \(x\ne -1\).
Step 3
Exam Tip
भाग देने पर \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\) मिलता है। मूल (g) के लिए \(x\ne -1\) भी जरूरी है।
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यदि (f(x)=x-2 ) और (g(x)=|x|) हों, तो ((f-g)(x)) का मान (0) कब होगा?
If (f(x)=x-2 ) and (g(x)=|x|), when is ((f-g)(x)) equal to (0)?
#relations-functions
#modulus
#equation
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A \(x\in{-1,0,1}\)
B \(x\in{0,1}\)
C \(x\in{-1,1}\)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(x\in{-1,0,1}\)
Step 1
Concept
In \(x^2=|x|\), put \(t=|x|\ge 0\), so \(t^2=t\). Thus (t=0) or (t=1), giving (x=-1,0,1).
Step 2
Why this answer is correct
The correct answer is A. \(x\in{-1,0,1}\). In \(x^2=|x|\), put \(t=|x|\ge 0\), so \(t^2=t\). Thus (t=0) or (t=1), giving (x=-1,0,1).
Step 3
Exam Tip
\(x^2=|x|\) में \(t=|x|\ge 0\) रखने पर \(t^2=t\), इसलिए (t=0) या (t=1)। अतः (x=-1,0,1) हैं।
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यदि (f(x)=\sqrt{x+1}) और (g(x)=\sqrt{2-x}) हों, तो ((fg)(x)) का अधिकतम मान क्या है?
If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{2-x}), what is the maximum value of ((fg)(x))?
#relations-functions
#maximum-value
#radical-product
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A \(\frac{3}{2}\)
B (3)
C \(\sqrt{3}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{2}\)
Step 1
Concept
((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum \(\frac{9}{4}\). Therefore the maximum is \(\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{2}\). ((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum \(\frac{9}{4}\). Therefore the maximum is \(\frac{3}{2}\).
Step 3
Exam Tip
((fg)(x)=\sqrt{(x+1)(2-x)}) और अंदर का द्विघात अधिकतम \(\frac{9}{4}\) है। इसलिए अधिकतम \(\frac{3}{2}\) है।
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यदि (f(x)=x-2 -6x+10) और (g(x)=x-1) हों, तो ((f-g)(x)) का न्यूनतम मान क्या है?
If (f(x)=x-2 -6x+10) and (g(x)=x-1), what is the minimum value of ((f-g)(x))?
#relations-functions
#minimum-value
#error-check
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A \(-\frac{9}{4}\)
B (0)
C \(\frac{9}{4}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{9}{4}\)
Step 1
Concept
((f-g)(x)=x-2 -7x+11=\(x-\frac{7}{2}\)2 -\frac{5}{4}), so the minimum is \(-\frac{5}{4}\). Check options carefully.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{9}{4}\). ((f-g)(x)=x-2 -7x+11=\(x-\frac{7}{2}\)2 -\frac{5}{4}), so the minimum is \(-\frac{5}{4}\). Check options carefully.
Step 3
Exam Tip
((f-g)(x)=x-2 -7x+11=\(x-\frac{7}{2}\)2 -\frac{5}{4}) नहीं, सही स्थिर मान से न्यूनतम \(-\frac{5}{4}\) है। विकल्पों में त्रुटि जांचें।
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यदि (f(x)=\frac{x-2 -1}{x+1}) और (g(x)=x-1) हों, तो (f) और (g) कब समान मान देते हैं?
If (f(x)=\frac{x-2 -1}{x+1}) and (g(x)=x-1), when do (f) and (g) give equal values?
#relations-functions
#equality-functions
#domain
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A हर \(x\ne -1\) पर / For every \(x\ne -1\)
B हर \(x\in\mathbb{R}\) पर / For every \(x\in\mathbb{R}\)
C केवल (x=1) पर / Only at (x=1)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
A. हर \(x\ne -1\) पर / For every \(x\ne -1\)
Step 1
Concept
(f(x)=x-1), but (x=-1) is not defined. Hence equal values occur only in the domain of (f).
Step 2
Why this answer is correct
The correct answer is A. हर \(x\ne -1\) पर / For every \(x\ne -1\). (f(x)=x-1), but (x=-1) is not defined. Hence equal values occur only in the domain of (f).
Step 3
Exam Tip
(f(x)=x-1) पर (x=-1) परिभाषित नहीं है। इसलिए समान मान केवल (f) के प्रांत में मिलते हैं।
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यदि (f(x)=\frac{x}{x+2}) और (g(x)=\frac{2}{x}) हों, तो ((f+g)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{x}{x+2}) and (g(x)=\frac{2}{x}), what is the domain of ((f+g)(x))?
#relations-functions
#sum-domain
#rational
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A \( \mathbb{R}\setminus{-2,0} \)
B \( \mathbb{R}\setminus{0} \)
C \( \mathbb{R}\setminus{-2} \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{-2,0} \)
Step 1
Concept
The first denominator gives \(x\ne -2\), and the second gives \(x\ne 0\). The domain of the sum keeps both restrictions.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-2,0} \). The first denominator gives \(x\ne -2\), and the second gives \(x\ne 0\). The domain of the sum keeps both restrictions.
Step 3
Exam Tip
पहले हर से \(x\ne -2\) और दूसरे हर से \(x\ne 0\) मिलता है। योग का प्रांत दोनों प्रतिबंधों को साथ रखता है।
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यदि (f(x)=x-2 +4x) और (g(x)=4) हों, तो ((f+g)(x)) का शून्य किस (x) पर है?
If (f(x)=x-2 +4x) and (g(x)=4), for which (x) is ((f+g)(x)) zero?
#relations-functions
#zeros
#quadratic
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A (x=-2)
B (x=2)
C (x=0)
D (x=-4)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=x-2 +4x+4=(x+2)2 ), so the zero is at (x=-2). Perfect square identity is useful in exams.
Step 2
Why this answer is correct
The correct answer is A. (x=-2). ((f+g)(x)=x-2 +4x+4=(x+2)2 ), so the zero is at (x=-2). Perfect square identity is useful in exams.
Step 3
Exam Tip
((f+g)(x)=x-2 +4x+4=(x+2)2 ), इसलिए शून्य (x=-2) पर है। पूर्ण वर्ग पहचान परीक्षा में उपयोगी है।
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यदि (f(x)=\frac{x-1}{x+1}) और (g(x)=\frac{x+1}{x-1}) हों, तो ((f-g)(x)) का सरल रूप क्या है?
If (f(x)=\frac{x-1}{x+1}) and (g(x)=\frac{x+1}{x-1}), what is the simplified form of ((f-g)(x))?
#relations-functions
#rational-difference
#algebra
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A \(\frac{-4x}{x^2-1},\ x\ne \pm1\)
B \(\frac{4x}{x^2-1},\ x\ne \pm1\)
C \(\frac{-2x}{x^2-1},\ x\ne \pm1\)
D \(\frac{2}{x^2-1},\ x\ne \pm1\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{-4x}{x^2-1},\ x\ne \pm1\)
Step 1
Concept
The numerator is ((x-1)2 -(x+1)2 =-4x). Handle the negative sign and square expansion carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{-4x}{x^2-1},\ x\ne \pm1\). The numerator is ((x-1)2 -(x+1)2 =-4x). Handle the negative sign and square expansion carefully.
Step 3
Exam Tip
अंश ((x-1)2 -(x+1)2 =-4x) है। ऋण चिह्न और वर्ग विस्तार ध्यान से करें।
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यदि (f(x)=\sqrt{x-1}) और (g(x)=x-3) हों, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{x-1}) and (g(x)=x-3), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
#relations-functions
#quotient-domain
#radical
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A ( \(1,\infty\) )
B \( [1,\infty\) )
C \( [3,\infty\) )
D \( \mathbb{R}\setminus{1} \)
Explanation opens after your attempt
Correct Answer
A. ( \(1,\infty\) )
Step 1
Concept
The denominator is \(\sqrt{x-1}\), so (x-1>0). In a quotient, the denominator can never be (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(1,\infty\) ). The denominator is \(\sqrt{x-1}\), so (x-1>0). In a quotient, the denominator can never be (0).
Step 3
Exam Tip
भाजक \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। भागफल में हर कभी (0) नहीं हो सकता।
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यदि (f(x)=x-2 -2x) और (g(x)=x+3) हों, तो ((fg)(1)) क्या है?
If (f(x)=x-2 -2x) and (g(x)=x+3), what is ((fg)(1))?
#relations-functions
#function-values
#product
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A (-4)
B (4)
C (0)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(1)=-1) and (g(1)=4), so ((fg)(1)=-4). For a product function, multiply values, do not add formulas.
Step 2
Why this answer is correct
The correct answer is A. (-4). (f(1)=-1) and (g(1)=4), so ((fg)(1)=-4). For a product function, multiply values, do not add formulas.
Step 3
Exam Tip
(f(1)=-1) और (g(1)=4), इसलिए ((fg)(1)=-4)। उत्पाद फलन में मानों को गुणा करें, सूत्रों को नहीं जोड़ें।
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यदि (f(x)=\frac{1}{x-2 +1}) और (g(x)=x-2 +1) हों, तो ((fg)(x)) का परिसर क्या होगा?
If (f(x)=\frac{1}{x-2 +1}) and (g(x)=x-2 +1), what is the range of ((fg)(x))?
#relations-functions
#range
#product
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A ({1})
B ( (0,1] )
C \( [1,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Step 1
Concept
((fg)(x)=1) for every real (x). The range of a constant function contains only that one value.
Step 2
Why this answer is correct
The correct answer is A. ({1}). ((fg)(x)=1) for every real (x). The range of a constant function contains only that one value.
Step 3
Exam Tip
((fg)(x)=1) हर वास्तविक (x) के लिए है। स्थिर फलन का परिसर केवल वही एक मान होता है।
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यदि (f(x)=\sqrt{x}) और (g(x)=\sqrt{1-x}) हों, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\sqrt{x}) and (g(x)=\sqrt{1-x}), what is the domain of ((f+g)(x))?
#relations-functions
#domain
#radical
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A ( [0,1] )
B ( (0,1) )
C \( [0,\infty\) )
D ( \(-\infty,1] \)
Explanation opens after your attempt
Correct Answer
A. ( [0,1] )
Step 1
Concept
The first root gives \(x\ge 0\), and the second gives \(x\le 1\). Their intersection is ( [0,1] ).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1] ). The first root gives \(x\ge 0\), and the second gives \(x\le 1\). Their intersection is ( [0,1] ).
Step 3
Exam Tip
पहले मूल से \(x\ge 0\) और दूसरे से \(x\le 1\) मिलता है। दोनों शर्तों का प्रतिच्छेद ( [0,1] ) है।
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यदि (f(x)=x-2 +px+4) और (g(x)=2x+1) हों तथा ((f+g)(x)) का न्यूनतम मान (0) हो, तो (p) का मान क्या होगा?
If (f(x)=x-2 +px+4) and (g(x)=2x+1), and the minimum value of ((f+g)(x)) is (0), what is (p)?
#relations-functions
#parameters
#minimum
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A \(p=\pm2\sqrt{5}-2\)
B \(p=2\sqrt{5}\)
C (p=-2)
D \(p=\pm\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(p=\pm2\sqrt{5}-2\)
Step 1
Concept
For ((f+g)(x)=x-2 +(p+2)x+5) to have minimum (0), its discriminant must be (0). Hence ((p+2)2 =20).
Step 2
Why this answer is correct
The correct answer is A. \(p=\pm2\sqrt{5}-2\). For ((f+g)(x)=x-2 +(p+2)x+5) to have minimum (0), its discriminant must be (0). Hence ((p+2)2 =20).
Step 3
Exam Tip
((f+g)(x)=x-2 +(p+2)x+5) का न्यूनतम (0) होने के लिए विविक्तकर (0) होगा। इसलिए ((p+2)2 =20)।
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यदि (f(x)=\frac{x-2 -4}{x+2}) और (g(x)=x-2) हों, तो ((f-g)(x)) किस पर (0) है?
If (f(x)=\frac{x-2 -4}{x+2}) and (g(x)=x-2), where is ((f-g)(x)) equal to (0)?
#relations-functions
#difference
#equality
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A हर \(x\ne -2\) पर / For every \(x\ne -2\)
B केवल (x=2) पर / Only at (x=2)
C हर \(x\in\mathbb{R}\) पर / For every \(x\in\mathbb{R}\)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
A. हर \(x\ne -2\) पर / For every \(x\ne -2\)
Step 1
Concept
(f(x)=x-2), but (x=-2) is excluded, so the difference is (0) for all \(x\ne -2\). Domain restriction is part of the answer.
Step 2
Why this answer is correct
The correct answer is A. हर \(x\ne -2\) पर / For every \(x\ne -2\). (f(x)=x-2), but (x=-2) is excluded, so the difference is (0) for all \(x\ne -2\). Domain restriction is part of the answer.
Step 3
Exam Tip
(f(x)=x-2) पर (x=-2) हटता है, इसलिए अंतर (0) सभी \(x\ne -2\) पर है। प्रांत प्रतिबंध उत्तर का हिस्सा है।
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यदि (f(x)=2x-2 -1) और (g(x)=x-2 +3) हों, तो ((f-g)(x)>0) कब होगा?
If (f(x)=2x-2 -1) and (g(x)=x-2 +3), when is ((f-g)(x)>0)?
#relations-functions
#inequality
#difference
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A (x<-2) या (x>2) / (x<-2) or (x>2)
B ( -2<x<2 )
C \(x\ne \pm2\)
D हर \(x\in\mathbb{R}\) / Every \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (x<-2) या (x>2) / (x<-2) or (x>2)
Step 1
Concept
((f-g)(x)=x-2 -4), so \(x^2-4>0\) gives (|x|>2). A number line is useful for quadratic inequalities.
Step 2
Why this answer is correct
The correct answer is A. (x<-2) या (x>2) / (x<-2) or (x>2). ((f-g)(x)=x-2 -4), so \(x^2-4>0\) gives (|x|>2). A number line is useful for quadratic inequalities.
Step 3
Exam Tip
((f-g)(x)=x-2 -4), इसलिए \(x^2-4>0\) से (|x|>2)। द्विघात असमिका में संख्या रेखा उपयोगी है।
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यदि (f(x)=\frac{x}{x-1}) और (g(x)=\frac{1}{x-1}) हों, तो ((f-g)(x)) का मान क्या होगा?
If (f(x)=\frac{x}{x-1}) and (g(x)=\frac{1}{x-1}), what is ((f-g)(x))?
#relations-functions
#rational-difference
#domain
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A \(1,\ x\ne 1\)
B \(1,\ x\in\mathbb{R}\)
C \(\frac{x-1}{x},\ x\ne 0\)
D \(\frac{x+1}{x-1},\ x\ne 1\)
Explanation opens after your attempt
Correct Answer
A. \(1,\ x\ne 1\)
Step 1
Concept
\(\frac{x}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}=1\), but \(x\ne 1\). The original restriction remains after simplification.
Step 2
Why this answer is correct
The correct answer is A. \(1,\ x\ne 1\). \(\frac{x}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}=1\), but \(x\ne 1\). The original restriction remains after simplification.
Step 3
Exam Tip
\(\frac{x}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}=1\), पर \(x\ne 1\)। सरलीकरण के बाद भी पुराना प्रतिबंध रहता है।
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यदि (f(x)=x-2 +1) और (g(x)=x-2 -1) हों, तो (\left\(\frac{f}{g}\right\)(x)) (1) से अधिक कब है?
If (f(x)=x-2 +1) and (g(x)=x-2 -1), when is (\left\(\frac{f}{g}\right\)(x)) greater than (1)?
#relations-functions
#quotient-inequality
#domain
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A (x<-1) या (x>1) / (x<-1) or (x>1)
B ( -1<x<1 )
C \(x\ne \pm1\)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
A. (x<-1) या (x>1) / (x<-1) or (x>1)
Step 1
Concept
\(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.
Step 2
Why this answer is correct
The correct answer is A. (x<-1) या (x>1) / (x<-1) or (x>1). \(\frac{x^2+1}{x^2-1}>1\) gives \(\frac{2}{x^2-1}>0\), so \(x^2-1>0\). Checking the denominator sign is essential.
Step 3
Exam Tip
\(\frac{x^2+1}{x^2-1}>1\) से \(\frac{2}{x^2-1}>0\), अतः \(x^2-1>0\)। हर का संकेत जांचना जरूरी है।
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यदि (f(x)=|x-2|) और (g(x)=|x+2|) हों, तो ((f-g)(0)) क्या है?
If (f(x)=|x-2|) and (g(x)=|x+2|), what is ((f-g)(0))?
#relations-functions
#modulus
#function-values
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A (0)
B (4)
C (-4)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(0)=2) and (g(0)=2), so the difference is (0). In modulus, substitute first and then take positive distance.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(0)=2) and (g(0)=2), so the difference is (0). In modulus, substitute first and then take positive distance.
Step 3
Exam Tip
(f(0)=2) और (g(0)=2), इसलिए अंतर (0) है। मापांक में अंदर का मान पहले रखें फिर धनात्मक दूरी लें।
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यदि (f(x)=\frac{1}{x-1}) और (g(x)=x+1) हों, तो ((fg)(x)=1) के हल कौन-से हैं?
If (f(x)=\frac{1}{x-1}) and (g(x)=x+1), what are the solutions of ((fg)(x)=1)?
#relations-functions
#equation-solving
#product
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A कोई वास्तविक हल नहीं / No real solution
B (x=0)
C (x=1)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक हल नहीं / No real solution
Step 1
Concept
\(\frac{x+1}{x-1}=1\) gives (x+1=x-1), which is impossible. Also, the condition \(x\ne 1\) remains.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक हल नहीं / No real solution. \(\frac{x+1}{x-1}=1\) gives (x+1=x-1), which is impossible. Also, the condition \(x\ne 1\) remains.
Step 3
Exam Tip
\(\frac{x+1}{x-1}=1\) से (x+1=x-1), जो असंभव है। साथ ही \(x\ne 1\) शर्त रहती है।
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यदि (f(x)=x-2 -4) और (g(x)=\sqrt{x-2}) हों, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या है?
If (f(x)=x-2 -4) and (g(x)=\sqrt{x-2}), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-domain
#radical
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A ( \(2,\infty\) )
B \( [2,\infty\) )
C \( \mathbb{R} \)
D ( \(-\infty,-2]\cup[2,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( \(2,\infty\) )
Step 1
Concept
The denominator is \(\sqrt{x-2}\), so (x-2>0). In a quotient, the denominator cannot be zero.
Step 2
Why this answer is correct
The correct answer is A. ( \(2,\infty\) ). The denominator is \(\sqrt{x-2}\), so (x-2>0). In a quotient, the denominator cannot be zero.
Step 3
Exam Tip
भाजक \(\sqrt{x-2}\) है, इसलिए (x-2>0) चाहिए। भागफल में हर शून्य नहीं हो सकता।
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यदि (f(x)=3x+2) और (g(x)=x-4) हों, तो ((f+g)(x)=0) के लिए (x) क्या है?
If (f(x)=3x+2) and (g(x)=x-4), what is (x) for ((f+g)(x)=0)?
#relations-functions
#sum
#equation-solving
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A \(x=\frac{1}{2}\)
B \(x=-\frac{1}{2}\)
C (x=2)
D (x=-2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{2}\)
Step 1
Concept
((f+g)(x)=4x-2), so (4x-2=0) gives \(x=\frac{1}{2}\). Check signs quickly in linear equations.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{2}\). ((f+g)(x)=4x-2), so (4x-2=0) gives \(x=\frac{1}{2}\). Check signs quickly in linear equations.
Step 3
Exam Tip
((f+g)(x)=4x-2), इसलिए (4x-2=0) से \(x=\frac{1}{2}\)। रैखिक समीकरण में संकेत जल्दी जांचें।
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यदि (f(x)=x-2 +2x+2) और (g(x)=x+1) हों, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या है?
If (f(x)=x-2 +2x+2) and (g(x)=x+1), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#quotient-domain
#polynomial
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A \( \mathbb{R}\setminus{-1} \)
B \( \mathbb{R} \)
C \( [0,\infty\) )
D \( \mathbb{R}\setminus{1} \)
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{-1} \)
Step 1
Concept
The denominator is (g(x)=x+1), so \(x\ne -1\). The numerator gives no additional real restriction.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-1} \). The denominator is (g(x)=x+1), so \(x\ne -1\). The numerator gives no additional real restriction.
Step 3
Exam Tip
भाजक (g(x)=x+1) है, इसलिए \(x\ne -1\)। अंश के लिए कोई अतिरिक्त वास्तविक प्रतिबंध नहीं है।
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यदि (f(x)=\sqrt{x+4}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{x+4}) and (g(x)=\frac{1}{x-1}), what is the domain of ((f+g)(x))?
#relations-functions
#sum-domain
#radical-rational
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A \( [-4,\infty\)\setminus{1} )
B \( [-4,\infty\) )
C \( \mathbb{R}\setminus{1} \)
D ( (-4,1) )
Explanation opens after your attempt
Correct Answer
A. \( [-4,\infty\)\setminus{1} )
Step 1
Concept
The root gives \(x\ge -4\), and the denominator gives \(x\ne 1\). Both conditions apply together for the sum.
Step 2
Why this answer is correct
The correct answer is A. \( [-4,\infty\)\setminus{1} ). The root gives \(x\ge -4\), and the denominator gives \(x\ne 1\). Both conditions apply together for the sum.
Step 3
Exam Tip
मूल से \(x\ge -4\) और हर से \(x\ne 1\) मिलता है। योग के लिए दोनों शर्तें साथ लागू होती हैं।
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यदि (f(x)=\sqrt{2x-3}) और (g(x)=\sqrt{7-x}) हों, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या होगा?
If (f(x)=\sqrt{2x-3}) and (g(x)=\sqrt{7-x}), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient-domain
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A \( \left[\frac{3}{2},7\right\) )
B \( \left[\frac{3}{2},7\right] \)
C ( \left\(\frac{3}{2},7\right\) )
D ( \left\(-\infty,7\right\) )
Explanation opens after your attempt
Correct Answer
A. \( \left[\frac{3}{2},7\right\) )
Step 1
Concept
The numerator root needs \(2x-3\ge 0\), and the denominator root needs (7-x>0). Therefore the domain is \( \left[\frac{3}{2},7\right\) ).
Step 2
Why this answer is correct
The correct answer is A. \( \left[\frac{3}{2},7\right\) ). The numerator root needs \(2x-3\ge 0\), and the denominator root needs (7-x>0). Therefore the domain is \( \left[\frac{3}{2},7\right\) ).
Step 3
Exam Tip
अंश के मूल से \(2x-3\ge 0\) और हर के मूल से (7-x>0) चाहिए। इसलिए प्रांत \( \left[\frac{3}{2},7\right\) ) है।
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यदि (f(x)=x-2 -4x+4) और (g(x)=2x-5) हों, तो ((f+g)(x)) का न्यूनतम मान क्या होगा?
If (f(x)=x-2 -4x+4) and (g(x)=2x-5), what is the minimum value of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#minimum-value
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A ( -2 )
B ( -1 )
C ( 0 )
D ( 2 )
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=x-2 -2x-1=(x-1)2 -2), so the minimum value is (-2). Completing the square is fast for such questions.
Step 2
Why this answer is correct
The correct answer is A. ( -2 ). ((f+g)(x)=x-2 -2x-1=(x-1)2 -2), so the minimum value is (-2). Completing the square is fast for such questions.
Step 3
Exam Tip
((f+g)(x)=x-2 -2x-1=(x-1)2 -2), इसलिए न्यूनतम मान (-2) है। पूर्ण वर्ग विधि ऐसे प्रश्नों में तेज होती है।
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