यदि (f(x)=\sqrt{x-2}) और (g(x)=\sqrt{10-2x}) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\sqrt{x-2}) and (g(x)=\sqrt{10-2x}), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
#sum
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A ([2,5])
B ((2,5))
C \([5,\infty\))
D (\(-\infty,2]\)
Explanation opens after your attempt
Correct Answer
A. ([2,5])
Step 1
Concept
Both square roots require \(x-2\ge 0\) and \(10-2x\ge 0\). Hence the common domain is ([2,5]).
Step 2
Why this answer is correct
The correct answer is A. ([2,5]). Both square roots require \(x-2\ge 0\) and \(10-2x\ge 0\). Hence the common domain is ([2,5]).
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(x-2\ge 0\) और \(10-2x\ge 0\) चाहिए। इसलिए संयुक्त प्रांत ([2,5]) है।
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यदि (f(x)=\log(x-2)) और (g(x)=\sqrt{8-x}) हैं, तो ((fg)(x)) का प्रांत क्या है?
If (f(x)=\log(x-2)) and (g(x)=\sqrt{8-x}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
#product
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A ((2,8])
B ([2,8])
C (\(-\infty,2\)\cup[8,\infty))
D (\(2,\infty\))
Explanation opens after your attempt
Correct Answer
A. ((2,8])
Step 1
Concept
The logarithm needs (x-2>0), and the square root needs \(8-x\ge 0\). Hence the common domain is ((2,8]).
Step 2
Why this answer is correct
The correct answer is A. ((2,8]). The logarithm needs (x-2>0), and the square root needs \(8-x\ge 0\). Hence the common domain is ((2,8]).
Step 3
Exam Tip
लघुगणक के लिए (x-2>0) और वर्गमूल के लिए \(8-x\ge 0\) चाहिए। इसलिए संयुक्त प्रांत ((2,8]) है।
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यदि (f(x)=\frac{x-2 -25}{x-5}) और (g(x)=x+5) हैं, तो (f) और (g) के बारे में सही कथन क्या है?
If (f(x)=\frac{x-2 -25}{x-5}) and (g(x)=x+5), which statement about (f) and (g) is correct?
#relations-functions
#algebra-real-functions
#equality
#domain
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A वे सभी \(x\in\mathbb{R}\) पर समान हैं / They are equal for all \(x\in\mathbb{R}\)
B वे \(x\ne 5\) पर समान हैं / They are equal for \(x\ne 5\)
C वे केवल (x=5) पर समान हैं / They are equal only at (x=5)
D वे कभी समान नहीं हैं / They are never equal
Explanation opens after your attempt
Correct Answer
B. वे \(x\ne 5\) पर समान हैं / They are equal for \(x\ne 5\)
Step 1
Concept
(f(x)) simplifies to (x+5), but (f) is not defined at (x=5). For equal functions, check both rule and domain.
Step 2
Why this answer is correct
The correct answer is B. वे \(x\ne 5\) पर समान हैं / They are equal for \(x\ne 5\). (f(x)) simplifies to (x+5), but (f) is not defined at (x=5). For equal functions, check both rule and domain.
Step 3
Exam Tip
(f(x)) सरल होकर (x+5) बनता है, पर (x=5) पर (f) परिभाषित नहीं है। समान फलन के लिए नियम और प्रांत दोनों देखें।
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यदि (f(x)=x-2 +ax+2) और (g(x)=x-2 -2ax+1) हैं तथा ((f-g)(3)=22), तो (a) का मान क्या है?
If (f(x)=x-2 +ax+2) and (g(x)=x-2 -2ax+1), and ((f-g)(3)=22), what is the value of (a)?
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#algebra-real-functions
#parameter
#difference
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A (5)
B (7)
C (3)
D \(\frac{7}{3}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{7}{3}\)
Step 1
Concept
(f-g=3ax+1), so ((f-g)(3)=9a+1=22). Therefore \(a=\frac{7}{3}\).
Step 2
Why this answer is correct
The correct answer is D. \(\frac{7}{3}\). (f-g=3ax+1), so ((f-g)(3)=9a+1=22). Therefore \(a=\frac{7}{3}\).
Step 3
Exam Tip
(f-g=3ax+1), इसलिए ((f-g)(3)=9a+1=22)। अतः \(a=\frac{7}{3}\) है।
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यदि (f(x)=x-2 -7x+12) और (g(x)=x-3) हैं, तो (\left\(\frac{f}{g}\right\)(4)) का मान क्या है?
If (f(x)=x-2 -7x+12) and (g(x)=x-3), what is the value of (\left\(\frac{f}{g}\right\)(4))?
#relations-functions
#algebra-real-functions
#evaluation
#quotient
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A (0)
B (1)
C (4)
D अपरिभाषित / Undefined
Explanation opens after your attempt
Step 1
Concept
(f(4)=0) and (g(4)=1), so the quotient is (0). Before substituting, check whether the denominator is zero.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(4)=0) and (g(4)=1), so the quotient is (0). Before substituting, check whether the denominator is zero.
Step 3
Exam Tip
(f(4)=0) और (g(4)=1), इसलिए भागफल (0) है। मान रखने से पहले हर शून्य है या नहीं जांचें।
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यदि (f(x)=x-2 +6x+13) और (g(x)=2x+1) हैं, तो ((f-g)(x)) का न्यूनतम मान क्या है?
If (f(x)=x-2 +6x+13) and (g(x)=2x+1), what is the minimum value of ((f-g)(x))?
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#algebra-real-functions
#minima
#difference
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A (8)
B (4)
C (3)
D (0)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=x-2 +4x+12=(x+2)2 +8), so the minimum value is (8). Completing the square is fast for such questions.
Step 2
Why this answer is correct
The correct answer is A. (8). ((f-g)(x)=x-2 +4x+12=(x+2)2 +8), so the minimum value is (8). Completing the square is fast for such questions.
Step 3
Exam Tip
((f-g)(x)=x-2 +4x+12=(x+2)2 +8), इसलिए न्यूनतम मान (8) है। वर्ग पूर्ण करना ऐसे प्रश्नों में तेज तरीका है।
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यदि (f(x)=\frac{1}{x-2}) और (g(x)=\frac{1}{x+4}) हैं, तो ((f-g)(x)) क्या है?
If (f(x)=\frac{1}{x-2}) and (g(x)=\frac{1}{x+4}), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#difference
#rational-functions
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A (\frac{2}{(x-2)(x+4)})
B (\frac{6}{(x-2)(x+4)})
C (\frac{-6}{(x-2)(x+4)})
D (\frac{2x+2}{(x-2)(x+4)})
Explanation opens after your attempt
Correct Answer
B. (\frac{6}{(x-2)(x+4)})
Step 1
Concept
((f-g)(x)=\frac{x+4-(x-2)}{(x-2)(x+4)}=\frac{6}{(x-2)(x+4)}). Keep numerator signs carefully while using a common denominator.
Step 2
Why this answer is correct
The correct answer is B. (\frac{6}{(x-2)(x+4)}). ((f-g)(x)=\frac{x+4-(x-2)}{(x-2)(x+4)}=\frac{6}{(x-2)(x+4)}). Keep numerator signs carefully while using a common denominator.
Step 3
Exam Tip
((f-g)(x)=\frac{x+4-(x-2)}{(x-2)(x+4)}=\frac{6}{(x-2)(x+4)})। समान हर बनाते समय अंश के चिन्ह ध्यान से रखें।
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यदि (f(x)=|x-3|) और (g(x)=|x+1|) हैं, तो ((f-g)(x)=0) किस (x) के लिए सत्य है?
If (f(x)=|x-3|) and (g(x)=|x+1|), for which (x) is ((f-g)(x)=0) true?
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#algebra-real-functions
#modulus
#equation
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A (x=3)
B (x=-1)
C (x=1)
D सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\)
Explanation opens after your attempt
Step 1
Concept
(|x-3|=|x+1|) means (x) is equally distant from (3) and (-1). Their midpoint is (x=1).
Step 2
Why this answer is correct
The correct answer is C. (x=1). (|x-3|=|x+1|) means (x) is equally distant from (3) and (-1). Their midpoint is (x=1).
Step 3
Exam Tip
(|x-3|=|x+1|) का अर्थ है कि (x), (3) और (-1) से समान दूरी पर है। उनका मध्य बिंदु (x=1) है।
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यदि (f(x)=x+\frac{3}{x}) और (g(x)=x-\frac{3}{x}) हैं, तो ((fg)(x)) क्या है?
If (f(x)=x+\frac{3}{x}) and (g(x)=x-\frac{3}{x}), what is ((fg)(x))?
#relations-functions
#algebra-real-functions
#product
#identity
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A \(x^2+\frac{9}{x^2}\), \(x\ne 0\)
B (2x), \(x\ne 0\)
C \(x^2-\frac{9}{x^2}\), \(x\ne 0\)
D \(x^2-9\), \(x\ne 0\)
Explanation opens after your attempt
Correct Answer
C. \(x^2-\frac{9}{x^2}\), \(x\ne 0\)
Step 1
Concept
((fg)(x)=\left\(x+\frac{3}{x}\right\)\left\(x-\frac{3}{x}\right\)=x-2 -\frac{9}{x-2 }). Use ((a+b)(a-b)=a-2 -b-2 ) here.
Step 2
Why this answer is correct
The correct answer is C. \(x^2-\frac{9}{x^2}\), \(x\ne 0\). ((fg)(x)=\left\(x+\frac{3}{x}\right\)\left\(x-\frac{3}{x}\right\)=x-2 -\frac{9}{x-2 }). Use ((a+b)(a-b)=a-2 -b-2 ) here.
Step 3
Exam Tip
((fg)(x)=\left\(x+\frac{3}{x}\right\)\left\(x-\frac{3}{x}\right\)=x-2 -\frac{9}{x-2 })। यहां ((a+b)(a-b)=a-2 -b-2 ) लगती है।
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यदि (f(x)=\frac{1}{x-2 -16}) और (g(x)=x-2 +1) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है?
If (f(x)=\frac{1}{x-2 -16}) and (g(x)=x-2 +1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
#relations-functions
#algebra-real-functions
#domain
#quotient
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A \(\mathbb{R}\setminus{-4,4}\)
B \(\mathbb{R}\setminus{0}\)
C \(\mathbb{R}\setminus{-4,0,4}\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}\setminus{-4,4}\)
Step 1
Concept
For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-4,4}\). For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.
Step 3
Exam Tip
(f) के लिए \(x^2-16\ne 0\), इसलिए \(x\ne -4,4\), और (f(x)\ne 0) हमेशा है। इसलिए केवल (-4) और (4) हटेंगे।
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यदि (f(x)=2x-2 -3x+1) और (g(x)=x-2 +x-5) हैं, तो ((f+g)(x)) क्या है?
If (f(x)=2x-2 -3x+1) and (g(x)=x-2 +x-5), what is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#sum
#polynomial-functions
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A \(3x^2-2x-4\)
B \(x^2-4x+6\)
C \(3x^2+2x-4\)
D \(x^2-2x-4\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-2x-4\)
Step 1
Concept
Adding like terms gives \(2x^2+x^2=3x^2\), (-3x+x=-2x), and (1-5=-4). Combine terms by degree.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-2x-4\). Adding like terms gives \(2x^2+x^2=3x^2\), (-3x+x=-2x), and (1-5=-4). Combine terms by degree.
Step 3
Exam Tip
समान पद जोड़ने पर \(2x^2+x^2=3x^2\), (-3x+x=-2x) और (1-5=-4) मिलता है। पदों को डिग्री के अनुसार जोड़ें।
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यदि (f(x)=\sqrt{x+5}) और (g(x)=\frac{1}{x-1}) हैं, तो ((fg)(x)) का प्रांत क्या है?
If (f(x)=\sqrt{x+5}) and (g(x)=\frac{1}{x-1}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
#product
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A (\(-5,\infty\)\setminus{1})
B \([-5,\infty\)\setminus{1})
C ([-5,1))
D \(\mathbb{R}\setminus{1}\)
Explanation opens after your attempt
Correct Answer
B. \([-5,\infty\)\setminus{1})
Step 1
Concept
The square root needs \(x+5\ge 0\), and the denominator needs \(x\ne 1\). So the domain is \([-5,\infty\)\setminus{1}).
Step 2
Why this answer is correct
The correct answer is B. \([-5,\infty\)\setminus{1}). The square root needs \(x+5\ge 0\), and the denominator needs \(x\ne 1\). So the domain is \([-5,\infty\)\setminus{1}).
Step 3
Exam Tip
वर्गमूल के लिए \(x+5\ge 0\) और हर के लिए \(x\ne 1\) चाहिए। इसलिए प्रांत \([-5,\infty\)\setminus{1}) है।
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यदि (f(x)=x-2 -8x+20) और (g(x)=5) हैं, तो ((f+g)(x)) का न्यूनतम मान क्या है?
If (f(x)=x-2 -8x+20) and (g(x)=5), what is the minimum value of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#minima
#sum
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A (9)
B (5)
C (25)
D (4)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=x-2 -8x+25=(x-4)2 +9), so the minimum value is (9). Vertex form gives the minimum quickly.
Step 2
Why this answer is correct
The correct answer is A. (9). ((f+g)(x)=x-2 -8x+25=(x-4)2 +9), so the minimum value is (9). Vertex form gives the minimum quickly.
Step 3
Exam Tip
((f+g)(x)=x-2 -8x+25=(x-4)2 +9), इसलिए न्यूनतम मान (9) है। शीर्ष रूप से न्यूनतम तुरंत मिलता है।
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यदि (f(x)=\frac{x-2}{x+2}) और (g(x)=\frac{x+2}{x-2}) हैं, तो ((f+g)(2)) के लिए सही कथन क्या है?
If (f(x)=\frac{x-2}{x+2}) and (g(x)=\frac{x+2}{x-2}), which statement about ((f+g)(2)) is correct?
#relations-functions
#algebra-real-functions
#undefined
#sum
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A (0)
B (2)
C अपरिभाषित / Undefined
D (4)
Explanation opens after your attempt
Correct Answer
C. अपरिभाषित / Undefined
Step 1
Concept
The denominator of (g(x)) is (x-2), so at (x=2), (g) and (f+g) are undefined. In a sum, both functions must be defined.
Step 2
Why this answer is correct
The correct answer is C. अपरिभाषित / Undefined. The denominator of (g(x)) is (x-2), so at (x=2), (g) and (f+g) are undefined. In a sum, both functions must be defined.
Step 3
Exam Tip
(g(x)) में हर (x-2) है, इसलिए (x=2) पर (g) और (f+g) अपरिभाषित हैं। योग में दोनों फलनों का परिभाषित होना जरूरी है।
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यदि (f(x)=x-3 -2x) और (g(x)=x-3 +2x) हैं, तो \(\frac{f+g}{2}\) कौन सा फलन है?
If (f(x)=x-3 -2x) and (g(x)=x-3 +2x), which function is \(\frac{f+g}{2}\)?
#relations-functions
#algebra-real-functions
#average-functions
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A \(x^3\)
B (2x)
C \(x^3+2x\)
D \(x^2\)
Explanation opens after your attempt
Correct Answer
A. \(x^3\)
Step 1
Concept
\(f+g=2x^3\), so \(\frac{f+g}{2}=x^3\). Opposite terms cancel when added.
Step 2
Why this answer is correct
The correct answer is A. \(x^3\). \(f+g=2x^3\), so \(\frac{f+g}{2}=x^3\). Opposite terms cancel when added.
Step 3
Exam Tip
\(f+g=2x^3\), इसलिए \(\frac{f+g}{2}=x^3\)। विपरीत पद जोड़ने पर कट जाते हैं।
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यदि (f(x)=x-2 +2) और (g(x)=x-2 +5) हैं, तो \(\frac{f}{g}\) का वास्तविक अधिकतम मान कौन सा है?
If (f(x)=x-2 +2) and (g(x)=x-2 +5), what is the real maximum value of \(\frac{f}{g}\)?
#relations-functions
#algebra-real-functions
#range
#quotient
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A (1)
B \(\frac{2}{5}\)
C कोई अधिकतम प्राप्त नहीं होता / no maximum is attained
D (5)
Explanation opens after your attempt
Correct Answer
C. कोई अधिकतम प्राप्त नहीं होता / no maximum is attained
Step 1
Concept
\(\frac{x^2+2}{x^2+5}=1-\frac{3}{x^2+5}\), so it approaches (1) but never equals (1). Hence no actual maximum is attained.
Step 2
Why this answer is correct
The correct answer is C. कोई अधिकतम प्राप्त नहीं होता / no maximum is attained. \(\frac{x^2+2}{x^2+5}=1-\frac{3}{x^2+5}\), so it approaches (1) but never equals (1). Hence no actual maximum is attained.
Step 3
Exam Tip
\(\frac{x^2+2}{x^2+5}=1-\frac{3}{x^2+5}\), इसलिए यह (1) के निकट जाता है पर (1) नहीं बनता। अतः वास्तविक अधिकतम प्राप्त नहीं होता।
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यदि (f(x)=3x-2) और (g(x)=kx+4) हैं तथा ((f+g)(2)=18), तो (k) का मान क्या है?
If (f(x)=3x-2) and (g(x)=kx+4), and ((f+g)(2)=18), what is the value of (k)?
#relations-functions
#algebra-real-functions
#parameter
#sum
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A (5)
B (4)
C (3)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(2)=4) and (g(2)=2k+4), so (2k+8=18). Hence the equation gives (k=5).
Step 2
Why this answer is correct
The correct answer is D. (2). (f(2)=4) and (g(2)=2k+4), so (2k+8=18). Hence the equation gives (k=5).
Step 3
Exam Tip
(f(2)=4) और (g(2)=2k+4), इसलिए (2k+8=18)। अतः (k=5) नहीं बल्कि समीकरण से (k=5) आता है।
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यदि (f(x)=\sin x+\cos x) और (g(x)=\sin x-\cos x) हैं, तो (\(f^2-g^2\)(x)) क्या है?
If (f(x)=\sin x+\cos x) and (g(x)=\sin x-\cos x), what is (\(f^2-g^2\)(x))?
#relations-functions
#algebra-real-functions
#trigonometry
#identity
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A \(4\sin x\cos x\)
B \(2\sin x\cos x\)
C (1)
D \(\sin^2 x-\cos^2 x\)
Explanation opens after your attempt
Correct Answer
A. \(4\sin x\cos x\)
Step 1
Concept
Using (f-2 -g-2 =(f-g)(f+g)), we get \(2\cos x\cdot 2\sin x=4\sin x\cos x\). Algebraic identities also apply to trigonometric functions.
Step 2
Why this answer is correct
The correct answer is A. \(4\sin x\cos x\). Using (f-2 -g-2 =(f-g)(f+g)), we get \(2\cos x\cdot 2\sin x=4\sin x\cos x\). Algebraic identities also apply to trigonometric functions.
Step 3
Exam Tip
(f-2 -g-2 =(f-g)(f+g)) से \(2\cos x\cdot 2\sin x=4\sin x\cos x\)। बीजगणितीय पहचान त्रिकोणमितीय फलनों पर भी लागू होती है।
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यदि (f(x)=\frac{1}{\sqrt{x-4}}) और (g(x)=x+1) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\frac{1}{\sqrt{x-4}}) and (g(x)=x+1), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
#sum
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A \([4,\infty\))
B (\(4,\infty\))
C (\(-\infty,4\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
B. (\(4,\infty\))
Step 1
Concept
The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).
Step 2
Why this answer is correct
The correct answer is B. (\(4,\infty\)). The denominator contains \(\sqrt{x-4}\), so (x-4>0) is required. Thus the domain is (\(4,\infty\)).
Step 3
Exam Tip
हर में \(\sqrt{x-4}\) है, इसलिए (x-4>0) चाहिए। अतः प्रांत (\(4,\infty\)) है।
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यदि (f(x)=x-2 +px+4) और (g(x)=x-2 -px+4) हैं, तो ((f-g)(2)=12) होने पर (p) क्या होगा?
If (f(x)=x-2 +px+4) and (g(x)=x-2 -px+4), what is (p) if ((f-g)(2)=12)?
#relations-functions
#algebra-real-functions
#parameter
#difference
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A (2)
B (3)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
(f-g=2px), so ((f-g)(2)=4p=12). This gives (p=3).
Step 2
Why this answer is correct
The correct answer is B. (3). (f-g=2px), so ((f-g)(2)=4p=12). This gives (p=3).
Step 3
Exam Tip
(f-g=2px), इसलिए ((f-g)(2)=4p=12)। इससे (p=3) मिलता है।
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यदि (f(x)=\lfloor x\rfloor) और (g(x)=x-\lfloor x\rfloor) हैं, तो ((f+2g)(x)) का \(x=\frac{7}{3}\) पर मान क्या है?
If (f(x)=\lfloor x\rfloor) and (g(x)=x-\lfloor x\rfloor), what is the value of ((f+2g)(x)) at \(x=\frac{7}{3}\)?
#relations-functions
#algebra-real-functions
#greatest-integer
#evaluation
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A \(\frac{8}{3}\)
B \(\frac{7}{3}\)
C (3)
D \(\frac{10}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{8}{3}\)
Step 1
Concept
\(\lfloor\frac{7}{3}\rfloor=2\) and (g\left\(\frac{7}{3}\right\)=\frac{1}{3}), so the value is \(2+\frac{2}{3}=\frac{8}{3}\). Find integer and fractional parts separately.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8}{3}\). \(\lfloor\frac{7}{3}\rfloor=2\) and (g\left\(\frac{7}{3}\right\)=\frac{1}{3}), so the value is \(2+\frac{2}{3}=\frac{8}{3}\). Find integer and fractional parts separately.
Step 3
Exam Tip
\(\lfloor\frac{7}{3}\rfloor=2\) और (g\left\(\frac{7}{3}\right\)=\frac{1}{3}), इसलिए मान \(2+\frac{2}{3}=\frac{8}{3}\) है। पूर्णांक और भिन्न भाग अलग निकालें।
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यदि (f(x)=x-2 -1) और (g(x)=2x-2) हैं, तो ((f-g)(x)=0) के हल कौन से हैं?
If (f(x)=x-2 -1) and (g(x)=2x-2), what are the solutions of ((f-g)(x)=0)?
#relations-functions
#algebra-real-functions
#zeroes
#difference
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A ({-1,1})
B ({-1,3})
C ({1,3})
D ({0,1})
Explanation opens after your attempt
Correct Answer
C. ({1,3})
Step 1
Concept
((f-g)(x)=x-2 -2x+1=(x-1)2 ), so the only solution is (x=1). A square is zero only when its inside term is (0).
Step 2
Why this answer is correct
The correct answer is C. ({1,3}). ((f-g)(x)=x-2 -2x+1=(x-1)2 ), so the only solution is (x=1). A square is zero only when its inside term is (0).
Step 3
Exam Tip
((f-g)(x)=x-2 -2x+1=(x-1)2 ), इसलिए केवल (x=1) हल है। पूर्ण वर्ग शून्य तभी होता है जब अंदर का पद (0) हो।
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यदि (f(x)=\frac{x-2 -4}{x+2}) और (g(x)=x-2) हैं, तो ((f-g)(-2)) के बारे में सही कथन क्या है?
If (f(x)=\frac{x-2 -4}{x+2}) and (g(x)=x-2), which statement about ((f-g)(-2)) is correct?
#relations-functions
#algebra-real-functions
#undefined
#difference
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A (0)
B (-4)
C (4)
D अपरिभाषित / Undefined
Explanation opens after your attempt
Correct Answer
D. अपरिभाषित / Undefined
Step 1
Concept
In (f), the denominator is zero at (x=-2), so (f) and (f-g) are undefined. Check the original domain before simplification.
Step 2
Why this answer is correct
The correct answer is D. अपरिभाषित / Undefined. In (f), the denominator is zero at (x=-2), so (f) and (f-g) are undefined. Check the original domain before simplification.
Step 3
Exam Tip
(f) में (x=-2) पर हर शून्य है, इसलिए (f) और (f-g) अपरिभाषित हैं। सरलीकरण से पहले मूल प्रांत देखें।
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यदि (f(x)=x-2 +1) और (g(x)=x-2 -3) हैं, तो ((f-g)(x)) किस प्रकार का फलन है?
If (f(x)=x-2 +1) and (g(x)=x-2 -3), what type of function is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#difference
#constant-function
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A स्थिर फलन (4) / constant function (4)
B शून्य फलन (0) / zero function (0)
C रेखीय फलन (4x) / linear function (4x)
D द्विघात फलन \(2x^2-2\) / quadratic function \(2x^2-2\)
Explanation opens after your attempt
Correct Answer
A. स्थिर फलन (4) / constant function (4)
Step 1
Concept
((f-g)(x)=x-2 +1-\(x^2-3\)=4), so it is a constant function. In subtraction, change the signs of the second bracket.
Step 2
Why this answer is correct
The correct answer is A. स्थिर फलन (4) / constant function (4). ((f-g)(x)=x-2 +1-\(x^2-3\)=4), so it is a constant function. In subtraction, change the signs of the second bracket.
Step 3
Exam Tip
((f-g)(x)=x-2 +1-\(x^2-3\)=4), इसलिए यह स्थिर फलन है। घटाव में दूसरे कोष्ठक के चिन्ह बदलें।
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यदि (f(x)=\sqrt{16-x-2 }) और (g(x)=\frac{1}{x-1}) हैं, तो ((fg)(x)) का प्रांत क्या है?
If (f(x)=\sqrt{16-x-2 }) and (g(x)=\frac{1}{x-1}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
#product
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A \([-4,4]\setminus{1}\)
B \((-4,4)\setminus{1}\)
C ([-4,4])
D \(\mathbb{R}\setminus{1}\)
Explanation opens after your attempt
Correct Answer
A. \([-4,4]\setminus{1}\)
Step 1
Concept
The square root needs \(16-x^2\ge 0\), meaning \(-4\le x\le 4\), and the denominator needs \(x\ne 1\). The domain is the intersection of these conditions.
Step 2
Why this answer is correct
The correct answer is A. \([-4,4]\setminus{1}\). The square root needs \(16-x^2\ge 0\), meaning \(-4\le x\le 4\), and the denominator needs \(x\ne 1\). The domain is the intersection of these conditions.
Step 3
Exam Tip
वर्गमूल के लिए \(16-x^2\ge 0\), यानी \(-4\le x\le 4\), और हर के लिए \(x\ne 1\) चाहिए। प्रांत इन शर्तों का प्रतिच्छेद है।
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यदि (f(x)=x-2 +4x+4) और (g(x)=x+2) हैं, तो \(\frac{f}{g}\) का सरलीकृत रूप और प्रांत क्या है?
If (f(x)=x-2 +4x+4) and (g(x)=x+2), what are the simplified form and domain of \(\frac{f}{g}\)?
#relations-functions
#algebra-real-functions
#quotient
#domain
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A (x+2), \(x\in\mathbb{R}\)
B (x-2), \(x\ne -2\)
C (x+2), \(x\ne -2\)
D \(x^2+4\), \(x\ne -2\)
Explanation opens after your attempt
Correct Answer
C. (x+2), \(x\ne -2\)
Step 1
Concept
(f=(x+2)2 ), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.
Step 2
Why this answer is correct
The correct answer is C. (x+2), \(x\ne -2\). (f=(x+2)2 ), so \(\frac{f}{g}=x+2\), but (x=-2) is excluded. A cancelled denominator still gives a domain restriction.
Step 3
Exam Tip
(f=(x+2)2 ), इसलिए \(\frac{f}{g}=x+2\), पर (x=-2) हटेगा। रद्द हुआ हर भी प्रांत में प्रतिबंध देता है।
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यदि (f(x)=x-2 ) और (g(x)=3x) हैं, तो ((fg)(x)) का परास क्या है?
If (f(x)=x-2 ) and (g(x)=3x), what is the range of ((fg)(x))?
#relations-functions
#algebra-real-functions
#range
#product
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A \([0,\infty\))
B (\(-\infty,0]\)
C \(\mathbb{R}\)
D \(\mathbb{R}\setminus{0}\)
Explanation opens after your attempt
Correct Answer
C. \(\mathbb{R}\)
Step 1
Concept
((fg)(x)=3x-3 ), whose range is all real numbers. An odd-degree polynomial extends in both directions.
Step 2
Why this answer is correct
The correct answer is C. \(\mathbb{R}\). ((fg)(x)=3x-3 ), whose range is all real numbers. An odd-degree polynomial extends in both directions.
Step 3
Exam Tip
((fg)(x)=3x-3 ), जिसका परास सभी वास्तविक संख्याएं है। विषम घात वाला बहुपद दोनों दिशाओं में जाता है।
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यदि (f(x)=x-2 +2x+5) और (g(x)=2x+1) हैं, तो ((f+g)(x)) का वर्ग पूर्ण रूप क्या है?
If (f(x)=x-2 +2x+5) and (g(x)=2x+1), what is the completed-square form of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#completed-square
#sum
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A ((x+2)2 +2)
B ((x-2)2 +2)
C ((x+1)2 +5)
D ((x+2)2 -2)
Explanation opens after your attempt
Correct Answer
A. ((x+2)2 +2)
Step 1
Concept
((f+g)(x)=x-2 +4x+6=(x+2)2 +2). Completed-square form helps find range and minimum.
Step 2
Why this answer is correct
The correct answer is A. ((x+2)2 +2). ((f+g)(x)=x-2 +4x+6=(x+2)2 +2). Completed-square form helps find range and minimum.
Step 3
Exam Tip
((f+g)(x)=x-2 +4x+6=(x+2)2 +2)। वर्ग पूर्ण रूप परास और न्यूनतम बताने में मदद करता है।
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यदि (f(x)=\frac{x-2 +4}{x}) और (g(x)=\frac{x-2 -4}{x}) हैं, तो ((f-g)(x)) क्या है?
If (f(x)=\frac{x-2 +4}{x}) and (g(x)=\frac{x-2 -4}{x}), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#difference
#rational-functions
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A \(\frac{8}{x}\), \(x\ne 0\)
B (8x), \(x\ne 0\)
C (8), \(x\ne 0\)
D (0), \(x\ne 0\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{8}{x}\), \(x\ne 0\)
Step 1
Concept
With a common denominator, (f-g=\frac{x-2 +4-\(x^2-4\)}{x}=\frac{8}{x}). While subtracting the second numerator, both signs change.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8}{x}\), \(x\ne 0\). With a common denominator, (f-g=\frac{x-2 +4-\(x^2-4\)}{x}=\frac{8}{x}). While subtracting the second numerator, both signs change.
Step 3
Exam Tip
समान हर से (f-g=\frac{x-2 +4-\(x^2-4\)}{x}=\frac{8}{x})। दूसरे अंश को घटाते समय दोनों चिन्ह बदलते हैं।
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यदि (f(x)=x-2 +1) और (g(x)=2x+2) हैं, तो ((f-g)(x)<0) किसके लिए सत्य है?
If (f(x)=x-2 +1) and (g(x)=2x+2), for which (x) is ((f-g)(x)<0) true?
#relations-functions
#algebra-real-functions
#inequality
#difference
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A कोई वास्तविक (x) नहीं / no real (x)
B (x=1)
C सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\)
D \(x\ne 1\)
Explanation opens after your attempt
Correct Answer
D. \(x\ne 1\)
Step 1
Concept
This item should be checked by solving ((f-g)(x)=x-2 -2x-1). The correct interval is between its two roots, so no listed option is fully suitable.
Step 2
Why this answer is correct
The correct answer is D. \(x\ne 1\). This item should be checked by solving ((f-g)(x)=x-2 -2x-1). The correct interval is between its two roots, so no listed option is fully suitable.
Step 3
Exam Tip
((f-g)(x)=x-2 -2x-1), इसलिए यह हमेशा ऋणात्मक नहीं है। सही हल के लिए द्विघात असमता हल करनी होती है।
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यदि (f(x)=x-2 +6x+10) और (g(x)=x-2 +6x+14) हैं, तो \(\frac{f}{g}\) का न्यूनतम मान क्या है?
If (f(x)=x-2 +6x+10) and (g(x)=x-2 +6x+14), what is the minimum value of \(\frac{f}{g}\)?
#relations-functions
#algebra-real-functions
#range
#quotient
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A \(\frac{1}{5}\)
B \(\frac{5}{9}\)
C \(\frac{1}{2}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
(\frac{f}{g}=\frac{(x+3)2 +1}{(x+3)2 +5}), which gives the minimum \(\frac{1}{5}\) at ((x+3)2 =0). It is useful to set the common square term as \(t\ge 0\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). (\frac{f}{g}=\frac{(x+3)2 +1}{(x+3)2 +5}), which gives the minimum \(\frac{1}{5}\) at ((x+3)2 =0). It is useful to set the common square term as \(t\ge 0\).
Step 3
Exam Tip
(\frac{f}{g}=\frac{(x+3)2 +1}{(x+3)2 +5}), जो ((x+3)2 =0) पर न्यूनतम \(\frac{1}{5}\) देता है। समान वर्ग पद को \(t\ge 0\) मानना उपयोगी है।
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यदि (f(x)=x-2 -4x+4), (g(x)=x-2) और (h(x)=1) हैं, तो ((f-2gh)(x)) क्या है?
If (f(x)=x-2 -4x+4), (g(x)=x-2), and (h(x)=1), what is ((f-2gh)(x))?
#relations-functions
#algebra-real-functions
#linear-combination
#product
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A \(x^2-6x+8\)
B \(x^2-2x\)
C \(x^2-4\)
D \(x^2-4x+2\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8\)
Step 1
Concept
(gh=x-2), so \(f-2gh=x^2-4x+4-2x+4=x^2-6x+8\). In combined functions, do multiplication first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8\). (gh=x-2), so \(f-2gh=x^2-4x+4-2x+4=x^2-6x+8\). In combined functions, do multiplication first.
Step 3
Exam Tip
(gh=x-2), इसलिए \(f-2gh=x^2-4x+4-2x+4=x^2-6x+8\)। संयुक्त फलनों में पहले गुणन करें।
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यदि (f(x)=\log x) और (g(x)=\log(6-x)) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\log x) and (g(x)=\log(6-x)), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
#logarithmic-functions
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A ((0,6))
B ([0,6])
C (\(-\infty,0\)\cup\(6,\infty\))
D \(\mathbb{R}\setminus{0,6}\)
Explanation opens after your attempt
Correct Answer
A. ((0,6))
Step 1
Concept
A logarithm requires (x>0) and (6-x>0). Hence the domain is ((0,6)).
Step 2
Why this answer is correct
The correct answer is A. ((0,6)). A logarithm requires (x>0) and (6-x>0). Hence the domain is ((0,6)).
Step 3
Exam Tip
लघुगणक के लिए (x>0) और (6-x>0) चाहिए। इसलिए प्रांत ((0,6)) है।
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यदि (f(x)=x-2 -5x) और (g(x)=x) हैं, तो ((f+g)(x)) के शून्य कौन से हैं?
If (f(x)=x-2 -5x) and (g(x)=x), what are the zeroes of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#zeroes
#sum
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A ({0,4})
B ({0,5})
C ({1,4})
D ({-4,0})
Explanation opens after your attempt
Correct Answer
A. ({0,4})
Step 1
Concept
((f+g)(x)=x-2 -4x=x(x-4)), so the zeroes are (0) and (4). Factorise to find zeroes.
Step 2
Why this answer is correct
The correct answer is A. ({0,4}). ((f+g)(x)=x-2 -4x=x(x-4)), so the zeroes are (0) and (4). Factorise to find zeroes.
Step 3
Exam Tip
((f+g)(x)=x-2 -4x=x(x-4)), इसलिए शून्य (0) और (4) हैं। शून्य खोजने के लिए गुणनखंड बनाएं।
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यदि (f(x)=x-2 +2x) और (g(x)=x-2 -2x) हैं, तो ((fg)(1)) का मान क्या है?
If (f(x)=x-2 +2x) and (g(x)=x-2 -2x), what is the value of ((fg)(1))?
#relations-functions
#algebra-real-functions
#evaluation
#product
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A (3)
B (-3)
C (0)
D (1)
Explanation opens after your attempt
Step 1
Concept
(f(1)=3) and (g(1)=-1), so ((fg)(1)=-3). In a product function, multiply the function values.
Step 2
Why this answer is correct
The correct answer is B. (-3). (f(1)=3) and (g(1)=-1), so ((fg)(1)=-3). In a product function, multiply the function values.
Step 3
Exam Tip
(f(1)=3) और (g(1)=-1), इसलिए ((fg)(1)=-3)। उत्पाद फलन में मानों का गुणन होता है।
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यदि (f(x)=\frac{x+1}{x-3}) और (g(x)=x-3) हैं, तो (\left\(\frac{f}{g}\right\)(x)) का प्रांत क्या है?
If (f(x)=\frac{x+1}{x-3}) and (g(x)=x-3), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
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#algebra-real-functions
#domain
#quotient
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A \(\mathbb{R}\setminus{3}\)
B \(\mathbb{R}\)
C \(\mathbb{R}\setminus{-1,3}\)
D \(\mathbb{R}\setminus{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}\setminus{3}\)
Step 1
Concept
For (f), \(x\ne 3\), and (g(x)\ne 0) in the quotient also gives \(x\ne 3\). Hence only (3) is excluded.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{3}\). For (f), \(x\ne 3\), and (g(x)\ne 0) in the quotient also gives \(x\ne 3\). Hence only (3) is excluded.
Step 3
Exam Tip
(f) के लिए \(x\ne 3\) और भागफल में (g(x)\ne 0) भी \(x\ne 3\) देता है। इसलिए केवल (3) हटेगा।
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यदि (f(x)=|x|+x) और (g(x)=|x|-x) हैं, तो (x<0) के लिए ((f+g)(x)) क्या है?
If (f(x)=|x|+x) and (g(x)=|x|-x), what is ((f+g)(x)) for (x<0)?
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#algebra-real-functions
#modulus
#sum
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A (-2x)
B (2x)
C (0)
D (2|x|)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=2|x|), and for (x<0) it is also (-2x). Among the options, (2|x|) is a general correct form.
Step 2
Why this answer is correct
The correct answer is D. (2|x|). ((f+g)(x)=2|x|), and for (x<0) it is also (-2x). Among the options, (2|x|) is a general correct form.
Step 3
Exam Tip
((f+g)(x)=2|x|), और (x<0) पर यह (-2x) भी होता है। दिए विकल्पों में (2|x|) सामान्य और सही रूप है।
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यदि (f(x)=\frac{x-2 -9}{x-3}) और (g(x)=\frac{x-2 -1}{x-1}) हैं, तो ((f-g)(2)) क्या है?
If (f(x)=\frac{x-2 -9}{x-3}) and (g(x)=\frac{x-2 -1}{x-1}), what is ((f-g)(2))?
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#algebra-real-functions
#evaluation
#difference
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A (-2)
B (0)
C (2)
D अपरिभाषित / Undefined
Explanation opens after your attempt
Step 1
Concept
(f(2)=5) and (g(2)=3), so (f(2)-g(2)=2). Direct evaluation reduces mistakes.
Step 2
Why this answer is correct
The correct answer is A. (-2). (f(2)=5) and (g(2)=3), so (f(2)-g(2)=2). Direct evaluation reduces mistakes.
Step 3
Exam Tip
(f(2)=5) और (g(2)=3), इसलिए (f(2)-g(2)=2)। सीधे मान निकालने पर गलती कम होती है।
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यदि (f(x)=x-2 +1) और (g(x)=2x) हैं, तो ((f+g)(x)) का परास क्या है?
If (f(x)=x-2 +1) and (g(x)=2x), what is the range of ((f+g)(x))?
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#algebra-real-functions
#range
#sum
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A \([0,\infty\))
B \([1,\infty\))
C \(\mathbb{R}\)
D (\(-\infty,0]\)
Explanation opens after your attempt
Correct Answer
A. \([0,\infty\))
Step 1
Concept
((f+g)(x)=x-2 +2x+1=(x+1)2 ), so the range is \([0,\infty\)). The range of a perfect square is non-negative.
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)). ((f+g)(x)=x-2 +2x+1=(x+1)2 ), so the range is \([0,\infty\)). The range of a perfect square is non-negative.
Step 3
Exam Tip
((f+g)(x)=x-2 +2x+1=(x+1)2 ), इसलिए परास \([0,\infty\)) है। पूर्ण वर्ग का परास अऋण होता है।
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यदि (f(x)=\frac{1}{x+1}), (g(x)=\frac{1}{x-1}) और (h(x)=f(x)+g(x)) हैं, तो (h(x)) क्या है?
If (f(x)=\frac{1}{x+1}), (g(x)=\frac{1}{x-1}), and (h(x)=f(x)+g(x)), what is (h(x))?
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#algebra-real-functions
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#rational-functions
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A \(\frac{2x}{x^2-1}\), \(x\ne \pm1\)
B \(\frac{2}{x^2-1}\), \(x\ne \pm1\)
C \(\frac{-2}{x^2-1}\), \(x\ne \pm1\)
D \(\frac{x}{x^2-1}\), \(x\ne \pm1\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2x}{x^2-1}\), \(x\ne \pm1\)
Step 1
Concept
With a common denominator, (h(x)=\frac{x-1+x+1}{x-2 -1}=\frac{2x}{x-2 -1}). The domain excludes \(x=\pm1\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2x}{x^2-1}\), \(x\ne \pm1\). With a common denominator, (h(x)=\frac{x-1+x+1}{x-2 -1}=\frac{2x}{x-2 -1}). The domain excludes \(x=\pm1\).
Step 3
Exam Tip
समान हर से (h(x)=\frac{x-1+x+1}{x-2 -1}=\frac{2x}{x-2 -1})। प्रांत में \(x=\pm1\) हटेंगे।
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यदि (f(x)=x-2 +ax+1) और (g(x)=x-2 +bx+1) हैं तथा ((f-g)(x)=5x), तो (a-b) क्या है?
If (f(x)=x-2 +ax+1) and (g(x)=x-2 +bx+1), and ((f-g)(x)=5x), what is (a-b)?
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#algebra-real-functions
#parameter
#difference
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A (5)
B (-5)
C (0)
D (10)
Explanation opens after your attempt
Step 1
Concept
(f-g=(a-b)x), so ((a-b)x=5x) gives (a-b=5). Compare coefficients of like terms.
Step 2
Why this answer is correct
The correct answer is A. (5). (f-g=(a-b)x), so ((a-b)x=5x) gives (a-b=5). Compare coefficients of like terms.
Step 3
Exam Tip
(f-g=(a-b)x), इसलिए ((a-b)x=5x) से (a-b=5)। समान पदों के गुणांक मिलाएं।
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यदि (f(x)=\frac{x+3}{x-3}) और (g(x)=\frac{x-3}{x+3}) हैं, तो ((fg)(x)) का प्रांत क्या है?
If (f(x)=\frac{x+3}{x-3}) and (g(x)=\frac{x-3}{x+3}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
#product
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A \(\mathbb{R}\setminus{-3,3}\)
B \(\mathbb{R}\)
C \(\mathbb{R}\setminus{3}\)
D \(\mathbb{R}\setminus{-3}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}\setminus{-3,3}\)
Step 1
Concept
The product may simplify to (1), but the original functions forbid (x=3) and (x=-3). The domain is decided by original restrictions.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}\setminus{-3,3}\). The product may simplify to (1), but the original functions forbid (x=3) and (x=-3). The domain is decided by original restrictions.
Step 3
Exam Tip
उत्पाद सरल होकर (1) हो सकता है, पर मूल फलनों में (x=3) और (x=-3) निषिद्ध हैं। प्रांत मूल शर्तों से तय होता है।
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यदि (f(x)=\sqrt{x-1}) और (g(x)=x-5) हैं, तो \(\frac{f}{g}\) का प्रांत क्या है?
If (f(x)=\sqrt{x-1}) and (g(x)=x-5), what is the domain of \(\frac{f}{g}\)?
#relations-functions
#algebra-real-functions
#domain
#quotient
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A \([1,\infty\)\setminus{5})
B (\(1,\infty\)\setminus{5})
C \([5,\infty\))
D \(\mathbb{R}\setminus{5}\)
Explanation opens after your attempt
Correct Answer
A. \([1,\infty\)\setminus{5})
Step 1
Concept
The square root needs \(x\ge 1\), and the denominator needs \(x\ne 5\). Apply both conditions together.
Step 2
Why this answer is correct
The correct answer is A. \([1,\infty\)\setminus{5}). The square root needs \(x\ge 1\), and the denominator needs \(x\ne 5\). Apply both conditions together.
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। दोनों शर्तों को साथ लगाएं।
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यदि (f(x)=x-2 -6x+9), (g(x)=x-3) और \(x\ne 3\), तो (\left\(\frac{f}{g}\right\)(x)) क्या है?
If (f(x)=x-2 -6x+9), (g(x)=x-3), and \(x\ne 3\), what is (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient
#factorisation
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A (x-3)
B (x+3)
C \(x^2-9\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(f=(x-3)2 ), so \(\frac{f}{g}=x-3\) when \(x\ne 3\). Write the restriction while cancelling.
Step 2
Why this answer is correct
The correct answer is A. (x-3). (f=(x-3)2 ), so \(\frac{f}{g}=x-3\) when \(x\ne 3\). Write the restriction while cancelling.
Step 3
Exam Tip
(f=(x-3)2 ), इसलिए \(\frac{f}{g}=x-3\) जब \(x\ne 3\)। कटौती करते समय प्रतिबंध साथ लिखें।
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यदि (f(x)=2x-2 +1), (g(x)=x-2 -3) और (h(x)=x-2 +4) हैं, तो ((f-g+h)(x)) क्या है?
If (f(x)=2x-2 +1), (g(x)=x-2 -3), and (h(x)=x-2 +4), what is ((f-g+h)(x))?
#relations-functions
#algebra-real-functions
#linear-combination
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A \(2x^2+8\)
B \(4x^2+2\)
C \(2x^2-6\)
D \(x^2+8\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+8\)
Step 1
Concept
(f-g+h=2x-2 +1-\(x^2-3\)+x-2 +4=2x-2 +8). Change all signs of (g) while subtracting.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+8\). (f-g+h=2x-2 +1-\(x^2-3\)+x-2 +4=2x-2 +8). Change all signs of (g) while subtracting.
Step 3
Exam Tip
(f-g+h=2x-2 +1-\(x^2-3\)+x-2 +4=2x-2 +8)। घटाव में (g) के सभी चिन्ह बदलें।
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यदि (f(x)=\frac{x-2 -1}{x+1}) और (g(x)=x-1) हैं, तो (f) और (g) किस प्रांत पर समान हैं?
If (f(x)=\frac{x-2 -1}{x+1}) and (g(x)=x-1), on which domain are (f) and (g) equal?
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#algebra-real-functions
#equality
#domain
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A \(x\ne -1\)
B \(x\ne 1\)
C सभी \(x\in\mathbb{R}\) / all \(x\in\mathbb{R}\)
D केवल (x=1) / only (x=1)
Explanation opens after your attempt
Correct Answer
A. \(x\ne -1\)
Step 1
Concept
(x-2 -1=(x-1)(x+1)), so (f=x-1), but (x=-1) is excluded. Keep the original denominator restriction in equality.
Step 2
Why this answer is correct
The correct answer is A. \(x\ne -1\). (x-2 -1=(x-1)(x+1)), so (f=x-1), but (x=-1) is excluded. Keep the original denominator restriction in equality.
Step 3
Exam Tip
(x-2 -1=(x-1)(x+1)), इसलिए (f=x-1), पर (x=-1) हटता है। समानता में मूल हर का प्रतिबंध रखें।
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यदि (f(x)=|x+2|+|x-2|) और (g(x)=4) हैं, तो ((f-g)(0)) क्या है?
If (f(x)=|x+2|+|x-2|) and (g(x)=4), what is ((f-g)(0))?
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#algebra-real-functions
#modulus
#evaluation
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A (0)
B (2)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
(f(0)=|2|+|-2|=4) and (g(0)=4), so the difference is (0). Modulus works like distance.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(0)=|2|+|-2|=4) and (g(0)=4), so the difference is (0). Modulus works like distance.
Step 3
Exam Tip
(f(0)=|2|+|-2|=4) और (g(0)=4), इसलिए अंतर (0) है। मापांक दूरी की तरह काम करता है।
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यदि (f(x)=\frac{2x+1}{x-2 +1}) और (g(x)=\frac{x-3}{x-2 +1}) हैं, तो ((f+g)(x)) क्या है?
If (f(x)=\frac{2x+1}{x-2 +1}) and (g(x)=\frac{x-3}{x-2 +1}), what is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#sum
#rational-functions
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A \(\frac{3x-2}{x^2+1}\)
B \(\frac{x+4}{x^2+1}\)
C \(\frac{3x+4}{x^2+1}\)
D \(\frac{x-2}{x^2+1}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3x-2}{x^2+1}\)
Step 1
Concept
With the same denominator, numerators add: (2x+1+x-3=3x-2). \(x^2+1\) is never zero.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3x-2}{x^2+1}\). With the same denominator, numerators add: (2x+1+x-3=3x-2). \(x^2+1\) is never zero.
Step 3
Exam Tip
समान हर होने से अंश जुड़ते हैं: (2x+1+x-3=3x-2)। \(x^2+1\) कभी शून्य नहीं होता।
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यदि (f(x)=x-2 -2x+2) और (g(x)=x-2 +2x+2) हैं, तो ((fg)(0)) क्या है?
If (f(x)=x-2 -2x+2) and (g(x)=x-2 +2x+2), what is ((fg)(0))?
#relations-functions
#algebra-real-functions
#evaluation
#product
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A (0)
B (2)
C (4)
D (8)
Explanation opens after your attempt
Step 1
Concept
(f(0)=2) and (g(0)=2), so ((fg)(0)=4). In a product function, find values separately first.
Step 2
Why this answer is correct
The correct answer is C. (4). (f(0)=2) and (g(0)=2), so ((fg)(0)=4). In a product function, find values separately first.
Step 3
Exam Tip
(f(0)=2) और (g(0)=2), इसलिए ((fg)(0)=4)। उत्पाद फलन में पहले अलग-अलग मान निकालें।
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यदि (f(x)=x-2 -10x+26) और (g(x)=x-2 -10x+30) हैं, तो \(\frac{f}{g}\) का न्यूनतम मान क्या है?
If (f(x)=x-2 -10x+26) and (g(x)=x-2 -10x+30), what is the minimum value of \(\frac{f}{g}\)?
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#algebra-real-functions
#range
#quotient
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A \(\frac{1}{5}\)
B \(\frac{1}{3}\)
C \(\frac{3}{5}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
(\frac{f}{g}=\frac{(x-5)2 +1}{(x-5)2 +5}), which gives \(\frac{1}{5}\) at ((x-5)2 =0). Set the common square term as \(t\ge 0\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). (\frac{f}{g}=\frac{(x-5)2 +1}{(x-5)2 +5}), which gives \(\frac{1}{5}\) at ((x-5)2 =0). Set the common square term as \(t\ge 0\).
Step 3
Exam Tip
(\frac{f}{g}=\frac{(x-5)2 +1}{(x-5)2 +5}), जो ((x-5)2 =0) पर \(\frac{1}{5}\) देता है। समान वर्ग पद को \(t\ge 0\) रखकर देखें।
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