यदि (f(x)=x-2 -4) और (g(x)=x-2) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सरल रूप क्या है, जहाँ \(x\ne 2\)?
If (f(x)=x-2 -4) and (g(x)=x-2), what is the simplified form of (\left\(\frac{f}{g}\right\)(x)), where \(x\ne 2\)?
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A (x+2)
B (x-2)
C \(x^2+4\)
D (2x)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 -4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2), but \(x\ne 2\). Keep the original denominator restriction even after simplification.
Step 2
Why this answer is correct
The correct answer is A. (x+2). (\frac{x-2 -4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2), but \(x\ne 2\). Keep the original denominator restriction even after simplification.
Step 3
Exam Tip
(\frac{x-2 -4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2), पर \(x\ne 2\)। सरल करने के बाद भी मूल हर की शर्त याद रखें।
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यदि (f(x)=2x-3) हो, तो ((3f)(x)) क्या होगा?
If (f(x)=2x-3), what is ((3f)(x))?
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#algebra-real-functions
#scalar-multiple
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A (6x-9)
B (5x-3)
C (6x-3)
D (2x-9)
Explanation opens after your attempt
Step 1
Concept
((3f)(x)=3f(x)=3(2x-3)=6x-9). Multiply the constant by every term.
Step 2
Why this answer is correct
The correct answer is A. (6x-9). ((3f)(x)=3f(x)=3(2x-3)=6x-9). Multiply the constant by every term.
Step 3
Exam Tip
((3f)(x)=3f(x)=3(2x-3)=6x-9)। स्थिर गुणक को सभी पदों से गुणा करें।
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यदि (f(x)=7-x) और (g(x)=x+1) हों, तो ((f-g)(3)) कितना होगा?
If (f(x)=7-x) and (g(x)=x+1), what is ((f-g)(3))?
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#function-value
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A (0)
B (4)
C (6)
D (-2)
Explanation opens after your attempt
Step 1
Concept
(f(3)=4) and (g(3)=4), so ((f-g)(3)=0). Substitute the same (x)-value in both functions.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(3)=4) and (g(3)=4), so ((f-g)(3)=0). Substitute the same (x)-value in both functions.
Step 3
Exam Tip
(f(3)=4) और (g(3)=4), इसलिए ((f-g)(3)=0)। मान निकालते समय दोनों फलनों में वही (x) रखें।
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यदि (f(x)=x+3) और (g(x)=2x) हों, तो ((fg)(1)) का मान ज्ञात करें।
If (f(x)=x+3) and (g(x)=2x), find the value of ((fg)(1)).
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#function-product-value
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A (8)
B (6)
C (4)
D (10)
Explanation opens after your attempt
Step 1
Concept
(f(1)=4) and (g(1)=2), so ((fg)(1)=4\cdot 2=8). For products, multiply the function values.
Step 2
Why this answer is correct
The correct answer is A. (8). (f(1)=4) and (g(1)=2), so ((fg)(1)=4\cdot 2=8). For products, multiply the function values.
Step 3
Exam Tip
(f(1)=4) और (g(1)=2), इसलिए ((fg)(1)=4\cdot 2=8)। गुणन में फलनों के मानों को गुणा करें।
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यदि (f(x)=6x) और (g(x)=3x) हों, तो (\left\(\frac{f}{g}\right\)(2)) क्या होगा?
If (f(x)=6x) and (g(x)=3x), what is (\left\(\frac{f}{g}\right\)(2))?
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A (2)
B (3)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{f}{g}\right\)(2)=\frac{f(2)}{g(2)}=\frac{12}{6}=2). The denominator value must not be (0).
Step 2
Why this answer is correct
The correct answer is A. (2). (\left\(\frac{f}{g}\right\)(2)=\frac{f(2)}{g(2)}=\frac{12}{6}=2). The denominator value must not be (0).
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(2)=\frac{f(2)}{g(2)}=\frac{12}{6}=2)। हर का मान (0) नहीं होना चाहिए।
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यदि (f(x)=x-2 +1) और (g(x)=x) हों, तो ((f+g)(-1)) का मान क्या है?
If (f(x)=x-2 +1) and (g(x)=x), what is the value of ((f+g)(-1))?
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#negative-input
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A (1)
B (0)
C (2)
D (-1)
Explanation opens after your attempt
Step 1
Concept
(f(-1)=2) and (g(-1)=-1), 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(-1)=2) and (g(-1)=-1), so the sum is (1). The square of a negative number is positive.
Step 3
Exam Tip
(f(-1)=2) और (g(-1)=-1), इसलिए योग (1) है। ऋणात्मक संख्या का वर्ग धनात्मक होता है।
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यदि (f(x)=x+5) और (g(x)=2) हों, तो ((fg)(x)) क्या होगा?
If (f(x)=x+5) and (g(x)=2), what is ((fg)(x))?
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#algebra-real-functions
#constant-function
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A (2x+10)
B (x+7)
C (2x+5)
D \(x^2+10\)
Explanation opens after your attempt
Correct Answer
A. (2x+10)
Step 1
Concept
((fg)(x)=(x+5)\cdot 2=2x+10). Use the distributive law when multiplying by a constant function.
Step 2
Why this answer is correct
The correct answer is A. (2x+10). ((fg)(x)=(x+5)\cdot 2=2x+10). Use the distributive law when multiplying by a constant function.
Step 3
Exam Tip
((fg)(x)=(x+5)\cdot 2=2x+10)। स्थिर फलन से गुणा करते समय वितरण नियम लगाएं।
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यदि (f(x)=4x+1) और (g(x)=x-2) हों, तो ((f+g)(x)) में (x) का गुणांक क्या है?
If (f(x)=4x+1) and (g(x)=x-2), what is the coefficient of (x) in ((f+g)(x))?
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#coefficient
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A (5)
B (4)
C (3)
D (-1)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=5x-1), so the coefficient of (x) is (5). Simplify by combining like terms first.
Step 2
Why this answer is correct
The correct answer is A. (5). ((f+g)(x)=5x-1), so the coefficient of (x) is (5). Simplify by combining like terms first.
Step 3
Exam Tip
((f+g)(x)=5x-1), इसलिए (x) का गुणांक (5) है। पहले समान पदों को जोड़कर सरल करें।
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यदि (f(x)=x-4) और (g(x)=x+4) हों, तो ((fg)(x)) का सही विस्तार कौन सा है?
If (f(x)=x-4) and (g(x)=x+4), which is the correct expansion of ((fg)(x))?
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#identity
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A \(x^2-16\)
B \(x^2+16\)
C \(x^2+8x+16\)
D (2x)
Explanation opens after your attempt
Correct Answer
A. \(x^2-16\)
Step 1
Concept
((x-4)(x+4)=x-2 -16). Identify ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16\). ((x-4)(x+4)=x-2 -16). Identify ((a-b)(a+b)=a-2 -b-2 ).
Step 3
Exam Tip
((x-4)(x+4)=x-2 -16)। ((a-b)(a+b)=a-2 -b-2 ) को पहचानें।
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यदि (f(x)=\frac{1}{x}) और (g(x)=x) हों, तो ((fg)(x)) का मान क्या है, जहाँ \(x\ne 0\)?
If (f(x)=\frac{1}{x}) and (g(x)=x), what is ((fg)(x)), where \(x\ne 0\)?
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A (1)
B \(x^2\)
C \(\frac{1}{x^2}\)
D (0)
Explanation opens after your attempt
Step 1
Concept
((fg)(x)=\frac{1}{x}\cdot x=1), but \(x\ne 0\). Exclude zero when (x) is in the denominator.
Step 2
Why this answer is correct
The correct answer is A. (1). ((fg)(x)=\frac{1}{x}\cdot x=1), but \(x\ne 0\). Exclude zero when (x) is in the denominator.
Step 3
Exam Tip
((fg)(x)=\frac{1}{x}\cdot x=1), लेकिन \(x\ne 0\)। हर में (x) हो तो शून्य को हटाएं।
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यदि (f(x)=\sqrt{x}) और (g(x)=x+1) हों, तो ((f+g)(x)) का वास्तविक डोमेन क्या है?
If (f(x)=\sqrt{x}) and (g(x)=x+1), what is the real domain of ((f+g)(x))?
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#domain-sum
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A \([0,\infty\))
B \(\mathbb{R}\)
C (\(-\infty,0\))
D (\(0,\infty\))
Explanation opens after your attempt
Correct Answer
A. \([0,\infty\))
Step 1
Concept
For \(\sqrt{x}\), we need \(x\ge 0\), while (x+1) is defined for all real (x). Hence the domain is \([0,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)). For \(\sqrt{x}\), we need \(x\ge 0\), while (x+1) is defined for all real (x). Hence the domain is \([0,\infty\)).
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) चाहिए और (x+1) सभी वास्तविक (x) पर परिभाषित है। अतः डोमेन \([0,\infty\)) है।
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यदि (f(x)=\frac{1}{x-3}) और (g(x)=x-2 ) हों, तो ((f+g)(x)) का डोमेन क्या होगा?
If (f(x)=\frac{1}{x-3}) and (g(x)=x-2 ), what is the domain of ((f+g)(x))?
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#algebra-real-functions
#domain-restriction
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A \(\mathbb{R}-{3}\)
B \(\mathbb{R}\)
C \(\mathbb{R}-{0}\)
D ({3})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{3}\)
Step 1
Concept
In \(\frac{1}{x-3}\), the denominator becomes (0) at (x=3). So the domain of the sum is \(\mathbb{R}-{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{3}\). In \(\frac{1}{x-3}\), the denominator becomes (0) at (x=3). So the domain of the sum is \(\mathbb{R}-{3}\).
Step 3
Exam Tip
\(\frac{1}{x-3}\) में (x=3) पर हर (0) हो जाता है। इसलिए योग का डोमेन \(\mathbb{R}-{3}\) है।
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यदि (f(x)=x+2) और (g(x)=x-5) हों, तो (\left\(\frac{f}{g}\right\)(x)) का डोमेन क्या है?
If (f(x)=x+2) and (g(x)=x-5), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
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#algebra-real-functions
#quotient-domain
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A \(\mathbb{R}-{5}\)
B \(\mathbb{R}-{-2}\)
C \(\mathbb{R}\)
D ({5})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{5}\)
Step 1
Concept
For a quotient, (g(x)\ne 0) is required. Since (x-5=0) gives (x=5), exclude (5).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{5}\). For a quotient, (g(x)\ne 0) is required. Since (x-5=0) gives (x=5), exclude (5).
Step 3
Exam Tip
भागफल में (g(x)\ne 0) होना चाहिए। (x-5=0) से (x=5), इसलिए (5) हटेगा।
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यदि (f(x)=x-2 +3) और (g(x)=2x-1) हों, तो ((2f+g)(x)) क्या है?
If (f(x)=x-2 +3) and (g(x)=2x-1), what is ((2f+g)(x))?
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#algebra-real-functions
#linear-combination
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A \(2x^2+2x+5\)
B \(2x^2+2x+2\)
C \(x^2+2x+5\)
D \(2x^2-2x+5\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+2x+5\)
Step 1
Concept
((2f+g)(x)=2\(x^2+3\)+(2x-1)=2x-2 +2x+5). Multiply first and then add.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+2x+5\). ((2f+g)(x)=2\(x^2+3\)+(2x-1)=2x-2 +2x+5). Multiply first and then add.
Step 3
Exam Tip
((2f+g)(x)=2\(x^2+3\)+(2x-1)=2x-2 +2x+5)। पहले गुणा करें फिर जोड़ें।
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यदि (f(x)=3x+2) और (g(x)=x-2 ) हों, तो ((f+2g)(1)) का मान क्या है?
If (f(x)=3x+2) and (g(x)=x-2 ), what is the value of ((f+2g)(1))?
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#combined-value
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A (7)
B (5)
C (9)
D (3)
Explanation opens after your attempt
Step 1
Concept
(f(1)=5) and (g(1)=1), so (f(1)+2g(1)=7). Do not forget the multiplier in a combined function.
Step 2
Why this answer is correct
The correct answer is A. (7). (f(1)=5) and (g(1)=1), so (f(1)+2g(1)=7). Do not forget the multiplier in a combined function.
Step 3
Exam Tip
(f(1)=5) और (g(1)=1), इसलिए (f(1)+2g(1)=7)। संयुक्त फलन में गुणक को न भूलें।
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यदि (f(x)=x) और (g(x)=x-2 ) हों, तो ((f+g)(x)) किसके बराबर है?
If (f(x)=x) and (g(x)=x-2 ), then ((f+g)(x)) is equal to which expression?
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A \(x+x^2\)
B \(x^3\)
C \(2x^2\)
D \(x^2-x\)
Explanation opens after your attempt
Correct Answer
A. \(x+x^2\)
Step 1
Concept
In addition, function values are added, not multiplied. Hence ((f+g)(x)=x+x-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(x+x^2\). In addition, function values are added, not multiplied. Hence ((f+g)(x)=x+x-2 ).
Step 3
Exam Tip
योग में फलनों के मान जोड़े जाते हैं, गुणा नहीं किए जाते। अतः ((f+g)(x)=x+x-2 )।
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यदि (f(x)=2x) और (g(x)=x+3) हों, तो ((f-g)(0)) क्या है?
If (f(x)=2x) and (g(x)=x+3), what is ((f-g)(0))?
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#zero-input
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A (-3)
B (3)
C (0)
D (6)
Explanation opens after your attempt
Step 1
Concept
(f(0)=0) and (g(0)=3), so ((f-g)(0)=0-3=-3). At (x=0), the constant term remains.
Step 2
Why this answer is correct
The correct answer is A. (-3). (f(0)=0) and (g(0)=3), so ((f-g)(0)=0-3=-3). At (x=0), the constant term remains.
Step 3
Exam Tip
(f(0)=0) और (g(0)=3), इसलिए ((f-g)(0)=0-3=-3)। (x=0) रखने पर स्थिर पद बचता है।
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यदि (f(x)=x-2 -1) और (g(x)=x+1) हों, तो (\left\(\frac{f}{g}\right\)(0)) का मान क्या है?
If (f(x)=x-2 -1) and (g(x)=x+1), what is the value of (\left\(\frac{f}{g}\right\)(0))?
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#algebra-real-functions
#quotient-value
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A (-1)
B (1)
C (0)
D अपरिभाषित / Undefined
Explanation opens after your attempt
Step 1
Concept
(f(0)=-1) and (g(0)=1), so the quotient is (-1). It is undefined only when the denominator is (0).
Step 2
Why this answer is correct
The correct answer is A. (-1). (f(0)=-1) and (g(0)=1), so the quotient is (-1). It is undefined only when the denominator is (0).
Step 3
Exam Tip
(f(0)=-1) और (g(0)=1), इसलिए भागफल (-1) है। केवल तब अपरिभाषित होगा जब हर (0) हो।
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यदि (f(x)=x+6) और (g(x)=x-6) हों, तो ((f+g)(x)) क्या होगा?
If (f(x)=x+6) and (g(x)=x-6), what is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#simplification
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A (2x)
B \(x^2-36\)
C (12)
D (0)
Explanation opens after your attempt
Step 1
Concept
((x+6)+(x-6)=2x) because (6) and (-6) cancel. Careful signs can earn easy marks.
Step 2
Why this answer is correct
The correct answer is A. (2x). ((x+6)+(x-6)=2x) because (6) and (-6) cancel. Careful signs can earn easy marks.
Step 3
Exam Tip
((x+6)+(x-6)=2x) क्योंकि (6) और (-6) कट जाते हैं। संकेतों पर ध्यान देना आसान अंक दिलाता है।
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यदि (f(x)=2x+1) और (g(x)=x+2) हों, तो ((fg)(0)) कितना है?
If (f(x)=2x+1) and (g(x)=x+2), what is ((fg)(0))?
#relations-functions
#algebra-real-functions
#evaluation
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A (2)
B (0)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
(f(0)=1) and (g(0)=2), so ((fg)(0)=2). Finding values separately is a safe method.
Step 2
Why this answer is correct
The correct answer is A. (2). (f(0)=1) and (g(0)=2), so ((fg)(0)=2). Finding values separately is a safe method.
Step 3
Exam Tip
(f(0)=1) और (g(0)=2), इसलिए ((fg)(0)=2)। पहले अलग-अलग मान निकालना सुरक्षित तरीका है।
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यदि (f(x)=x-2 ) और (g(x)=2x) हों, तो ((f-g)(2)) क्या होगा?
If (f(x)=x-2 ) and (g(x)=2x), what is ((f-g)(2))?
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#value-comparison
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A (0)
B (4)
C (8)
D (-4)
Explanation opens after your attempt
Step 1
Concept
(f(2)=4) and (g(2)=4), so the difference is (0). If the values are equal, their difference is zero.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(2)=4) and (g(2)=4), so the difference is (0). If the values are equal, their difference is zero.
Step 3
Exam Tip
(f(2)=4) और (g(2)=4), इसलिए अंतर (0) है। समान मान मिलने पर घटाव शून्य होगा।
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यदि (f(x)=\sqrt{x-1}) और (g(x)=x) हों, तो ((fg)(x)) का वास्तविक डोमेन क्या है?
If (f(x)=\sqrt{x-1}) and (g(x)=x), what is the real domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain-product
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A \([1,\infty\))
B \([0,\infty\))
C \(\mathbb{R}\)
D (\(-\infty,1]\)
Explanation opens after your attempt
Correct Answer
A. \([1,\infty\))
Step 1
Concept
For \(\sqrt{x-1}\), \(x-1\ge 0\), so \(x\ge 1\). The product domain is the intersection of both domains.
Step 2
Why this answer is correct
The correct answer is A. \([1,\infty\)). For \(\sqrt{x-1}\), \(x-1\ge 0\), so \(x\ge 1\). The product domain is the intersection of both domains.
Step 3
Exam Tip
\(\sqrt{x-1}\) के लिए \(x-1\ge 0\), यानी \(x\ge 1\)। गुणन का डोमेन दोनों डोमेन का प्रतिच्छेद होता है।
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यदि (f(x)=x+1) और (g(x)=\frac{1}{x}) हों, तो ((f+g)(x)) का डोमेन क्या है?
If (f(x)=x+1) and (g(x)=\frac{1}{x}), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain-sum
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A \(\mathbb{R}-{0}\)
B \(\mathbb{R}\)
C (\(0,\infty\))
D ({0})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{0}\)
Step 1
Concept
In \(\frac{1}{x}\), (x=0) is not allowed. Therefore the domain of the sum is \(\mathbb{R}-{0}\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{0}\). In \(\frac{1}{x}\), (x=0) is not allowed. Therefore the domain of the sum is \(\mathbb{R}-{0}\).
Step 3
Exam Tip
\(\frac{1}{x}\) में (x=0) मान्य नहीं है। इसलिए योग का डोमेन \(\mathbb{R}-{0}\) है।
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यदि (f(x)=\frac{x}{x+2}) और (g(x)=x-1) हों, तो ((fg)(x)) के डोमेन से कौन सा मान हटेगा?
If (f(x)=\frac{x}{x+2}) and (g(x)=x-1), which value is excluded from the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#excluded-value
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A (-2)
B (2)
C (1)
D (0)
Explanation opens after your attempt
Step 1
Concept
Since (x+2=0) gives (x=-2), (f(x)) is undefined there. This value is also excluded in the product.
Step 2
Why this answer is correct
The correct answer is A. (-2). Since (x+2=0) gives (x=-2), (f(x)) is undefined there. This value is also excluded in the product.
Step 3
Exam Tip
(x+2=0) से (x=-2) पर (f(x)) अपरिभाषित है। गुणन में भी यह मान हटेगा।
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यदि (f(x)=2x-2 ) और (g(x)=x-2 +3) हों, तो ((f+g)(x)) क्या है?
If (f(x)=2x-2 ) and (g(x)=x-2 +3), what is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#quadratic-sum
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A \(3x^2+3\)
B \(2x^4+6x^2\)
C \(x^2+3\)
D \(3x^4\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+3\)
Step 1
Concept
((f+g)(x)=2x-2 +\(x^2+3\)=3x-2 +3). Only terms with the same power are combined.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+3\). ((f+g)(x)=2x-2 +\(x^2+3\)=3x-2 +3). Only terms with the same power are combined.
Step 3
Exam Tip
((f+g)(x)=2x-2 +\(x^2+3\)=3x-2 +3)। समान घात वाले पद ही जोड़े जाते हैं।
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यदि (f(x)=x-2 +5) और (g(x)=x-2 -2) हों, तो ((f-g)(x)) क्या होगा?
If (f(x)=x-2 +5) and (g(x)=x-2 -2), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#subtraction-signs
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A (7)
B \(2x^2+3\)
C (3)
D \(x^2+7\)
Explanation opens after your attempt
Step 1
Concept
(\(x^2+5\)-\(x^2-2\)=7). Both signs inside the second bracket change.
Step 2
Why this answer is correct
The correct answer is A. (7). (\(x^2+5\)-\(x^2-2\)=7). Both signs inside the second bracket change.
Step 3
Exam Tip
(\(x^2+5\)-\(x^2-2\)=7)। दूसरे कोष्ठक के दोनों पदों के चिह्न बदलते हैं।
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यदि (f(x)=x+2) और (g(x)=x+3) हों, तो ((fg)(x)) में स्थिर पद क्या है?
If (f(x)=x+2) and (g(x)=x+3), what is the constant term in ((fg)(x))?
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#algebra-real-functions
#constant-term
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A (6)
B (5)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
((x+2)(x+3)=x-2 +5x+6), so the constant term is (6). The constant term often comes from multiplying constants.
Step 2
Why this answer is correct
The correct answer is A. (6). ((x+2)(x+3)=x-2 +5x+6), so the constant term is (6). The constant term often comes from multiplying constants.
Step 3
Exam Tip
((x+2)(x+3)=x-2 +5x+6), इसलिए स्थिर पद (6) है। स्थिर पद अक्सर स्थिरों के गुणन से मिलता है।
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यदि (f(x)=4) और (g(x)=x-2 -1) हों, तो ((f-g)(x)) क्या है?
If (f(x)=4) and (g(x)=x-2 -1), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#constant-minus-function
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A \(5-x^2\)
B \(x^2+3\)
C \(4x^2-4\)
D \(x^2-5\)
Explanation opens after your attempt
Correct Answer
A. \(5-x^2\)
Step 1
Concept
((f-g)(x)=4-\(x^2-1\)=5-x-2 ). Watch the minus sign while removing brackets.
Step 2
Why this answer is correct
The correct answer is A. \(5-x^2\). ((f-g)(x)=4-\(x^2-1\)=5-x-2 ). Watch the minus sign while removing brackets.
Step 3
Exam Tip
((f-g)(x)=4-\(x^2-1\)=5-x-2 )। कोष्ठक हटाते समय ऋण चिह्न का ध्यान रखें।
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यदि (f(x)=x-1) और (g(x)=3x+2) हों, तो ((g-f)(x)) क्या होगा?
If (f(x)=x-1) and (g(x)=3x+2), what is ((g-f)(x))?
#relations-functions
#algebra-real-functions
#order-of-subtraction
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A (2x+3)
B (4x+1)
C (2x+1)
D \(3x^2-x-2\)
Explanation opens after your attempt
Step 1
Concept
((g-f)(x)=g(x)-f(x)=(3x+2)-(x-1)=2x+3). Changing the order can change the difference.
Step 2
Why this answer is correct
The correct answer is A. (2x+3). ((g-f)(x)=g(x)-f(x)=(3x+2)-(x-1)=2x+3). Changing the order can change the difference.
Step 3
Exam Tip
((g-f)(x)=g(x)-f(x)=(3x+2)-(x-1)=2x+3)। क्रम बदलने से अंतर बदल सकता है।
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यदि (f(x)=x-2 ) और (g(x)=x) हों, तो ((f-g)(-2)) का मान क्या है?
If (f(x)=x-2 ) and (g(x)=x), what is the value of ((f-g)(-2))?
#relations-functions
#algebra-real-functions
#negative-value
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A (6)
B (2)
C (4)
D (-6)
Explanation opens after your attempt
Step 1
Concept
(f(-2)=4) and (g(-2)=-2), so (4-(-2)=6). Subtracting a negative number becomes addition.
Step 2
Why this answer is correct
The correct answer is A. (6). (f(-2)=4) and (g(-2)=-2), so (4-(-2)=6). Subtracting a negative number becomes addition.
Step 3
Exam Tip
(f(-2)=4) और (g(-2)=-2), इसलिए (4-(-2)=6)। ऋणात्मक संख्या घटाते समय जोड़ बनता है।
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यदि (f(x)=x+1), (g(x)=x+2) और (h(x)=x+3) हों, तो ((f+g+h)(x)) क्या है?
If (f(x)=x+1), (g(x)=x+2), and (h(x)=x+3), what is ((f+g+h)(x))?
#relations-functions
#algebra-real-functions
#three-functions
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A (3x+6)
B (x+6)
C (3x+3)
D \(x^3+6\)
Explanation opens after your attempt
Step 1
Concept
Adding the three functions gives (x+x+x=3x) and (1+2+3=6). So the answer is (3x+6).
Step 2
Why this answer is correct
The correct answer is A. (3x+6). Adding the three functions gives (x+x+x=3x) and (1+2+3=6). So the answer is (3x+6).
Step 3
Exam Tip
तीनों फलनों को जोड़ने पर (x+x+x=3x) और (1+2+3=6)। इसलिए उत्तर (3x+6) है।
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यदि (f(x)=2x+5) और (g(x)=x) हों, तो (\left\(\frac{f}{g}\right\)(1)) क्या होगा?
If (f(x)=2x+5) and (g(x)=x), what is (\left\(\frac{f}{g}\right\)(1))?
#relations-functions
#algebra-real-functions
#quotient-evaluation
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A (7)
B (5)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{f}{g}\right\)(1)=\frac{f(1)}{g(1)}=\frac{7}{1}=7). Always check the denominator value in a quotient.
Step 2
Why this answer is correct
The correct answer is A. (7). (\left\(\frac{f}{g}\right\)(1)=\frac{f(1)}{g(1)}=\frac{7}{1}=7). Always check the denominator value in a quotient.
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(1)=\frac{f(1)}{g(1)}=\frac{7}{1}=7)। भागफल में हर का मान जांचना जरूरी है।
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यदि (f(x)=x-2 -9) और (g(x)=x+3) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सरल रूप क्या है, जहाँ \(x\ne -3\)?
If (f(x)=x-2 -9) and (g(x)=x+3), what is the simplified form of (\left\(\frac{f}{g}\right\)(x)), where \(x\ne -3\)?
#relations-functions
#algebra-real-functions
#simplified-quotient
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A (x-3)
B (x+3)
C \(x^2+9\)
D (2x)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 -9}{x+3}=\frac{(x-3)(x+3)}{x+3}=x-3). The condition \(x\ne -3\) comes from the original denominator.
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). The condition \(x\ne -3\) comes from the original denominator.
Step 3
Exam Tip
(\frac{x-2 -9}{x+3}=\frac{(x-3)(x+3)}{x+3}=x-3)। शर्त \(x\ne -3\) मूल हर से आती है।
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यदि (f(x)=\frac{1}{x-1}) और (g(x)=\frac{1}{x+1}) हों, तो ((f+g)(x)) के डोमेन से कौन से मान हटेंगे?
If (f(x)=\frac{1}{x-1}) and (g(x)=\frac{1}{x+1}), which values are excluded from the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#multiple-restrictions
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A (1,-1)
B (0,1)
C (1,2)
D केवल (0) / Only (0)
Explanation opens after your attempt
Step 1
Concept
(x-1=0) gives (x=1), and (x+1=0) gives (x=-1). Both restrictions apply to the domain of the sum.
Step 2
Why this answer is correct
The correct answer is A. (1,-1). (x-1=0) gives (x=1), and (x+1=0) gives (x=-1). Both restrictions apply to the domain of the sum.
Step 3
Exam Tip
(x-1=0) से (x=1) और (x+1=0) से (x=-1)। योग के डोमेन में दोनों प्रतिबंध लागू होंगे।
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यदि (f(x)=x+4) और (g(x)=2x-3) हों, तो ((f+g)(5)) का मान क्या है?
If (f(x)=x+4) and (g(x)=2x-3), what is the value of ((f+g)(5))?
#relations-functions
#algebra-real-functions
#easy-numerical
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A (16)
B (14)
C (18)
D (10)
Explanation opens after your attempt
Step 1
Concept
(f(5)=9) and (g(5)=7), so the sum is (16). You can also form ((f+g)(x)=3x+1) first.
Step 2
Why this answer is correct
The correct answer is A. (16). (f(5)=9) and (g(5)=7), so the sum is (16). You can also form ((f+g)(x)=3x+1) first.
Step 3
Exam Tip
(f(5)=9) और (g(5)=7), इसलिए योग (16) है। सीधे ((f+g)(x)=3x+1) बनाकर भी मान मिल सकता है।
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यदि (f(x)=3x-2) और (g(x)=x+5) हों, तो ((fg)(2)) का मान क्या है?
If (f(x)=3x-2) and (g(x)=x+5), what is the value of ((fg)(2))?
#relations-functions
#algebra-real-functions
#product-evaluation
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A (28)
B (14)
C (21)
D (35)
Explanation opens after your attempt
Step 1
Concept
(f(2)=4) and (g(2)=7), so ((fg)(2)=28). For a product, multiply, do not add.
Step 2
Why this answer is correct
The correct answer is A. (28). (f(2)=4) and (g(2)=7), so ((fg)(2)=28). For a product, multiply, do not add.
Step 3
Exam Tip
(f(2)=4) और (g(2)=7), इसलिए ((fg)(2)=28)। उत्पाद में जोड़ नहीं, गुणा करें।
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यदि (f(x)=x-2 +2) और (g(x)=2x) हों, तो ((f+g)(x)) किस पूर्ण वर्ग जैसा लिखा जा सकता है?
If (f(x)=x-2 +2) and (g(x)=2x), which completed-square-like form represents ((f+g)(x))?
#relations-functions
#algebra-real-functions
#completed-square
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A ((x+1)2 +1)
B ((x-1)2 +1)
C ((x+2)2 )
D ((x+1)2 -1)
Explanation opens after your attempt
Correct Answer
A. ((x+1)2 +1)
Step 1
Concept
((f+g)(x)=x-2 +2x+2=(x+1)2 +1). Recognizing a completed square also helps in graphs and range.
Step 2
Why this answer is correct
The correct answer is A. ((x+1)2 +1). ((f+g)(x)=x-2 +2x+2=(x+1)2 +1). Recognizing a completed square also helps in graphs and range.
Step 3
Exam Tip
((f+g)(x)=x-2 +2x+2=(x+1)2 +1)। पूर्ण वर्ग पहचानना ग्राफ और रेंज में भी मदद करता है।
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यदि (f(x)=x-2) और (g(x)=x-2) हों, तो ((f-g)(x)) क्या होगा?
If (f(x)=x-2) and (g(x)=x-2), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#zero-function
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A (0)
B (2x-4)
C \(x^2-4x+4\)
D (x-2)
Explanation opens after your attempt
Step 1
Concept
The two functions are identical, so their difference is (0) for every (x). The difference of identical functions is the zero function.
Step 2
Why this answer is correct
The correct answer is A. (0). The two functions are identical, so their difference is (0) for every (x). The difference of identical functions is the zero function.
Step 3
Exam Tip
दोनों फलन समान हैं, इसलिए उनका अंतर हर (x) पर (0) है। समान फलनों का अंतर शून्य फलन होता है।
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यदि (f(x)=1) और (g(x)=x-2 +2x) हों, तो ((fg)(x)) क्या है?
If (f(x)=1) and (g(x)=x-2 +2x), what is ((fg)(x))?
#relations-functions
#algebra-real-functions
#identity-function-product
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A \(x^2+2x\)
B \(x^2+2x+1\)
C (1)
D \(x^2\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x\)
Step 1
Concept
Multiplying by (1) does not change the function. Therefore ((fg)(x)=x-2 +2x).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x\). Multiplying by (1) does not change the function. Therefore ((fg)(x)=x-2 +2x).
Step 3
Exam Tip
(1) से गुणा करने पर फलन नहीं बदलता। इसलिए ((fg)(x)=x-2 +2x) है।
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यदि (f(x)=0) और (g(x)=x+7) हों, तो ((f+g)(x)) क्या है?
If (f(x)=0) and (g(x)=x+7), what is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#zero-function-addition
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A (x+7)
B (0)
C (7x)
D (x)
Explanation opens after your attempt
Step 1
Concept
Adding the zero function leaves the other function unchanged. Hence ((f+g)(x)=x+7).
Step 2
Why this answer is correct
The correct answer is A. (x+7). Adding the zero function leaves the other function unchanged. Hence ((f+g)(x)=x+7).
Step 3
Exam Tip
शून्य फलन जोड़ने से दूसरा फलन वैसा ही रहता है। इसलिए ((f+g)(x)=x+7)।
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यदि (f(x)=x-2 -2x) और (g(x)=x) हों, तो ((f+3g)(x)) क्या होगा?
If (f(x)=x-2 -2x) and (g(x)=x), what is ((f+3g)(x))?
#relations-functions
#algebra-real-functions
#linear-combination
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A \(x^2+x\)
B \(x^2-5x\)
C \(3x^2+x\)
D \(x^2+3\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x\)
Step 1
Concept
((f+3g)(x)=x-2 -2x+3x=x-2 +x). Apply the multiplier correctly in a linear combination.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x\). ((f+3g)(x)=x-2 -2x+3x=x-2 +x). Apply the multiplier correctly in a linear combination.
Step 3
Exam Tip
((f+3g)(x)=x-2 -2x+3x=x-2 +x)। रैखिक संयोजन में गुणक सही लगाएं।
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यदि (f(x)=\frac{1}{x+4}) और (g(x)=x-2 ) हों, तो ((f-g)(x)) का डोमेन क्या है?
If (f(x)=\frac{1}{x+4}) and (g(x)=x-2 ), what is the domain of ((f-g)(x))?
#relations-functions
#algebra-real-functions
#domain-difference
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A \(\mathbb{R}-{-4}\)
B \(\mathbb{R}-{4}\)
C \(\mathbb{R}\)
D (\(4,\infty\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-4}\)
Step 1
Concept
In \(\frac{1}{x+4}\), the denominator is (0) at (x=-4). The domain of the difference is also \(\mathbb{R}-{-4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-4}\). In \(\frac{1}{x+4}\), the denominator is (0) at (x=-4). The domain of the difference is also \(\mathbb{R}-{-4}\).
Step 3
Exam Tip
\(\frac{1}{x+4}\) में (x=-4) पर हर (0) है। अंतर का डोमेन भी \(\mathbb{R}-{-4}\) होगा।
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यदि (f(x)=x+1) और (g(x)=x-2 +1) हों, तो (\left\(\frac{g}{f}\right\)(x)) का डोमेन क्या है?
If (f(x)=x+1) and (g(x)=x-2 +1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient-domain
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A \(\mathbb{R}-{-1}\)
B \(\mathbb{R}-{1}\)
C \(\mathbb{R}\)
D (\(-\infty,-1\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-1}\)
Step 1
Concept
In \(\frac{g}{f}\), the denominator is (f(x)=x+1), so (x=-1) is excluded. Identify the denominator function carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-1}\). In \(\frac{g}{f}\), the denominator is (f(x)=x+1), so (x=-1) is excluded. Identify the denominator function carefully.
Step 3
Exam Tip
\(\frac{g}{f}\) में हर (f(x)=x+1) है, इसलिए (x=-1) हटेगा। भागफल में हर वाले फलन को ध्यान से देखें।
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यदि (f(x)=2x+3) और (g(x)=2x-3) हों, तो ((fg)(x)) क्या है?
If (f(x)=2x+3) and (g(x)=2x-3), what is ((fg)(x))?
#relations-functions
#algebra-real-functions
#product-identity
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A \(4x^2-9\)
B \(4x^2+9\)
C \(2x^2-9\)
D (4x-9)
Explanation opens after your attempt
Correct Answer
A. \(4x^2-9\)
Step 1
Concept
((2x+3)(2x-3)=(2x)2 -32 =4x-2 -9). This is a direct use of difference of squares.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-9\). ((2x+3)(2x-3)=(2x)2 -32 =4x-2 -9). This is a direct use of difference of squares.
Step 3
Exam Tip
((2x+3)(2x-3)=(2x)2 -32 =4x-2 -9)। यह अंतर वर्ग का सीधा प्रयोग 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)) किसके बराबर है?
If (f(x)=x-2 +1) and (g(x)=x-2 +1), then (\left\(\frac{f}{g}\right\)(x)) equals which expression?
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#algebra-real-functions
#quotient-identical-functions
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A (1)
B (0)
C \(x^2+1\)
D \(2x^2+2\)
Explanation opens after your attempt
Step 1
Concept
Since (f(x)=g(x)), (\frac{f(x)}{g(x)}=1). Here \(x^2+1\) is never (0), so all real (x) are valid.
Step 2
Why this answer is correct
The correct answer is A. (1). Since (f(x)=g(x)), (\frac{f(x)}{g(x)}=1). Here \(x^2+1\) is never (0), so all real (x) are valid.
Step 3
Exam Tip
क्योंकि (f(x)=g(x)), इसलिए (\frac{f(x)}{g(x)}=1)। यहाँ \(x^2+1\) कभी (0) नहीं होता, इसलिए सभी वास्तविक (x) मान्य हैं।
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यदि (f(x)=x-2 +4x) और (g(x)=2x-2 -x) हों, तो ((f+g)(x)) क्या है?
If (f(x)=x-2 +4x) and (g(x)=2x-2 -x), what is ((f+g)(x))?
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#algebra-real-functions
#function-addition
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A \(3x^2+3x\)
B \(x^2+5x\)
C \(3x^2-3x\)
D \(2x^4-4x^2\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2+3x\)
Step 1
Concept
((f+g)(x)=x-2 +4x+2x-2 -x=3x-2 +3x). Add only like terms with the same power.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2+3x\). ((f+g)(x)=x-2 +4x+2x-2 -x=3x-2 +3x). Add only like terms with the same power.
Step 3
Exam Tip
((f+g)(x)=x-2 +4x+2x-2 -x=3x-2 +3x)। समान घात वाले पदों को ही जोड़ें।
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यदि (f(x)=5x+2) और (g(x)=2x-7) हों, तो ((f-g)(1)) का मान क्या होगा?
If (f(x)=5x+2) and (g(x)=2x-7), what is the value of ((f-g)(1))?
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#algebra-real-functions
#function-subtraction
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A (12)
B (7)
C (9)
D (4)
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Step 1
Concept
(f(1)=7) and (g(1)=-5), so ((f-g)(1)=7-(-5)=12). Be careful while subtracting a negative value.
Step 2
Why this answer is correct
The correct answer is A. (12). (f(1)=7) and (g(1)=-5), so ((f-g)(1)=7-(-5)=12). Be careful while subtracting a negative value.
Step 3
Exam Tip
(f(1)=7) और (g(1)=-5), इसलिए ((f-g)(1)=7-(-5)=12)। ऋणात्मक मान घटाते समय सावधानी रखें।
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यदि (f(x)=x+2) और (g(x)=x-2 -2x) हों, तो ((fg)(2)) क्या है?
If (f(x)=x+2) and (g(x)=x-2 -2x), what is ((fg)(2))?
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#algebra-real-functions
#function-product
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A (0)
B (4)
C (8)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(2)=4) and (g(2)=0), so ((fg)(2)=4\cdot 0=0). In multiplication, if one value is (0), the product is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). (f(2)=4) and (g(2)=0), so ((fg)(2)=4\cdot 0=0). In multiplication, if one value is (0), the product is (0).
Step 3
Exam Tip
(f(2)=4) और (g(2)=0), इसलिए ((fg)(2)=4\cdot 0=0)। गुणन में किसी एक मान के (0) होने पर उत्पाद (0) होता है।
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यदि (f(x)=\sqrt{x+2}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का वास्तविक डोमेन क्या है?
If (f(x)=\sqrt{x+2}) and (g(x)=\frac{1}{x-1}), what is the real domain of ((f+g)(x))?
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#algebra-real-functions
#domain
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A \([-2,\infty\)-{1})
B \(\mathbb{R}-{1}\)
C \([-2,\infty\))
D (\(1,\infty\))
Explanation opens after your attempt
Correct Answer
A. \([-2,\infty\)-{1})
Step 1
Concept
For \(\sqrt{x+2}\), \(x\ge -2\), and for \(\frac{1}{x-1}\), \(x\ne 1\). Apply both conditions together.
Step 2
Why this answer is correct
The correct answer is A. \([-2,\infty\)-{1}). For \(\sqrt{x+2}\), \(x\ge -2\), and for \(\frac{1}{x-1}\), \(x\ne 1\). Apply both conditions together.
Step 3
Exam Tip
\(\sqrt{x+2}\) के लिए \(x\ge -2\) और \(\frac{1}{x-1}\) के लिए \(x\ne 1\) चाहिए। दोनों शर्तों को साथ लगाएं।
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यदि (f(x)=x-2 -1) और (g(x)=x-2 +1) हों, तो ((g-f)(x)) क्या होगा?
If (f(x)=x-2 -1) and (g(x)=x-2 +1), what is ((g-f)(x))?
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#algebra-real-functions
#function-difference
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A (2)
B \(2x^2\)
C (0)
D \(x^2+2\)
Explanation opens after your attempt
Step 1
Concept
((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2). While subtracting, the signs inside the second bracket change.
Step 2
Why this answer is correct
The correct answer is A. (2). ((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2). While subtracting, the signs inside the second bracket change.
Step 3
Exam Tip
((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2)। घटाते समय दूसरे कोष्ठक के चिह्न बदलते हैं।
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