यदि (f(x)=2x+3) और (g(x)=x-2 -1) हों, तो ((f+g)(x)) क्या होगा?
If (f(x)=2x+3) and (g(x)=x-2 -1), what is ((f+g)(x))?
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#algebra-real-functions
#function-addition
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A \(x^2+2x+2\)
B \(x^2+2x+4\)
C \(2x^2+3x-1\)
D \(x^2-2x-4\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x+2\)
Step 1
Concept
((f+g)(x)=f(x)+g(x)), so \(2x+3+x^2-1=x^2+2x+2\). In exams, combine like terms carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x+2\). ((f+g)(x)=f(x)+g(x)), so \(2x+3+x^2-1=x^2+2x+2\). In exams, combine like terms carefully.
Step 3
Exam Tip
((f+g)(x)=f(x)+g(x)), इसलिए \(2x+3+x^2-1=x^2+2x+2\)। परीक्षा में like terms को साथ जोड़ें।
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यदि (f(x)=x-2 +4x) और (g(x)=3x-5) हों, तो ((f-g)(x)) ज्ञात कीजिए।
If (f(x)=x-2 +4x) and (g(x)=3x-5), find ((f-g)(x)).
#relations-functions
#algebra-real-functions
#function-subtraction
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A \(x^2+x+5\)
B \(x^2+x-5\)
C \(x^2+7x-5\)
D \(x^2-7x+5\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x+5\)
Step 1
Concept
((f-g)(x)=f(x)-g(x)), hence (x-2 +4x-(3x-5)=x-2 +x+5). While subtracting, change the signs inside the bracket.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x+5\). ((f-g)(x)=f(x)-g(x)), hence (x-2 +4x-(3x-5)=x-2 +x+5). While subtracting, change the signs inside the bracket.
Step 3
Exam Tip
((f-g)(x)=f(x)-g(x)), अतः (x-2 +4x-(3x-5)=x-2 +x+5)। घटाते समय bracket का sign बदलता है।
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यदि (f(x)=x+2) और (g(x)=x-3) हों, तो ((fg)(x)) क्या होगा?
If (f(x)=x+2) and (g(x)=x-3), what is ((fg)(x))?
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#algebra-real-functions
#function-product
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A \(x^2-x-6\)
B \(x^2+x-6\)
C \(x^2-5x+6\)
D \(x^2+5x+6\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-x-6\)
Step 1
Concept
((fg)(x)=f(x)g(x)), so ((x+2)(x-3)=x-2 -x-6). In multiplication, distribute both terms properly.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-x-6\). ((fg)(x)=f(x)g(x)), so ((x+2)(x-3)=x-2 -x-6). In multiplication, distribute both terms properly.
Step 3
Exam Tip
((fg)(x)=f(x)g(x)), इसलिए ((x+2)(x-3)=x-2 -x-6)। गुणा में दोनों terms को distribute करें।
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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?
If (f(x)=x-2 -9) and (g(x)=x-3), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
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#function-quotient
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A \(x+3,\ x \neq 3\)
B \(x-3,\ x \neq -3\)
C \(x+3,\ x \neq -3\)
D \(x-3,\ x \neq 3\)
Explanation opens after your attempt
Correct Answer
A. \(x+3,\ x \neq 3\)
Step 1
Concept
(\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), but (x=3) is not allowed. Even after cancellation, remember the original denominator.
Step 2
Why this answer is correct
The correct answer is A. \(x+3,\ x \neq 3\). (\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), but (x=3) is not allowed. Even after cancellation, remember the original denominator.
Step 3
Exam Tip
(\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), पर (x=3) allowed नहीं है। cancel करने के बाद भी मूल denominator याद रखें।
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यदि (f(x)=\sqrt{x-1}) और (g(x)=\sqrt{5-x}) हों, तो ((f+g)(x)) का domain क्या है?
If (f(x)=\sqrt{x-1}) and (g(x)=\sqrt{5-x}), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
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A ([1,5])
B ((1,5))
C (\(-\infty,1]\)
D \([5,\infty\))
Explanation opens after your attempt
Correct Answer
A. ([1,5])
Step 1
Concept
Both square roots must be defined, so \(x-1\geq 0\) and \(5-x\geq 0\), giving \(x\in[1,5]\). For a sum, take the intersection of domains.
Step 2
Why this answer is correct
The correct answer is A. ([1,5]). Both square roots must be defined, so \(x-1\geq 0\) and \(5-x\geq 0\), giving \(x\in[1,5]\). For a sum, take the intersection of domains.
Step 3
Exam Tip
दोनों square roots defined होने चाहिए, इसलिए \(x-1\geq 0\) और \(5-x\geq 0\), जिससे \(x\in[1,5]\)। sum के लिए domains का intersection लें।
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यदि (f(x)=\frac{1}{x-2}) और (g(x)=x+4) हों, तो ((f+g)(x)) का domain क्या होगा?
If (f(x)=\frac{1}{x-2}) and (g(x)=x+4), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
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A \(\mathbb{R}-{2}\)
B \(\mathbb{R}-{-4}\)
C \(\mathbb{R}\)
D ({2})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{2}\)
Step 1
Concept
In (f(x)), the denominator (x-2) cannot be zero, so \(x\neq 2\). The polynomial (g(x)) is defined for all real (x).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{2}\). In (f(x)), the denominator (x-2) cannot be zero, so \(x\neq 2\). The polynomial (g(x)) is defined for all real (x).
Step 3
Exam Tip
(f(x)) में denominator (x-2) शून्य नहीं हो सकता, इसलिए \(x\neq 2\)। polynomial (g(x)) सभी real (x) पर defined है।
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यदि (f(x)=x-2 ) और (g(x)=2x+1) हों, तो ((f+g)(-1)) का मान क्या है?
If (f(x)=x-2 ) and (g(x)=2x+1), what is the value of ((f+g)(-1))?
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#algebra-real-functions
#evaluation
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A (0)
B (1)
C (2)
D (-1)
Explanation opens after your attempt
Step 1
Concept
((f+g)(-1)=f(-1)+g(-1)=1+(-1)=0). First substitute the value in each function, then add.
Step 2
Why this answer is correct
The correct answer is A. (0). ((f+g)(-1)=f(-1)+g(-1)=1+(-1)=0). First substitute the value in each function, then add.
Step 3
Exam Tip
((f+g)(-1)=f(-1)+g(-1)=1+(-1)=0)। पहले functions में value रखें, फिर जोड़ें।
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यदि (f(x)=3x-2) और (g(x)=x-2 +1) हों, तो ((g-f)(2)) क्या होगा?
If (f(x)=3x-2) and (g(x)=x-2 +1), what is ((g-f)(2))?
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#algebra-real-functions
#evaluation
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A (1)
B (3)
C (-1)
D (5)
Explanation opens after your attempt
Step 1
Concept
(g(2)=5) and (f(2)=4), so ((g-f)(2)=5-4=1). Read the order carefully because (g-f) and (f-g) are different.
Step 2
Why this answer is correct
The correct answer is A. (1). (g(2)=5) and (f(2)=4), so ((g-f)(2)=5-4=1). Read the order carefully because (g-f) and (f-g) are different.
Step 3
Exam Tip
(g(2)=5) और (f(2)=4), इसलिए ((g-f)(2)=5-4=1)। order को ध्यान से पढ़ें क्योंकि (g-f) और (f-g) अलग होते हैं।
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यदि (f(x)=x+1), (g(x)=x-1) और (h(x)=x-2 ) हों, तो ((f+g+h)(x)) क्या होगा?
If (f(x)=x+1), (g(x)=x-1), and (h(x)=x-2 ), what is ((f+g+h)(x))?
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#algebra-real-functions
#function-addition
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A \(x^2+2x\)
B \(x^2+2\)
C \(x^2-2x\)
D \(2x^2+x\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x\)
Step 1
Concept
((f+g+h)(x)=(x+1)+(x-1)+x-2 =x-2 +2x). The constants (1) and (-1) cancel out.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x\). ((f+g+h)(x)=(x+1)+(x-1)+x-2 =x-2 +2x). The constants (1) and (-1) cancel out.
Step 3
Exam Tip
((f+g+h)(x)=(x+1)+(x-1)+x-2 =x-2 +2x)। constants (1) और (-1) cancel हो जाते हैं।
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यदि (f(x)=2x-2 -1) हो, तो ((3f)(x)) क्या है?
If (f(x)=2x-2 -1), what is ((3f)(x))?
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#algebra-real-functions
#scalar-multiple
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A \(6x^2-3\)
B \(6x^2-1\)
C \(2x^2-3\)
D \(3x^2-1\)
Explanation opens after your attempt
Correct Answer
A. \(6x^2-3\)
Step 1
Concept
((3f)(x)=3f(x)=3\(2x^2-1\)=6x-2 -3). Multiply the scalar by the whole function.
Step 2
Why this answer is correct
The correct answer is A. \(6x^2-3\). ((3f)(x)=3f(x)=3\(2x^2-1\)=6x-2 -3). Multiply the scalar by the whole function.
Step 3
Exam Tip
((3f)(x)=3f(x)=3\(2x^2-1\)=6x-2 -3)। scalar को पूरी function से multiply करें।
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यदि (f(x)=x-2 +2x) और (g(x)=x) हों, तो (\left\(\frac{f}{g}\right\)(x)) का domain क्या है?
If (f(x)=x-2 +2x) and (g(x)=x), what is the domain of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient-domain
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A \(\mathbb{R}-{0}\)
B \(\mathbb{R}\)
C \(\mathbb{R}-{-2}\)
D ({0})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{0}\)
Step 1
Concept
In a quotient, the denominator (g(x)=x) must not be zero, so \(x\neq 0\). Apply the restriction before simplifying.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{0}\). In a quotient, the denominator (g(x)=x) must not be zero, so \(x\neq 0\). Apply the restriction before simplifying.
Step 3
Exam Tip
quotient में denominator (g(x)=x) शून्य नहीं होना चाहिए, इसलिए \(x\neq 0\)। simplify करने से पहले restriction लगाएं।
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यदि (f(x)=\frac{x+1}{x-4}) और (g(x)=x-2 ) हों, तो ((fg)(x)) का domain क्या है?
If (f(x)=\frac{x+1}{x-4}) and (g(x)=x-2 ), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#product-domain
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A \(\mathbb{R}-{4}\)
B \(\mathbb{R}-{-1}\)
C \(\mathbb{R}\)
D \(\mathbb{R}-{0}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{4}\)
Step 1
Concept
The domain of a product is the intersection of both domains, and in (f(x)), \(x\neq 4\). Since (g(x)) is a polynomial, it adds no new restriction.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{4}\). The domain of a product is the intersection of both domains, and in (f(x)), \(x\neq 4\). Since (g(x)) is a polynomial, it adds no new restriction.
Step 3
Exam Tip
product का domain दोनों domains का intersection होता है, और (f(x)) में \(x\neq 4\)। (g(x)) polynomial है, इसलिए कोई नई restriction नहीं है।
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यदि (f(x)=x-2 -4) और (g(x)=x+2) हों, तो (\left\(\frac{f}{g}\right\)(-2)) के बारे में सही कथन कौन सा है?
If (f(x)=x-2 -4) and (g(x)=x+2), which statement is correct about (\left\(\frac{f}{g}\right\)(-2))?
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#algebra-real-functions
#undefined-value
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A यह defined नहीं है / It is not defined
B इसका मान (0) है / Its value is (0)
C इसका मान (-4) है / Its value is (-4)
D इसका मान (2) है / Its value is (2)
Explanation opens after your attempt
Correct Answer
A. यह defined नहीं है / It is not defined
Step 1
Concept
(g(-2)=0), so the quotient (\left\(\frac{f}{g}\right\)(-2)) is not defined. Even if the numerator is (0), a zero denominator is not allowed.
Step 2
Why this answer is correct
The correct answer is A. यह defined नहीं है / It is not defined. (g(-2)=0), so the quotient (\left\(\frac{f}{g}\right\)(-2)) is not defined. Even if the numerator is (0), a zero denominator is not allowed.
Step 3
Exam Tip
(g(-2)=0), इसलिए quotient (\left\(\frac{f}{g}\right\)(-2)) defined नहीं है। numerator भी (0) हो तो भी denominator (0) allowed नहीं होता।
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यदि (f(x)=|x|) और (g(x)=x) हों, तो ((f-g)(-3)) का मान क्या है?
If (f(x)=|x|) and (g(x)=x), what is the value of ((f-g)(-3))?
#relations-functions
#algebra-real-functions
#modulus-function
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A (6)
B (0)
C (-6)
D (3)
Explanation opens after your attempt
Step 1
Concept
(f(-3)=3) and (g(-3)=-3), so ((f-g)(-3)=3-(-3)=6). A modulus gives a non-negative output.
Step 2
Why this answer is correct
The correct answer is A. (6). (f(-3)=3) and (g(-3)=-3), so ((f-g)(-3)=3-(-3)=6). A modulus gives a non-negative output.
Step 3
Exam Tip
(f(-3)=3) और (g(-3)=-3), इसलिए ((f-g)(-3)=3-(-3)=6)। modulus में negative input का output positive होता है।
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यदि (f(x)=2x+5) और (g(x)=5-2x) हों, तो ((f+g)(x)) किसके बराबर है?
If (f(x)=2x+5) and (g(x)=5-2x), what is ((f+g)(x)) equal to?
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#algebra-real-functions
#constant-function
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A (10)
B (4x)
C (0)
D (10-4x)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=2x+5+5-2x=10), so it is a constant function. When opposite terms cancel, write the remaining constant carefully.
Step 2
Why this answer is correct
The correct answer is A. (10). ((f+g)(x)=2x+5+5-2x=10), so it is a constant function. When opposite terms cancel, write the remaining constant carefully.
Step 3
Exam Tip
((f+g)(x)=2x+5+5-2x=10), इसलिए यह constant function है। opposite terms cancel हों तो ध्यान से लिखें।
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यदि (f(x)=x-2 ) और (g(x)=-x-2 ) हों, तो ((f+g)(x)) कौन सी function है?
If (f(x)=x-2 ) and (g(x)=-x-2 ), which function is ((f+g)(x))?
#relations-functions
#algebra-real-functions
#zero-function
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A zero function (0)
B identity function (x)
C constant function (1)
D quadratic function \(x^2\)
Explanation opens after your attempt
Correct Answer
A. zero function (0)
Step 1
Concept
((f+g)(x)=x-2 -x-2 =0), so it is the zero function. The sum of equal opposite functions gives zero.
Step 2
Why this answer is correct
The correct answer is A. zero function (0). ((f+g)(x)=x-2 -x-2 =0), so it is the zero function. The sum of equal opposite functions gives zero.
Step 3
Exam Tip
((f+g)(x)=x-2 -x-2 =0), इसलिए यह zero function है। समान opposite functions का sum zero देता है।
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यदि (f(x)=x+5) और (g(x)=2x-1) हों, तो ((2f-3g)(x)) क्या होगा?
If (f(x)=x+5) and (g(x)=2x-1), what is ((2f-3g)(x))?
#relations-functions
#algebra-real-functions
#linear-combination
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A (-4x+13)
B (4x+13)
C (-4x+7)
D (8x+13)
Explanation opens after your attempt
Correct Answer
A. (-4x+13)
Step 1
Concept
((2f-3g)(x)=2(x+5)-3(2x-1)=-4x+13). Apply coefficients to the whole function.
Step 2
Why this answer is correct
The correct answer is A. (-4x+13). ((2f-3g)(x)=2(x+5)-3(2x-1)=-4x+13). Apply coefficients to the whole function.
Step 3
Exam Tip
((2f-3g)(x)=2(x+5)-3(2x-1)=-4x+13)। coefficients को पूरी function पर लागू करें।
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यदि (f(x)=x-2 +1) और (g(x)=x-2 -1) हों, तो ((f-g)(x)) क्या होगा?
If (f(x)=x-2 +1) and (g(x)=x-2 -1), what is ((f-g)(x))?
#relations-functions
#algebra-real-functions
#function-subtraction
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A (2)
B \(2x^2\)
C (0)
D (-2)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=x-2 +1-\(x^2-1\)=2). While removing the bracket, the sign of (-1) changes to (+1).
Step 2
Why this answer is correct
The correct answer is A. (2). ((f-g)(x)=x-2 +1-\(x^2-1\)=2). While removing the bracket, the sign of (-1) changes to (+1).
Step 3
Exam Tip
((f-g)(x)=x-2 +1-\(x^2-1\)=2)। bracket हटाते समय (-1) का sign बदलकर (+1) होता है।
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यदि (f(x)=x+1) और (g(x)=\frac{1}{x+1}) हों, तो ((fg)(x)) का value किसके बराबर है?
If (f(x)=x+1) and (g(x)=\frac{1}{x+1}), what is the value of ((fg)(x))?
#relations-functions
#algebra-real-functions
#product-restriction
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A \(1,\ x\neq -1\)
B \(1,\ x\neq 1\)
C \(0,\ x\neq -1\)
D \(x+1,\ x\neq -1\)
Explanation opens after your attempt
Correct Answer
A. \(1,\ x\neq -1\)
Step 1
Concept
((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), but (x=-1) is not allowed. Always check the denominator in a reciprocal function.
Step 2
Why this answer is correct
The correct answer is A. \(1,\ x\neq -1\). ((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), but (x=-1) is not allowed. Always check the denominator in a reciprocal function.
Step 3
Exam Tip
((fg)(x)=(x+1)\cdot\frac{1}{x+1}=1), पर (x=-1) allowed नहीं है। reciprocal वाली function में denominator check करें।
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यदि (f(x)=\sqrt{x+2}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का domain क्या है?
If (f(x)=\sqrt{x+2}) and (g(x)=\frac{1}{x-1}), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain-intersection
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A \([-2,\infty\)-{1})
B (\(-2,\infty\))
C \(\mathbb{R}-{1}\)
D ([-2,1])
Explanation opens after your attempt
Correct Answer
A. \([-2,\infty\)-{1})
Step 1
Concept
For \(\sqrt{x+2}\), \(x\geq -2\), and for \(\frac{1}{x-1}\), \(x\neq 1\). Their intersection gives \([-2,\infty\)-{1}).
Step 2
Why this answer is correct
The correct answer is A. \([-2,\infty\)-{1}). For \(\sqrt{x+2}\), \(x\geq -2\), and for \(\frac{1}{x-1}\), \(x\neq 1\). Their intersection gives \([-2,\infty\)-{1}).
Step 3
Exam Tip
\(\sqrt{x+2}\) के लिए \(x\geq -2\) और \(\frac{1}{x-1}\) के लिए \(x\neq 1\)। intersection से domain \([-2,\infty\)-{1}) मिलता है।
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यदि (f(x)=x-2 -2x) और (g(x)=x-2) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सरल रूप क्या है?
If (f(x)=x-2 -2x) and (g(x)=x-2), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient-simplification
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A \(x,\ x\neq 2\)
B \(x-2,\ x\neq 0\)
C \(x,\ x\neq 0\)
D \(x+2,\ x\neq 2\)
Explanation opens after your attempt
Correct Answer
A. \(x,\ x\neq 2\)
Step 1
Concept
(\frac{x-2 -2x}{x-2}=\frac{x(x-2)}{x-2}=x), but \(x\neq 2\). A cancelled factor does not remove the original restriction.
Step 2
Why this answer is correct
The correct answer is A. \(x,\ x\neq 2\). (\frac{x-2 -2x}{x-2}=\frac{x(x-2)}{x-2}=x), but \(x\neq 2\). A cancelled factor does not remove the original restriction.
Step 3
Exam Tip
(\frac{x-2 -2x}{x-2}=\frac{x(x-2)}{x-2}=x), पर \(x\neq 2\)। cancelled factor से मूल restriction खत्म नहीं होती।
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यदि (f(x)=x-2 +3) और (g(x)=2x) हों, तो ((fg)(1)) क्या है?
If (f(x)=x-2 +3) and (g(x)=2x), what is ((fg)(1))?
#relations-functions
#algebra-real-functions
#evaluation
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A (8)
B (6)
C (4)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(1)=4) and (g(1)=2), so ((fg)(1)=4\cdot2=8). In a product, the function values are multiplied.
Step 2
Why this answer is correct
The correct answer is A. (8). (f(1)=4) and (g(1)=2), so ((fg)(1)=4\cdot2=8). In a product, the function values are multiplied.
Step 3
Exam Tip
(f(1)=4) और (g(1)=2), इसलिए ((fg)(1)=4\cdot2=8)। product में values multiply होती हैं।
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यदि (f(x)=x-4) और (g(x)=x-2 +2x) हों, तो ((g+f)(x)) क्या है?
If (f(x)=x-4) and (g(x)=x-2 +2x), what is ((g+f)(x))?
#relations-functions
#algebra-real-functions
#function-addition
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A \(x^2+3x-4\)
B \(x^2+x-4\)
C \(x^2+2x-4\)
D \(x^2+3x+4\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x-4\)
Step 1
Concept
((g+f)(x)=g(x)+f(x)=x-2 +2x+x-4=x-2 +3x-4). In addition, changing the order does not change the answer.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x-4\). ((g+f)(x)=g(x)+f(x)=x-2 +2x+x-4=x-2 +3x-4). In addition, changing the order does not change the answer.
Step 3
Exam Tip
((g+f)(x)=g(x)+f(x)=x-2 +2x+x-4=x-2 +3x-4)। addition में order बदलने से answer नहीं बदलता।
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यदि (f(x)=\frac{2}{x+3}) और (g(x)=\frac{1}{x-1}) हों, तो ((f+g)(x)) का domain क्या होगा?
If (f(x)=\frac{2}{x+3}) and (g(x)=\frac{1}{x-1}), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#rational-domain
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A \(\mathbb{R}-{-3,1}\)
B \(\mathbb{R}-{3,-1}\)
C \(\mathbb{R}-{-3}\)
D \(\mathbb{R}-{1}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-3,1}\)
Step 1
Concept
Both denominators must be non-zero, so \(x\neq -3\) and \(x\neq 1\). For rational functions, check all denominators.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-3,1}\). Both denominators must be non-zero, so \(x\neq -3\) and \(x\neq 1\). For rational functions, check all denominators.
Step 3
Exam Tip
दोनों denominators non-zero होने चाहिए, इसलिए \(x\neq -3\) और \(x\neq 1\)। rational functions में सभी denominators check करें।
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यदि (f(x)=x-2 ) और (g(x)=x+1) हों, तो ((f-g)(0)) क्या होगा?
If (f(x)=x-2 ) and (g(x)=x+1), what is ((f-g)(0))?
#relations-functions
#algebra-real-functions
#evaluation
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A (-1)
B (1)
C (0)
D (2)
Explanation opens after your attempt
Step 1
Concept
(f(0)=0) and (g(0)=1), so ((f-g)(0)=0-1=-1). At (0), do not forget the constant term.
Step 2
Why this answer is correct
The correct answer is A. (-1). (f(0)=0) and (g(0)=1), so ((f-g)(0)=0-1=-1). At (0), do not forget the constant term.
Step 3
Exam Tip
(f(0)=0) और (g(0)=1), इसलिए ((f-g)(0)=0-1=-1)। (0) पर value रखते समय constant term न भूलें।
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यदि (f(x)=2x-7) और (g(x)=x+5) हों, तो ((f+g)(3)) का मान क्या है?
If (f(x)=2x-7) and (g(x)=x+5), what is the value of ((f+g)(3))?
#relations-functions
#algebra-real-functions
#evaluation
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A (7)
B (9)
C (11)
D (5)
Explanation opens after your attempt
Step 1
Concept
(f(3)=-1) and (g(3)=8), so ((f+g)(3)=7). Direct substitution is the safest method.
Step 2
Why this answer is correct
The correct answer is A. (7). (f(3)=-1) and (g(3)=8), so ((f+g)(3)=7). Direct substitution is the safest method.
Step 3
Exam Tip
(f(3)=-1) और (g(3)=8), इसलिए ((f+g)(3)=7)। direct substitution सबसे सुरक्षित तरीका है।
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यदि (f(x)=x-2 -1) और (g(x)=x-2 +1) हों, तो ((fg)(x)) क्या है?
If (f(x)=x-2 -1) and (g(x)=x-2 +1), what is ((fg)(x))?
#relations-functions
#algebra-real-functions
#product
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A \(x^4-1\)
B \(x^4+1\)
C \(x^4-2x^2-1\)
D \(x^4+2x^2-1\)
Explanation opens after your attempt
Correct Answer
A. \(x^4-1\)
Step 1
Concept
(\(x^2-1\)\(x^2+1\)=x-4 -1) by difference of squares. Recognizing the formula makes calculation faster.
Step 2
Why this answer is correct
The correct answer is A. \(x^4-1\). (\(x^2-1\)\(x^2+1\)=x-4 -1) by difference of squares. Recognizing the formula makes calculation faster.
Step 3
Exam Tip
(\(x^2-1\)\(x^2+1\)=x-4 -1) difference of squares से मिलता है। formula पहचानने से calculation तेज होती है।
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यदि (f(x)=x+2) और (g(x)=x-2 +4x+4) हों, तो (\left\(\frac{g}{f}\right\)(x)) क्या है?
If (f(x)=x+2) and (g(x)=x-2 +4x+4), what is (\left\(\frac{g}{f}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient
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A \(x+2,\ x\neq -2\)
B \(x-2,\ x\neq 2\)
C \(x+2,\ x\neq 2\)
D \(x^2+4,\ x\neq -2\)
Explanation opens after your attempt
Correct Answer
A. \(x+2,\ x\neq -2\)
Step 1
Concept
(g(x)=(x+2)2 ), so (\frac{g(x)}{f(x)}=x+2), but \(x\neq -2\). In a quotient, remove the zero value of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(x+2,\ x\neq -2\). (g(x)=(x+2)2 ), so (\frac{g(x)}{f(x)}=x+2), but \(x\neq -2\). In a quotient, remove the zero value of the denominator.
Step 3
Exam Tip
(g(x)=(x+2)2 ), इसलिए (\frac{g(x)}{f(x)}=x+2), लेकिन \(x\neq -2\)। quotient में denominator की zero value हटानी जरूरी है।
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यदि (f(x)=\sqrt{x}) और (g(x)=x-2) हों, तो ((fg)(x)) का domain क्या है?
If (f(x)=\sqrt{x}) and (g(x)=x-2), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#domain
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A \([0,\infty\))
B (\(0,\infty\))
C \(\mathbb{R}\)
D (\(-\infty,0]\)
Explanation opens after your attempt
Correct Answer
A. \([0,\infty\))
Step 1
Concept
\(\sqrt{x}\) needs \(x\geq 0\), and (g(x)) is defined for all real (x). Therefore, the product domain is \([0,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)). \(\sqrt{x}\) needs \(x\geq 0\), and (g(x)) is defined for all real (x). Therefore, the product domain is \([0,\infty\)).
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\geq 0\) चाहिए और (g(x)) सभी real (x) पर defined है। इसलिए product का domain \([0,\infty\)) है।
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यदि (f(x)=x-2 +2) और (g(x)=4x) हों, तो ((f+g)(x)) को complete square form में कैसे लिखा जा सकता है?
If (f(x)=x-2 +2) and (g(x)=4x), how can ((f+g)(x)) be written in complete square form?
#relations-functions
#algebra-real-functions
#complete-square
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A ((x+2)2 -2)
B ((x-2)2 -2)
C ((x+2)2 +2)
D ((x-4)2 +2)
Explanation opens after your attempt
Correct Answer
A. ((x+2)2 -2)
Step 1
Concept
((f+g)(x)=x-2 +4x+2=(x+2)2 -2). In completing the square, add and subtract the square of half the coefficient.
Step 2
Why this answer is correct
The correct answer is A. ((x+2)2 -2). ((f+g)(x)=x-2 +4x+2=(x+2)2 -2). In completing the square, add and subtract the square of half the coefficient.
Step 3
Exam Tip
((f+g)(x)=x-2 +4x+2=(x+2)2 -2)। complete square में आधे coefficient का square जोड़कर घटाते हैं।
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यदि (f(x)=\frac{1}{x}) और (g(x)=x) हों, तो ((f+g)(x)) का domain क्या है?
If (f(x)=\frac{1}{x}) and (g(x)=x), what is the domain of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#domain
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A \(\mathbb{R}-{0}\)
B \(\mathbb{R}\)
C (\(0,\infty\))
D (\(-\infty,0\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{0}\)
Step 1
Concept
In \(\frac{1}{x}\), the denominator becomes zero at (x=0). 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 \(\frac{1}{x}\), the denominator becomes zero at (x=0). Therefore, the domain of the sum is \(\mathbb{R}-{0}\).
Step 3
Exam Tip
\(\frac{1}{x}\) में (x=0) पर denominator शून्य हो जाता है। इसलिए sum का domain \(\mathbb{R}-{0}\) है।
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यदि (f(x)=x-3 ) और (g(x)=x) हों, तो ((f-g)(x)) का factorized form क्या है?
If (f(x)=x-3 ) and (g(x)=x), what is the factorized form of ((f-g)(x))?
#relations-functions
#algebra-real-functions
#factorization
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A (x(x-1)(x+1))
B (x(x+1)2 )
C (x(x-1)2 )
D (x-2 (x-1))
Explanation opens after your attempt
Correct Answer
A. (x(x-1)(x+1))
Step 1
Concept
((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1)). Taking the common factor first makes it easier.
Step 2
Why this answer is correct
The correct answer is A. (x(x-1)(x+1)). ((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1)). Taking the common factor first makes it easier.
Step 3
Exam Tip
((f-g)(x)=x-3 -x=x\(x^2-1\)=x(x-1)(x+1))। पहले common factor निकालना आसान रहता है।
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यदि (f(x)=x+3) और (g(x)=2x-6) हों, तो ((fg)(0)) क्या है?
If (f(x)=x+3) and (g(x)=2x-6), what is ((fg)(0))?
#relations-functions
#algebra-real-functions
#evaluation
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A (-18)
B (18)
C (-9)
D (0)
Explanation opens after your attempt
Step 1
Concept
(f(0)=3) and (g(0)=-6), so ((fg)(0)=3\cdot(-6)=-18). Do not make a sign error in multiplication.
Step 2
Why this answer is correct
The correct answer is A. (-18). (f(0)=3) and (g(0)=-6), so ((fg)(0)=3\cdot(-6)=-18). Do not make a sign error in multiplication.
Step 3
Exam Tip
(f(0)=3) और (g(0)=-6), इसलिए ((fg)(0)=3\cdot(-6)=-18)। sign multiplication में गलती न करें।
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यदि (f(x)=x-2 +5x+6) और (g(x)=x+2) हों, तो (\left\(\frac{f}{g}\right\)(x)) का सरल रूप क्या है?
If (f(x)=x-2 +5x+6) and (g(x)=x+2), what is the simplified form of (\left\(\frac{f}{g}\right\)(x))?
#relations-functions
#algebra-real-functions
#quotient
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A \(x+3,\ x\neq -2\)
B \(x+2,\ x\neq -3\)
C \(x+3,\ x\neq -3\)
D \(x-3,\ x\neq -2\)
Explanation opens after your attempt
Correct Answer
A. \(x+3,\ x\neq -2\)
Step 1
Concept
(x-2 +5x+6=(x+2)(x+3)), so the quotient is (x+3), but \(x\neq -2\). Write the restriction along with the factorization.
Step 2
Why this answer is correct
The correct answer is A. \(x+3,\ x\neq -2\). (x-2 +5x+6=(x+2)(x+3)), so the quotient is (x+3), but \(x\neq -2\). Write the restriction along with the factorization.
Step 3
Exam Tip
(x-2 +5x+6=(x+2)(x+3)), इसलिए quotient (x+3) है, पर \(x\neq -2\)। factorization के साथ restriction लिखना जरूरी है।
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यदि (f(x)=2x+1) और (g(x)=x-4) हों, तो ((f+g)(x)=0) के लिए (x) का मान क्या है?
If (f(x)=2x+1) and (g(x)=x-4), what is the value of (x) for ((f+g)(x)=0)?
#relations-functions
#algebra-real-functions
#equation
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A (1)
B (-1)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=3x-3), so (3x-3=0) gives (x=1). First form the sum, then solve the equation.
Step 2
Why this answer is correct
The correct answer is A. (1). ((f+g)(x)=3x-3), so (3x-3=0) gives (x=1). First form the sum, then solve the equation.
Step 3
Exam Tip
((f+g)(x)=3x-3), इसलिए (3x-3=0) से (x=1)। पहले sum बनाएं, फिर equation solve करें।
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यदि (f(x)=x-2 -4x+4) और (g(x)=x-2) हों, तो ((fg)(x)) में (x=2) पर value क्या है?
If (f(x)=x-2 -4x+4) and (g(x)=x-2), what is the value of ((fg)(x)) at (x=2)?
#relations-functions
#algebra-real-functions
#zero-product
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A (0)
B (2)
C (4)
D (-2)
Explanation opens after your attempt
Step 1
Concept
(f(2)=0) and (g(2)=0), so ((fg)(2)=0\cdot0=0). If any factor is zero, the product is zero.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(2)=0) and (g(2)=0), so ((fg)(2)=0\cdot0=0). If any factor is zero, the product is zero.
Step 3
Exam Tip
(f(2)=0) और (g(2)=0), इसलिए ((fg)(2)=0\cdot0=0)। product में कोई factor zero हो तो product zero होता है।
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यदि (f(x)=\frac{x}{x+2}) और (g(x)=\frac{3}{x-5}) हों, तो ((fg)(x)) का domain क्या है?
If (f(x)=\frac{x}{x+2}) and (g(x)=\frac{3}{x-5}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#product-domain
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A \(\mathbb{R}-{-2,5}\)
B \(\mathbb{R}-{2,-5}\)
C \(\mathbb{R}-{-2}\)
D \(\mathbb{R}-{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-2,5}\)
Step 1
Concept
Both denominators (x+2) and (x-5) must be non-zero. Therefore, \(x\neq -2\) and \(x\neq 5\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-2,5}\). Both denominators (x+2) and (x-5) must be non-zero. Therefore, \(x\neq -2\) and \(x\neq 5\).
Step 3
Exam Tip
दोनों denominators (x+2) और (x-5) non-zero होने चाहिए। इसलिए \(x\neq -2\) और \(x\neq 5\)।
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यदि (f(x)=x-2 ) और (g(x)=2x) हों, तो ((f+g)(x)) का minimum value क्या है?
If (f(x)=x-2 ) and (g(x)=2x), what is the minimum value of ((f+g)(x))?
#relations-functions
#algebra-real-functions
#minimum-value
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A (-1)
B (0)
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 value is (-1). Completing the square helps find the minimum of a quadratic.
Step 2
Why this answer is correct
The correct answer is A. (-1). ((f+g)(x)=x-2 +2x=(x+1)2 -1), so the minimum value is (-1). Completing the square helps find the minimum of a quadratic.
Step 3
Exam Tip
((f+g)(x)=x-2 +2x=(x+1)2 -1), इसलिए minimum value (-1) है। complete square से quadratic का minimum आसानी से मिलता है।
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यदि (f(x)=x-2 +1) और (g(x)=x-2 -3) हों, तो ((f-g)(5)) क्या होगा?
If (f(x)=x-2 +1) and (g(x)=x-2 -3), what is ((f-g)(5))?
#relations-functions
#algebra-real-functions
#evaluation
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A (4)
B (0)
C (28)
D (-4)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=\(x^2+1\)-\(x^2-3\)=4), so at (x=5) the value is also (4). Simplifying first reduces calculation.
Step 2
Why this answer is correct
The correct answer is A. (4). ((f-g)(x)=\(x^2+1\)-\(x^2-3\)=4), so at (x=5) the value is also (4). Simplifying first reduces calculation.
Step 3
Exam Tip
((f-g)(x)=\(x^2+1\)-\(x^2-3\)=4), इसलिए (x=5) पर भी value (4) है। पहले simplify करने से calculation कम होती है।
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यदि (f(x)=\sqrt{x-4}) और (g(x)=\sqrt{x+1}) हों, तो ((fg)(x)) का domain क्या होगा?
If (f(x)=\sqrt{x-4}) and (g(x)=\sqrt{x+1}), what is the domain of ((fg)(x))?
#relations-functions
#algebra-real-functions
#radical-domain
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A \([4,\infty\))
B \([-1,\infty\))
C ([-1,4])
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([4,\infty\))
Step 1
Concept
For \(\sqrt{x-4}\), \(x\geq 4\), and for \(\sqrt{x+1}\), \(x\geq -1\). The intersection is \([4,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([4,\infty\)). For \(\sqrt{x-4}\), \(x\geq 4\), and for \(\sqrt{x+1}\), \(x\geq -1\). The intersection is \([4,\infty\)).
Step 3
Exam Tip
\(\sqrt{x-4}\) के लिए \(x\geq 4\) और \(\sqrt{x+1}\) के लिए \(x\geq -1\)। intersection \([4,\infty\)) है।
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यदि (f(x)=x-1) और (g(x)=x+1) हों, तो ((f+g)(x)) और ((f-g)(x)) के बारे में सही pair कौन सा है?
If (f(x)=x-1) and (g(x)=x+1), which pair is correct for ((f+g)(x)) and ((f-g)(x))?
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#algebra-real-functions
#combined-operations
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A (2x,\ -2)
B (2x,\ 2)
C (-2,\ 2x)
D \(x^2-1,\ -2\)
Explanation opens after your attempt
Correct Answer
A. (2x,\ -2)
Step 1
Concept
((f+g)(x)=2x) and ((f-g)(x)=(x-1)-(x+1)=-2). Perform plus and minus operations separately.
Step 2
Why this answer is correct
The correct answer is A. (2x,\ -2). ((f+g)(x)=2x) and ((f-g)(x)=(x-1)-(x+1)=-2). Perform plus and minus operations separately.
Step 3
Exam Tip
((f+g)(x)=2x) और ((f-g)(x)=(x-1)-(x+1)=-2)। plus और minus operations अलग-अलग करें।
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यदि (f(x)=\frac{1}{x-2 -1}) और (g(x)=x+2) हों, तो ((f+g)(x)) के domain से कौन से मान हटेंगे?
If (f(x)=\frac{1}{x-2 -1}) and (g(x)=x+2), which values are excluded from the domain of ((f+g)(x))?
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#algebra-real-functions
#domain-exclusion
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A (x=-1) और (x=1) / (x=-1) and (x=1)
B (x=0) और (x=1) / (x=0) and (x=1)
C (x=-2) और (x=2) / (x=-2) and (x=2)
D केवल (x=1) / only (x=1)
Explanation opens after your attempt
Correct Answer
A. (x=-1) और (x=1) / (x=-1) and (x=1)
Step 1
Concept
(x-2 -1=(x-1)(x+1)), so the denominator is zero at (x=1) and (x=-1). Values making the denominator zero are excluded from the domain.
Step 2
Why this answer is correct
The correct answer is A. (x=-1) और (x=1) / (x=-1) and (x=1). (x-2 -1=(x-1)(x+1)), so the denominator is zero at (x=1) and (x=-1). Values making the denominator zero are excluded from the domain.
Step 3
Exam Tip
(x-2 -1=(x-1)(x+1)), इसलिए denominator zero (x=1) और (x=-1) पर होता है। domain में denominator zero values हटती हैं।
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यदि (f(x)=2x-2 ) और (g(x)=3x-2 ) हों, तो ((5f-2g)(x)) क्या है?
If (f(x)=2x-2 ) and (g(x)=3x-2 ), what is ((5f-2g)(x))?
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#algebra-real-functions
#linear-combination
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A \(4x^2\)
B \(10x^2\)
C \(6x^2\)
D \(16x^2\)
Explanation opens after your attempt
Correct Answer
A. \(4x^2\)
Step 1
Concept
((5f-2g)(x)=5\(2x^2\)-2\(3x^2\)=10x-2 -6x-2 =4x-2 ). Combine like terms at the end.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2\). ((5f-2g)(x)=5\(2x^2\)-2\(3x^2\)=10x-2 -6x-2 =4x-2 ). Combine like terms at the end.
Step 3
Exam Tip
((5f-2g)(x)=5\(2x^2\)-2\(3x^2\)=10x-2 -6x-2 =4x-2 )। like terms को अंत में combine करें।
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यदि (f(x)=x-2 -5x+6) और (g(x)=x-3) हों, तो (\left\(\frac{f}{g}\right\)(2)) क्या होगा?
If (f(x)=x-2 -5x+6) and (g(x)=x-3), what is (\left\(\frac{f}{g}\right\)(2))?
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#algebra-real-functions
#evaluation
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A (0)
B (1)
C (-1)
D defined नहीं / not defined
Explanation opens after your attempt
Step 1
Concept
(f(2)=0) and (g(2)=-1), so the quotient is (0). A zero numerator is allowed when the denominator is non-zero.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(2)=0) and (g(2)=-1), so the quotient is (0). A zero numerator is allowed when the denominator is non-zero.
Step 3
Exam Tip
(f(2)=0) और (g(2)=-1), इसलिए quotient (0) है। denominator non-zero हो तो numerator (0) allowed है।
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यदि (f(x)=x+1) और (g(x)=x-2 +1) हों, तो ((f+g)(x)) की degree क्या है?
If (f(x)=x+1) and (g(x)=x-2 +1), what is the degree of ((f+g)(x))?
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#algebra-real-functions
#degree
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A (2)
B (1)
C (3)
D (0)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=x-2 +x+2), whose highest power is (2). The degree of a polynomial is given by its highest exponent.
Step 2
Why this answer is correct
The correct answer is A. (2). ((f+g)(x)=x-2 +x+2), whose highest power is (2). The degree of a polynomial is given by its highest exponent.
Step 3
Exam Tip
((f+g)(x)=x-2 +x+2), जिसकी highest power (2) है। polynomial की degree highest exponent से मिलती है।
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यदि (f(x)=x-2 -1) और (g(x)=x-1) हों, तो ((f-g)(x)) का factorized form क्या होगा?
If (f(x)=x-2 -1) and (g(x)=x-1), what is the factorized form of ((f-g)(x))?
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#algebra-real-functions
#factorization
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A (x(x-1))
B (x(x+1))
C ((x-1)2 )
D \(x^2-x+1\)
Explanation opens after your attempt
Correct Answer
A. (x(x-1))
Step 1
Concept
((f-g)(x)=x-2 -1-(x-1)=x-2 -x=x(x-1)). Keep the signs correct in subtraction.
Step 2
Why this answer is correct
The correct answer is A. (x(x-1)). ((f-g)(x)=x-2 -1-(x-1)=x-2 -x=x(x-1)). Keep the signs correct in subtraction.
Step 3
Exam Tip
((f-g)(x)=x-2 -1-(x-1)=x-2 -x=x(x-1))। subtraction में signs सही रखें।
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यदि (f(x)=\frac{x+2}{x-1}) और (g(x)=x-1) हों, तो ((fg)(x)) का simplified form क्या है?
If (f(x)=\frac{x+2}{x-1}) and (g(x)=x-1), what is the simplified form of ((fg)(x))?
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#algebra-real-functions
#product-simplification
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A \(x+2,\ x\neq 1\)
B \(x+2,\ x\neq -2\)
C \(x-1,\ x\neq 1\)
D \(x^2+x-2,\ x\neq 1\)
Explanation opens after your attempt
Correct Answer
A. \(x+2,\ x\neq 1\)
Step 1
Concept
((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), but because of the original (f(x)), \(x\neq 1\). Write the domain restriction with the simplification.
Step 2
Why this answer is correct
The correct answer is A. \(x+2,\ x\neq 1\). ((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), but because of the original (f(x)), \(x\neq 1\). Write the domain restriction with the simplification.
Step 3
Exam Tip
((fg)(x)=\frac{x+2}{x-1}(x-1)=x+2), पर मूल (f(x)) के कारण \(x\neq 1\)। simplification के साथ domain restriction लिखें।
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यदि (f(x)=x-2 +2x+1) और (g(x)=x+1) हों, तो (\left\(\frac{f}{g}\right\)(-1)) के लिए सही कथन क्या है?
If (f(x)=x-2 +2x+1) and (g(x)=x+1), which statement is correct for (\left\(\frac{f}{g}\right\)(-1))?
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#algebra-real-functions
#undefined-value
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A defined नहीं है / not defined
B value (0) है / value is (0)
C value (-1) है / value is (-1)
D value (1) है / value is (1)
Explanation opens after your attempt
Correct Answer
A. defined नहीं है / not defined
Step 1
Concept
(g(-1)=0), so the quotient is not defined at (-1). Check the original denominator before cancellation.
Step 2
Why this answer is correct
The correct answer is A. defined नहीं है / not defined. (g(-1)=0), so the quotient is not defined at (-1). Check the original denominator before cancellation.
Step 3
Exam Tip
(g(-1)=0), इसलिए quotient (-1) पर defined नहीं है। cancel करने से पहले original denominator check करें।
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यदि (f(x)=x-2 ), (g(x)=x) और (h(x)=1) हों, तो ((f+2g+h)(x)) क्या होगा?
If (f(x)=x-2 ), (g(x)=x), and (h(x)=1), what is ((f+2g+h)(x))?
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A ((x+1)2 )
B \(x^2+x+1\)
C \(x^2+2x\)
D ((x-1)2 )
Explanation opens after your attempt
Correct Answer
A. ((x+1)2 )
Step 1
Concept
((f+2g+h)(x)=x-2 +2x+1=(x+1)2 ). Attach each coefficient to the correct function.
Step 2
Why this answer is correct
The correct answer is A. ((x+1)2 ). ((f+2g+h)(x)=x-2 +2x+1=(x+1)2 ). Attach each coefficient to the correct function.
Step 3
Exam Tip
((f+2g+h)(x)=x-2 +2x+1=(x+1)2 )। coefficients को सही function से जोड़ें।
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यदि (f(x)=x-2) और (g(x)=x+2) हों, तो ((fg)(x)+4) किसके बराबर है?
If (f(x)=x-2) and (g(x)=x+2), what is ((fg)(x)+4) equal to?
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#product-identity
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A \(x^2\)
B \(x^2+4\)
C \(x^2-8\)
D (2x+4)
Explanation opens after your attempt
Correct Answer
A. \(x^2\)
Step 1
Concept
((fg)(x)=(x-2)(x+2)=x-2 -4), so ((fg)(x)+4=x-2 ). Use the difference of squares identity.
Step 2
Why this answer is correct
The correct answer is A. \(x^2\). ((fg)(x)=(x-2)(x+2)=x-2 -4), so ((fg)(x)+4=x-2 ). Use the difference of squares identity.
Step 3
Exam Tip
((fg)(x)=(x-2)(x+2)=x-2 -4), इसलिए ((fg)(x)+4=x-2 )। difference of squares का उपयोग करें।
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