यदि (f(x)=x-2 +1) और (g(x)=2x-3) हैं, तो ((f+g)(x)) क्या होगा?
If (f(x)=x-2 +1) and (g(x)=2x-3), what is ((f+g)(x))?
#functions
#algebra
#sum
#medium
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A \(x^2+2x-2\)
B \(x^2-2x+4\)
C \(2x^3-3\)
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)=x-2 +1+2x-3=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)=x-2 +1+2x-3=x-2 +2x-2). In exams, combine like terms carefully.
Step 3
Exam Tip
((f+g)(x)=f(x)+g(x)=x-2 +1+2x-3=x-2 +2x-2)। परीक्षा में समान पदों को ध्यान से जोड़ें।
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यदि (f(x)=3x+5) और (g(x)=x-2 -4) हैं, तो ((g-f)(x)) ज्ञात कीजिए।
If (f(x)=3x+5) and (g(x)=x-2 -4), find ((g-f)(x)).
#functions
#subtraction
#polynomial
#medium
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A \(x^2-3x-9\)
B \(x^2+3x+1\)
C \(x^2-3x+1\)
D \(x^2+3x-9\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-9\)
Step 1
Concept
((g-f)(x)=g(x)-f(x)=x-2 -4-(3x+5)=x-2 -3x-9). While subtracting, apply the minus sign to the whole (f(x)).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-9\). ((g-f)(x)=g(x)-f(x)=x-2 -4-(3x+5)=x-2 -3x-9). While subtracting, apply the minus sign to the whole (f(x)).
Step 3
Exam Tip
((g-f)(x)=g(x)-f(x)=x-2 -4-(3x+5)=x-2 -3x-9)। घटाते समय पूरे (f(x)) पर ऋण लगाएँ।
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यदि (f(x)=x+2) और (g(x)=x-5) हैं, तो ((fg)(x)) क्या है?
If (f(x)=x+2) and (g(x)=x-5), what is ((fg)(x))?
#functions
#product
#expansion
#medium
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A \(x^2-3x-10\)
B \(x^2+7x+10\)
C \(x^2-10\)
D (2x-3)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-10\)
Step 1
Concept
((fg)(x)=f(x)g(x)=(x+2)(x-5)=x-2 -3x-10). After expanding, combine like terms.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-10\). ((fg)(x)=f(x)g(x)=(x+2)(x-5)=x-2 -3x-10). After expanding, combine like terms.
Step 3
Exam Tip
((fg)(x)=f(x)g(x)=(x+2)(x-5)=x-2 -3x-10)। गुणा में विस्तार के बाद समान पद मिलाएँ।
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यदि (f(x)=x-2 -9) और (g(x)=x-3) हैं, तो \(x \neq 3\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या होगा?
If (f(x)=x-2 -9) and (g(x)=x-3), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq 3\)?
#functions
#quotient
#domain
#medium
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A (x+3)
B (x-3)
C \(x^2+3\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{f}{g}\right\)(x)=\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), where \(x \neq 3\). Always check that the denominator is not zero.
Step 2
Why this answer is correct
The correct answer is A. (x+3). (\left\(\frac{f}{g}\right\)(x)=\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), where \(x \neq 3\). Always check that the denominator is not zero.
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(x)=\frac{x-2 -9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3), जहाँ \(x \neq 3\)। हर शून्य न हो, यह हमेशा जाँचें।
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यदि (f(x)=\frac{1}{x-2}) और (g(x)=\frac{1}{x+4}) हैं, तो ((f+g)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{1}{x-2}) and (g(x)=\frac{1}{x+4}), what is the domain of ((f+g)(x))?
#functions
#rational
#domain
#medium
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A \(x \in \mathbb{R}, x \neq 2, x \neq -4\)
B \(x \in \mathbb{R}, x \neq 2\)
C \(x \in \mathbb{R}, x \neq -4\)
D \(x \in \mathbb{R}, x \neq -2, x \neq 4\)
Explanation opens after your attempt
Correct Answer
A. \(x \in \mathbb{R}, x \neq 2, x \neq -4\)
Step 1
Concept
The denominators (x-2) and (x+4) must not be zero, so \(x \neq 2\) and \(x \neq -4\). In rational functions, check denominators first.
Step 2
Why this answer is correct
The correct answer is A. \(x \in \mathbb{R}, x \neq 2, x \neq -4\). The denominators (x-2) and (x+4) must not be zero, so \(x \neq 2\) and \(x \neq -4\). In rational functions, check denominators first.
Step 3
Exam Tip
हरों (x-2) और (x+4) को शून्य नहीं होना चाहिए, इसलिए \(x \neq 2\) और \(x \neq -4\)। परिमेय फलनों में हर पहले जाँचें।
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यदि (f(x)=2x-2 ) और (g(x)=x+1) हैं, तो ((f-g)(-2)) का मान क्या है?
If (f(x)=2x-2 ) and (g(x)=x+1), what is the value of ((f-g)(-2))?
#functions
#value
#substitution
#medium
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A (9)
B (7)
C (5)
D (-9)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=2x-2 -(x+1)), so ((f-g)(-2)=8-(-1)=9). First form the function, then substitute the value.
Step 2
Why this answer is correct
The correct answer is A. (9). ((f-g)(x)=2x-2 -(x+1)), so ((f-g)(-2)=8-(-1)=9). First form the function, then substitute the value.
Step 3
Exam Tip
((f-g)(x)=2x-2 -(x+1)), इसलिए ((f-g)(-2)=8-(-1)=9)। पहले फलन बनाकर फिर मान रखें।
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यदि (f(x)=x-2 +2x) और (g(x)=x-2 -2x) हैं, तो ((f+g)(3)) क्या है?
If (f(x)=x-2 +2x) and (g(x)=x-2 -2x), what is ((f+g)(3))?
#functions
#sum
#value
#medium
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A (18)
B (12)
C (6)
D (36)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=2x-2 ), hence ((f+g)(3)=2\cdot 32 =18). Opposite terms may cancel, so simplify first.
Step 2
Why this answer is correct
The correct answer is A. (18). ((f+g)(x)=2x-2 ), hence ((f+g)(3)=2\cdot 32 =18). Opposite terms may cancel, so simplify first.
Step 3
Exam Tip
((f+g)(x)=2x-2 ), अतः ((f+g)(3)=2\cdot 32 =18)। विरोधी पद कट सकते हैं, इसलिए सरल रूप देखें।
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यदि (f(x)=x-2 -1) और (g(x)=2x+1) हैं, तो ((fg)(2)) ज्ञात करें।
If (f(x)=x-2 -1) and (g(x)=2x+1), find ((fg)(2)).
#functions
#product
#value
#medium
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A (15)
B (10)
C (12)
D (20)
Explanation opens after your attempt
Step 1
Concept
((fg)(2)=f(2)g(2)=(4-1)(5)=15). In a product, finding both values separately is often easier.
Step 2
Why this answer is correct
The correct answer is A. (15). ((fg)(2)=f(2)g(2)=(4-1)(5)=15). In a product, finding both values separately is often easier.
Step 3
Exam Tip
((fg)(2)=f(2)g(2)=(4-1)(5)=15)। उत्पाद में दोनों फलनों का मान अलग-अलग निकालना आसान है।
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यदि (f(x)=4x-1) और (g(x)=x-2 +2) हैं, तो ((g+f)(x)) और ((f+g)(x)) के बारे में सही कथन कौन-सा है?
If (f(x)=4x-1) and (g(x)=x-2 +2), which statement about ((g+f)(x)) and ((f+g)(x)) is correct?
#functions
#commutative
#sum
#medium
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A दोनों बराबर हैं / Both are equal
B दोनों हमेशा विपरीत हैं / Both are always opposites
C दोनों के प्रांत हमेशा अलग हैं / Their domains are always different
D दोनों केवल (x=0) पर बराबर हैं / They are equal only at (x=0)
Explanation opens after your attempt
Correct Answer
A. दोनों बराबर हैं / Both are equal
Step 1
Concept
Addition of functions is commutative, so ((f+g)(x)=(g+f)(x)). Changing order in addition does not change the value.
Step 2
Why this answer is correct
The correct answer is A. दोनों बराबर हैं / Both are equal. Addition of functions is commutative, so ((f+g)(x)=(g+f)(x)). Changing order in addition does not change the value.
Step 3
Exam Tip
फलनों का योग क्रमविनिमेय है, इसलिए ((f+g)(x)=(g+f)(x))। योग में क्रम बदलने से मान नहीं बदलता।
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यदि (f(x)=x+6) और (g(x)=x-6) हैं, तो ((fg)(x)) का सरल रूप क्या है?
If (f(x)=x+6) and (g(x)=x-6), what is the simplified form of ((fg)(x))?
#functions
#product
#identity
#medium
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A \(x^2-36\)
B \(x^2+36\)
C (2x-36)
D \(x^2-12x+36\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-36\)
Step 1
Concept
((fg)(x)=(x+6)(x-6)=x-2 -36). This follows the \(a^2-b^2\) pattern.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-36\). ((fg)(x)=(x+6)(x-6)=x-2 -36). This follows the \(a^2-b^2\) pattern.
Step 3
Exam Tip
((fg)(x)=(x+6)(x-6)=x-2 -36)। यह \(a^2-b^2\) का पैटर्न है।
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यदि (f(x)=\frac{x+1}{x-1}) और (g(x)=x-1) हैं, तो ((fg)(x)) का प्रांत क्या है?
If (f(x)=\frac{x+1}{x-1}) and (g(x)=x-1), what is the domain of ((fg)(x))?
#functions
#product
#domain
#medium
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A \(x \in \mathbb{R}, x \neq 1\)
B \(x \in \mathbb{R}\)
C \(x \in \mathbb{R}, x \neq -1\)
D (x>1)
Explanation opens after your attempt
Correct Answer
A. \(x \in \mathbb{R}, x \neq 1\)
Step 1
Concept
In the original (f(x)), \(x-1 \neq 0\), so \(x \neq 1\), even if simplification occurs after multiplication. Decide the domain from original expressions.
Step 2
Why this answer is correct
The correct answer is A. \(x \in \mathbb{R}, x \neq 1\). In the original (f(x)), \(x-1 \neq 0\), so \(x \neq 1\), even if simplification occurs after multiplication. Decide the domain from original expressions.
Step 3
Exam Tip
मूल (f(x)) में \(x-1 \neq 0\), इसलिए \(x \neq 1\), भले ही गुणा के बाद सरलीकरण हो। प्रांत मूल अभिव्यक्तियों से तय करें।
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यदि (f(x)=\frac{2}{x}) और (g(x)=x-2 ) हैं, तो ((fg)(x)) क्या है और उसका प्रांत क्या है?
If (f(x)=\frac{2}{x}) and (g(x)=x-2 ), what is ((fg)(x)) and its domain?
#functions
#domain
#product
#rational
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A \(2x, x \neq 0\)
B \(2x, x \in \mathbb{R}\)
C \(2x^2, x \neq 0\)
D \(\frac{2}{x^2}, x \neq 0\)
Explanation opens after your attempt
Correct Answer
A. \(2x, x \neq 0\)
Step 1
Concept
((fg)(x)=\frac{2}{x}\cdot x-2 =2x), but \(x \neq 0\) because of the original denominator. Keep original restrictions after simplification.
Step 2
Why this answer is correct
The correct answer is A. \(2x, x \neq 0\). ((fg)(x)=\frac{2}{x}\cdot x-2 =2x), but \(x \neq 0\) because of the original denominator. Keep original restrictions after simplification.
Step 3
Exam Tip
((fg)(x)=\frac{2}{x}\cdot x-2 =2x), पर मूल हर के कारण \(x \neq 0\)। सरलीकरण के बाद भी मूल प्रतिबंध रखें।
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यदि (f(x)=x-2 +4x+4) और (g(x)=x+2) हैं, तो \(x \neq -2\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या होगा?
If (f(x)=x-2 +4x+4) and (g(x)=x+2), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq -2\)?
#functions
#quotient
#factorisation
#medium
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A (x+2)
B (x-2)
C \(x^2+2\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 +4x+4}{x+2}=\frac{(x+2)2 }{x+2}=x+2), where \(x \neq -2\). Write the zero of the denominator separately.
Step 2
Why this answer is correct
The correct answer is A. (x+2). (\frac{x-2 +4x+4}{x+2}=\frac{(x+2)2 }{x+2}=x+2), where \(x \neq -2\). Write the zero of the denominator separately.
Step 3
Exam Tip
(\frac{x-2 +4x+4}{x+2}=\frac{(x+2)2 }{x+2}=x+2), जहाँ \(x \neq -2\)। हर का शून्य मान अलग से लिखें।
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यदि (f(x)=|x|) और (g(x)=x) हैं, तो ((f-g)(-5)) का मान क्या है?
If (f(x)=|x|) and (g(x)=x), what is the value of ((f-g)(-5))?
#functions
#absolute-value
#value
#medium
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A (10)
B (0)
C (-10)
D (5)
Explanation opens after your attempt
Step 1
Concept
((f-g)(-5)=|-5|-(-5)=5+5=10). Absolute value is always non-negative.
Step 2
Why this answer is correct
The correct answer is A. (10). ((f-g)(-5)=|-5|-(-5)=5+5=10). Absolute value is always non-negative.
Step 3
Exam Tip
((f-g)(-5)=|-5|-(-5)=5+5=10)। निरपेक्ष मान हमेशा गैर-ऋणात्मक होता है।
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यदि (f(x)=x-3 ) और (g(x)=2x-3 -1) हैं, तो ((2f-g)(x)) क्या होगा?
If (f(x)=x-3 ) and (g(x)=2x-3 -1), what is ((2f-g)(x))?
#functions
#linear-combination
#cubic
#medium
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A (1)
B (-1)
C \(4x^3-1\)
D \(x^3+1\)
Explanation opens after your attempt
Step 1
Concept
((2f-g)(x)=2x-3 -\(2x^3-1\)=1). Do not forget sign changes while opening brackets.
Step 2
Why this answer is correct
The correct answer is A. (1). ((2f-g)(x)=2x-3 -\(2x^3-1\)=1). Do not forget sign changes while opening brackets.
Step 3
Exam Tip
((2f-g)(x)=2x-3 -\(2x^3-1\)=1)। कोष्ठक खोलते समय चिह्न बदलना न भूलें।
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यदि (f(x)=x-2 ) और (g(x)=\sqrt{x}) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=x-2 ) and (g(x)=\sqrt{x}), what is the domain of ((f+g)(x))?
#functions
#domain
#square-root
#medium
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A \(x \geq 0\)
B \(x \in \mathbb{R}\)
C (x>0)
D \(x \leq 0\)
Explanation opens after your attempt
Correct Answer
A. \(x \geq 0\)
Step 1
Concept
\(x^2\) is defined for all real (x), but \(\sqrt{x}\) needs \(x \geq 0\). The domain of the sum is the intersection.
Step 2
Why this answer is correct
The correct answer is A. \(x \geq 0\). \(x^2\) is defined for all real (x), but \(\sqrt{x}\) needs \(x \geq 0\). The domain of the sum is the intersection.
Step 3
Exam Tip
\(x^2\) सभी वास्तविक (x) के लिए परिभाषित है, लेकिन \(\sqrt{x}\) के लिए \(x \geq 0\) चाहिए। योग का प्रांत प्रतिच्छेद है।
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यदि (f(x)=\frac{1}{x}) और (g(x)=\sqrt{x-2}) हैं, तो ((fg)(x)) का प्रांत क्या होगा?
If (f(x)=\frac{1}{x}) and (g(x)=\sqrt{x-2}), what is the domain of ((fg)(x))?
#functions
#domain
#product
#medium
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A \(x \geq 2\)
B (x>2)
C \(x \neq 0\)
D \(x \leq 2\)
Explanation opens after your attempt
Correct Answer
A. \(x \geq 2\)
Step 1
Concept
\(\sqrt{x-2}\) requires \(x \geq 2\) and \(\frac{1}{x}\) requires \(x \neq 0\); the intersection is \(x \geq 2\). Identify the strongest restriction.
Step 2
Why this answer is correct
The correct answer is A. \(x \geq 2\). \(\sqrt{x-2}\) requires \(x \geq 2\) and \(\frac{1}{x}\) requires \(x \neq 0\); the intersection is \(x \geq 2\). Identify the strongest restriction.
Step 3
Exam Tip
\(\sqrt{x-2}\) के लिए \(x \geq 2\) और \(\frac{1}{x}\) के लिए \(x \neq 0\); प्रतिच्छेद \(x \geq 2\) है। सबसे कड़ा प्रतिबंध पहचानें।
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यदि (f(x)=x-1) और (g(x)=x+1) हैं, तो (\left\(\frac{f}{g}\right\)(0)) का मान क्या है?
If (f(x)=x-1) and (g(x)=x+1), what is the value of (\left\(\frac{f}{g}\right\)(0))?
#functions
#quotient
#value
#medium
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A (-1)
B (1)
C (0)
D परिभाषित नहीं / Not defined
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{f}{g}\right\)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1). In division, first check whether the denominator value is zero.
Step 2
Why this answer is correct
The correct answer is A. (-1). (\left\(\frac{f}{g}\right\)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1). In division, first check whether the denominator value is zero.
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1)। भाग में हर का मान शून्य है या नहीं, पहले देखें।
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यदि (f(x)=x-2 -4) और (g(x)=x-2 +4) हैं, तो ((g-f)(x)) क्या होगा?
If (f(x)=x-2 -4) and (g(x)=x-2 +4), what is ((g-f)(x))?
#functions
#subtraction
#constant
#medium
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A (8)
B \(2x^2\)
C \(x^2+8\)
D (-8)
Explanation opens after your attempt
Step 1
Concept
((g-f)(x)=\(x^2+4\)-\(x^2-4\)=8). The like \(x^2\) terms cancel.
Step 2
Why this answer is correct
The correct answer is A. (8). ((g-f)(x)=\(x^2+4\)-\(x^2-4\)=8). The like \(x^2\) terms cancel.
Step 3
Exam Tip
((g-f)(x)=\(x^2+4\)-\(x^2-4\)=8)। समान \(x^2\) पद कट जाते हैं।
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यदि (f(x)=2x+3), (g(x)=x-4) और (h(x)=x-2 ) हैं, तो ((f+g-h)(x)) क्या है?
If (f(x)=2x+3), (g(x)=x-4), and (h(x)=x-2 ), what is ((f+g-h)(x))?
#functions
#linear-combination
#medium
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A \(-x^2+3x-1\)
B \(x^2+3x-1\)
C \(-x^2+x+7\)
D \(x^2-3x+1\)
Explanation opens after your attempt
Correct Answer
A. \(-x^2+3x-1\)
Step 1
Concept
((f+g-h)(x)=2x+3+x-4-x-2 =-x-2 +3x-1). With many functions, combine terms in order.
Step 2
Why this answer is correct
The correct answer is A. \(-x^2+3x-1\). ((f+g-h)(x)=2x+3+x-4-x-2 =-x-2 +3x-1). With many functions, combine terms in order.
Step 3
Exam Tip
((f+g-h)(x)=2x+3+x-4-x-2 =-x-2 +3x-1)। कई फलनों में पदों को क्रम से मिलाएँ।
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यदि (f(x)=x-2 +x) और (g(x)=x) हैं, तो \(x \neq 0\) के लिए (\left\(\frac{f}{g}\right\)(x)) क्या है?
If (f(x)=x-2 +x) and (g(x)=x), what is (\left\(\frac{f}{g}\right\)(x)) for \(x \neq 0\)?
#functions
#quotient
#simplification
#medium
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A (x+1)
B (x)
C \(x^2+1\)
D (1)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 +x}{x}=\frac{x(x+1)}{x}=x+1), where \(x \neq 0\). Do not forget the restriction while cancelling a common factor.
Step 2
Why this answer is correct
The correct answer is A. (x+1). (\frac{x-2 +x}{x}=\frac{x(x+1)}{x}=x+1), where \(x \neq 0\). Do not forget the restriction while cancelling a common factor.
Step 3
Exam Tip
(\frac{x-2 +x}{x}=\frac{x(x+1)}{x}=x+1), जहाँ \(x \neq 0\)। साझा गुणनखंड काटते समय प्रतिबंध न भूलें।
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यदि (f(x)=\frac{x}{x+2}) और (g(x)=\frac{2}{x+2}) हैं, तो ((f+g)(x)) का सरल रूप क्या है?
If (f(x)=\frac{x}{x+2}) and (g(x)=\frac{2}{x+2}), what is the simplified form of ((f+g)(x))?
#functions
#sum
#rational
#medium
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A (1), \(x \neq -2\)
B (1), \(x \in \mathbb{R}\)
C (x+2), \(x \neq -2\)
D \(\frac{x+2}{x}\), \(x \neq 0\)
Explanation opens after your attempt
Correct Answer
A. (1), \(x \neq -2\)
Step 1
Concept
((f+g)(x)=\frac{x+2}{x+2}=1), but \(x+2 \neq 0\). It is important to write the domain restriction with the simplified answer.
Step 2
Why this answer is correct
The correct answer is A. (1), \(x \neq -2\). ((f+g)(x)=\frac{x+2}{x+2}=1), but \(x+2 \neq 0\). It is important to write the domain restriction with the simplified answer.
Step 3
Exam Tip
((f+g)(x)=\frac{x+2}{x+2}=1), लेकिन \(x+2 \neq 0\)। सरल उत्तर के साथ प्रांत का प्रतिबंध लिखना जरूरी है।
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यदि (f(x)=3) और (g(x)=x-2 -7x) हैं, तो ((fg)(x)) क्या होगा?
If (f(x)=3) and (g(x)=x-2 -7x), what is ((fg)(x))?
#functions
#constant-function
#product
#medium
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A \(3x^2-21x\)
B \(x^2-7x+3\)
C \(3x^2-7x\)
D \(x^2-21x\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-21x\)
Step 1
Concept
((fg)(x)=3\(x^2-7x\)=3x-2 -21x). When multiplying by a constant function, multiply every term.
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-21x\). ((fg)(x)=3\(x^2-7x\)=3x-2 -21x). When multiplying by a constant function, multiply every term.
Step 3
Exam Tip
((fg)(x)=3\(x^2-7x\)=3x-2 -21x)। स्थिर फलन से गुणा करते समय हर पद पर गुणा करें।
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यदि (f(x)=x-2 +1) है, तो ((f+f)(x)) क्या होगा?
If (f(x)=x-2 +1), what is ((f+f)(x))?
#functions
#sum
#same-function
#medium
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A \(2x^2+2\)
B \(x^2+2\)
C \(2x^2+1\)
D \(x^4+1\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+2\)
Step 1
Concept
((f+f)(x)=2f(x)=2\(x^2+1\)=2x-2 +2). Adding the same function gives (2f(x)).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+2\). ((f+f)(x)=2f(x)=2\(x^2+1\)=2x-2 +2). Adding the same function gives (2f(x)).
Step 3
Exam Tip
((f+f)(x)=2f(x)=2\(x^2+1\)=2x-2 +2)। समान फलन जोड़ने पर (2f(x)) बनता है।
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यदि (f(x)=x-2) है, तो ((ff)(x)) क्या है?
If (f(x)=x-2), what is ((ff)(x))?
#functions
#product
#common-mistake
#medium
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A \(x^2-4x+4\)
B (2x-4)
C \(x^2-4\)
D \(x^2+4x+4\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x+4\)
Step 1
Concept
((ff)(x)=f(x)\cdot f(x)=(x-2)2 =x-2 -4x+4). Do not confuse ((ff)(x)) with (f(f(x))).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+4\). ((ff)(x)=f(x)\cdot f(x)=(x-2)2 =x-2 -4x+4). Do not confuse ((ff)(x)) with (f(f(x))).
Step 3
Exam Tip
((ff)(x)=f(x)\cdot f(x)=(x-2)2 =x-2 -4x+4)। ((ff)(x)) को (f(f(x))) न समझें।
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यदि (f(x)=\sin x) और (g(x)=\cos x) हैं, तो (\(f^2+g^2\)(x)) क्या होगा?
If (f(x)=\sin x) and (g(x)=\cos x), what is (\(f^2+g^2\)(x))?
#functions
#trigonometry
#identity
#medium
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A (1)
B \(\sin x+\cos x\)
C \(\sin^2 x-\cos^2 x\)
D \(2\sin x\cos x\)
Explanation opens after your attempt
Step 1
Concept
(\(f^2+g^2\)(x)=\sin-2 x+\cos-2 x=1). Remember the trigonometric identity.
Step 2
Why this answer is correct
The correct answer is A. (1). (\(f^2+g^2\)(x)=\sin-2 x+\cos-2 x=1). Remember the trigonometric identity.
Step 3
Exam Tip
(\(f^2+g^2\)(x)=\sin-2 x+\cos-2 x=1)। त्रिकोणमितीय पहचान याद रखें।
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यदि (f(x)=x-2 ) और (g(x)=2x) हैं, तो ((f+g)(a)) का मान क्या है?
If (f(x)=x-2 ) and (g(x)=2x), what is the value of ((f+g)(a))?
#functions
#parameter
#value
#medium
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A \(a^2+2a\)
B \(a^2+2x\)
C \(x^2+2a\)
D \(2a^2\)
Explanation opens after your attempt
Correct Answer
A. \(a^2+2a\)
Step 1
Concept
((f+g)(a)=f(a)+g(a)=a-2 +2a). When the variable changes, replace every (x) by that value.
Step 2
Why this answer is correct
The correct answer is A. \(a^2+2a\). ((f+g)(a)=f(a)+g(a)=a-2 +2a). When the variable changes, replace every (x) by that value.
Step 3
Exam Tip
((f+g)(a)=f(a)+g(a)=a-2 +2a)। चर बदलने पर हर (x) की जगह वही मान रखें।
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यदि (f(x)=x-2 -2x) और (g(x)=x-2 +2x) हैं, तो ((fg)(0)) क्या है?
If (f(x)=x-2 -2x) and (g(x)=x-2 +2x), what is ((fg)(0))?
#functions
#product
#value
#medium
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A (0)
B (4)
C (-4)
D (1)
Explanation opens after your attempt
Step 1
Concept
(f(0)=0) and (g(0)=0), so ((fg)(0)=0). If one value in a product is zero, the product is zero.
Step 2
Why this answer is correct
The correct answer is A. (0). (f(0)=0) and (g(0)=0), so ((fg)(0)=0). If one value in a product is zero, the product is zero.
Step 3
Exam Tip
(f(0)=0) और (g(0)=0), इसलिए ((fg)(0)=0)। उत्पाद में कोई एक मान शून्य हो तो उत्पाद शून्य होता है।
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यदि (f(x)=\frac{1}{x-1}) और (g(x)=x+3) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है?
If (f(x)=\frac{1}{x-1}) and (g(x)=x+3), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
#functions
#quotient
#domain
#medium
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A \(x \in \mathbb{R}, x \neq 1\)
B \(x \in \mathbb{R}, x \neq -3\)
C \(x \in \mathbb{R}, x \neq 1, x \neq -3\)
D (x>1)
Explanation opens after your attempt
Correct Answer
A. \(x \in \mathbb{R}, x \neq 1\)
Step 1
Concept
(f(x)) needs \(x \neq 1\), and (f(x)=\frac{1}{x-1}) is never zero. So the only restriction is \(x \neq 1\).
Step 2
Why this answer is correct
The correct answer is A. \(x \in \mathbb{R}, x \neq 1\). (f(x)) needs \(x \neq 1\), and (f(x)=\frac{1}{x-1}) is never zero. So the only restriction is \(x \neq 1\).
Step 3
Exam Tip
(f(x)) के लिए \(x \neq 1\) चाहिए और (f(x)=\frac{1}{x-1}) कभी शून्य नहीं होता। इसलिए केवल \(x \neq 1\) प्रतिबंध है।
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यदि (f(x)=x-2 -16) और (g(x)=x+4) हैं, तो \(x \neq -4\) के लिए (\left\(\frac{f}{g}\right\)(-1)) का मान क्या है?
If (f(x)=x-2 -16) and (g(x)=x+4), what is (\left\(\frac{f}{g}\right\)(-1)) for \(x \neq -4\)?
#functions
#quotient
#value
#medium
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A (3)
B (-5)
C (5)
D परिभाषित नहीं / Not defined
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{f}{g}\right\)(x)=\frac{(x-4)(x+4)}{x+4}=x-4), so at (x=-1) the value is (-5). Check directly as \(\frac{-15}{3}=-5\).
Step 2
Why this answer is correct
The correct answer is A. (3). (\left\(\frac{f}{g}\right\)(x)=\frac{(x-4)(x+4)}{x+4}=x-4), so at (x=-1) the value is (-5). Check directly as \(\frac{-15}{3}=-5\).
Step 3
Exam Tip
(\left\(\frac{f}{g}\right\)(x)=\frac{(x-4)(x+4)}{x+4}=x-4), इसलिए (x=-1) पर मान (-5) होना चाहिए; पर सही जाँच में \(\frac{-15}{3}=-5\)।
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यदि (f(x)=x-2 -16) और (g(x)=x+4) हैं, तो \(x \neq -4\) के लिए (\left\(\frac{f}{g}\right\)(-1)) का सही मान क्या है?
If (f(x)=x-2 -16) and (g(x)=x+4), what is the correct value of (\left\(\frac{f}{g}\right\)(-1)) for \(x \neq -4\)?
#functions
#quotient
#value
#medium
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A (-5)
B (3)
C (5)
D परिभाषित नहीं / Not defined
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 -16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4), hence (-1-4=-5). Substitute carefully after simplification.
Step 2
Why this answer is correct
The correct answer is A. (-5). (\frac{x-2 -16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4), hence (-1-4=-5). Substitute carefully after simplification.
Step 3
Exam Tip
(\frac{x-2 -16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4), अतः (-1-4=-5)। सरलीकरण के बाद सही मान रखें।
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यदि (f(x)=\sqrt{9-x-2 }) और (g(x)=x+1) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\sqrt{9-x-2 }) and (g(x)=x+1), what is the domain of ((f+g)(x))?
#functions
#domain
#square-root
#medium
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A \(-3 \leq x \leq 3\)
B \(x \geq 3\)
C \(x \leq -3\)
D \(x \in \mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(-3 \leq x \leq 3\)
Step 1
Concept
For the square root, \(9-x^2 \geq 0\), so \(-3 \leq x \leq 3\). The linear function gives no extra restriction.
Step 2
Why this answer is correct
The correct answer is A. \(-3 \leq x \leq 3\). For the square root, \(9-x^2 \geq 0\), so \(-3 \leq x \leq 3\). The linear function gives no extra restriction.
Step 3
Exam Tip
वर्गमूल के लिए \(9-x^2 \geq 0\), इसलिए \(-3 \leq x \leq 3\)। रैखिक फलन कोई अतिरिक्त प्रतिबंध नहीं देता।
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यदि (f(x)=2x-5) और (g(x)=5-2x) हैं, तो ((f+g)(x)) क्या होगा?
If (f(x)=2x-5) and (g(x)=5-2x), what is ((f+g)(x))?
#functions
#sum
#zero-function
#medium
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A (0)
B (4x-10)
C (10-4x)
D (2x)
Explanation opens after your attempt
Step 1
Concept
((f+g)(x)=2x-5+5-2x=0). Recognising opposite terms gives a quick solution.
Step 2
Why this answer is correct
The correct answer is A. (0). ((f+g)(x)=2x-5+5-2x=0). Recognising opposite terms gives a quick solution.
Step 3
Exam Tip
((f+g)(x)=2x-5+5-2x=0)। विपरीत पदों को पहचानना तेज समाधान देता है।
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यदि (f(x)=x-2 +3x+2) और (g(x)=x+1) हैं, तो \(x \neq -1\) के लिए (\left\(\frac{f}{g}\right\)(2)) क्या होगा?
If (f(x)=x-2 +3x+2) and (g(x)=x+1), what is (\left\(\frac{f}{g}\right\)(2)) for \(x \neq -1\)?
#functions
#quotient
#factorisation
#medium
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A (4)
B (6)
C (3)
D (12)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 +3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), so at (x=2) the value is (4). Factorisation saves time.
Step 2
Why this answer is correct
The correct answer is A. (4). (\frac{x-2 +3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), so at (x=2) the value is (4). Factorisation saves time.
Step 3
Exam Tip
(\frac{x-2 +3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2), इसलिए (x=2) पर मान (4) है। गुणनखंड बनाकर समय बचाएँ।
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यदि (f(x)=x-2 ) और (g(x)=x+2) हैं, तो ((f+g)(x)-(fg)(x)) का मान (x=1) पर क्या होगा?
If (f(x)=x-2 ) and (g(x)=x+2), what is the value of ((f+g)(x)-(fg)(x)) at (x=1)?
#functions
#mixed-operations
#value
#medium
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A (1)
B (3)
C (-1)
D (0)
Explanation opens after your attempt
Step 1
Concept
(f(1)=1) and (g(1)=3), so ((f+g)(1)-(fg)(1)=4-3=1). In such questions, find individual values first.
Step 2
Why this answer is correct
The correct answer is A. (1). (f(1)=1) and (g(1)=3), so ((f+g)(1)-(fg)(1)=4-3=1). In such questions, find individual values first.
Step 3
Exam Tip
(f(1)=1) और (g(1)=3), इसलिए ((f+g)(1)-(fg)(1)=4-3=1)। ऐसे प्रश्नों में पहले अलग-अलग मान निकालें।
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यदि (f(x)=x-2 +2) और (g(x)=2x) हैं, तो ((f-g)(x)=0) के हल क्या हैं?
If (f(x)=x-2 +2) and (g(x)=2x), what are the solutions of ((f-g)(x)=0)?
#functions
#equation
#quadratic
#medium
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A कोई वास्तविक हल नहीं / No real solution
B (x=1)
C (x=-1)
D (x=0)
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक हल नहीं / No real solution
Step 1
Concept
((f-g)(x)=x-2 -2x+2=(x-1)2 +1), which is never (0). Completing the square quickly shows real solutions.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक हल नहीं / No real solution. ((f-g)(x)=x-2 -2x+2=(x-1)2 +1), which is never (0). Completing the square quickly shows real solutions.
Step 3
Exam Tip
((f-g)(x)=x-2 -2x+2=(x-1)2 +1), जो कभी (0) नहीं होता। वर्ग पूर्ण करने से वास्तविक हल जल्दी दिखता है।
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यदि (f(x)=x-2 -1) और (g(x)=x-2 +1) हैं, तो ((g-f)(x)) का परिसर क्या है?
If (f(x)=x-2 -1) and (g(x)=x-2 +1), what is the range of ((g-f)(x))?
#functions
#range
#constant-function
#medium
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A ({2})
B ({0})
C \([0,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Step 1
Concept
((g-f)(x)=2), so its range is only ({2}). The range of a constant function is a singleton set.
Step 2
Why this answer is correct
The correct answer is A. ({2}). ((g-f)(x)=2), so its range is only ({2}). The range of a constant function is a singleton set.
Step 3
Exam Tip
((g-f)(x)=2), इसलिए इसका परिसर केवल ({2}) है। स्थिर फलन का परिसर एक-सदस्यीय होता है।
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यदि (f(x)=x+1) और (g(x)=x-2 ) हैं, तो ((fg)(x)) का घात क्या है?
If (f(x)=x+1) and (g(x)=x-2 ), what is the degree of ((fg)(x))?
#functions
#polynomial
#degree
#medium
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A (3)
B (2)
C (1)
D (4)
Explanation opens after your attempt
Step 1
Concept
((fg)(x)=(x+1)x-2 =x-3 +x-2 ), so the degree is (3). In polynomial products, check the highest power.
Step 2
Why this answer is correct
The correct answer is A. (3). ((fg)(x)=(x+1)x-2 =x-3 +x-2 ), so the degree is (3). In polynomial products, check the highest power.
Step 3
Exam Tip
((fg)(x)=(x+1)x-2 =x-3 +x-2 ), इसलिए घात (3) है। बहुपद गुणा में उच्चतम घात देखें।
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यदि (f(x)=x-2 -5x) और (g(x)=3x) हैं, तो ((f+g)(x)) के शून्यक कौन-से हैं?
If (f(x)=x-2 -5x) and (g(x)=3x), what are the zeroes of ((f+g)(x))?
#functions
#zeroes
#quadratic
#medium
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A (0) और (2) / (0) and (2)
B (0) और (5) / (0) and (5)
C (2) और (5) / (2) and (5)
D केवल (2) / Only (2)
Explanation opens after your attempt
Correct Answer
A. (0) और (2) / (0) and (2)
Step 1
Concept
((f+g)(x)=x-2 -2x=x(x-2)), so the zeroes are (0) and (2). For zeroes, equate the function to (0).
Step 2
Why this answer is correct
The correct answer is A. (0) और (2) / (0) and (2). ((f+g)(x)=x-2 -2x=x(x-2)), so the zeroes are (0) and (2). For zeroes, equate the function to (0).
Step 3
Exam Tip
((f+g)(x)=x-2 -2x=x(x-2)), इसलिए शून्यक (0) और (2) हैं। शून्यक के लिए फलन को (0) के बराबर रखें।
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यदि (f(x)=\frac{x-2}{x+1}) और (g(x)=\frac{x+3}{x+1}) हैं, तो ((f-g)(x)) क्या होगा?
If (f(x)=\frac{x-2}{x+1}) and (g(x)=\frac{x+3}{x+1}), what is ((f-g)(x))?
#functions
#rational
#subtraction
#medium
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A \(\frac{-5}{x+1}, x \neq -1\)
B \(\frac{5}{x+1}, x \neq -1\)
C \(\frac{2x+1}{x+1}, x \neq -1\)
D \(-5, x \neq -1\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{-5}{x+1}, x \neq -1\)
Step 1
Concept
With the same denominator, subtract numerators: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1}). Because of denominator (x+1), \(x \neq -1\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{-5}{x+1}, x \neq -1\). With the same denominator, subtract numerators: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1}). Because of denominator (x+1), \(x \neq -1\).
Step 3
Exam Tip
समान हर होने पर अंश घटाएँ: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1})। हर (x+1) के कारण \(x \neq -1\)।
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यदि (f(x)=2x+1) और (g(x)=x-2) हैं, तो ((3f+2g)(x)) क्या होगा?
If (f(x)=2x+1) and (g(x)=x-2), what is ((3f+2g)(x))?
#functions
#linear-combination
#medium
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A (8x-1)
B (8x+1)
C (5x-1)
D (4x+7)
Explanation opens after your attempt
Step 1
Concept
((3f+2g)(x)=3(2x+1)+2(x-2)=8x-1). Multiply the coefficient by the whole function.
Step 2
Why this answer is correct
The correct answer is A. (8x-1). ((3f+2g)(x)=3(2x+1)+2(x-2)=8x-1). Multiply the coefficient by the whole function.
Step 3
Exam Tip
((3f+2g)(x)=3(2x+1)+2(x-2)=8x-1)। गुणांक को पूरे फलन से गुणा करें।
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यदि (f(x)=x-2 +6x+9) और (g(x)=x+3) हैं, तो \(x \neq -3\) के लिए (\left\(\frac{f}{g}\right\)(0)) क्या है?
If (f(x)=x-2 +6x+9) and (g(x)=x+3), what is (\left\(\frac{f}{g}\right\)(0)) for \(x \neq -3\)?
#functions
#quotient
#perfect-square
#medium
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A (3)
B (0)
C (9)
D परिभाषित नहीं / Not defined
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 +6x+9}{x+3}=\frac{(x+3)2 }{x+3}=x+3), hence the value at (x=0) is (3). The perfect square identity is useful.
Step 2
Why this answer is correct
The correct answer is A. (3). (\frac{x-2 +6x+9}{x+3}=\frac{(x+3)2 }{x+3}=x+3), hence the value at (x=0) is (3). The perfect square identity is useful.
Step 3
Exam Tip
(\frac{x-2 +6x+9}{x+3}=\frac{(x+3)2 }{x+3}=x+3), अतः (x=0) पर मान (3) है। पूर्ण वर्ग पहचान उपयोगी है।
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यदि (f(x)=\frac{1}{\sqrt{x-1}}) और (g(x)=x) हैं, तो ((f+g)(x)) का प्रांत क्या है?
If (f(x)=\frac{1}{\sqrt{x-1}}) and (g(x)=x), what is the domain of ((f+g)(x))?
#functions
#domain
#radical-denominator
#medium
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A (x>1)
B \(x \geq 1\)
C \(x \in \mathbb{R}\)
D (x<1)
Explanation opens after your attempt
Step 1
Concept
The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).
Step 2
Why this answer is correct
The correct answer is A. (x>1). The denominator has \(\sqrt{x-1}\), so (x-1>0) is required. Hence the domain is (x>1).
Step 3
Exam Tip
हर में \(\sqrt{x-1}\) है, इसलिए (x-1>0) होना चाहिए। अतः प्रांत (x>1) है।
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यदि (f(x)=x-2 +1) और (g(x)=x-2 -1) हैं, तो ((f-g)(5)) क्या होगा?
If (f(x)=x-2 +1) and (g(x)=x-2 -1), what is ((f-g)(5))?
#functions
#subtraction
#value
#medium
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A (2)
B (10)
C (0)
D (50)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=\(x^2+1\)-\(x^2-1\)=2), so at any (x) the value is (2). Simplifying first reduces calculation.
Step 2
Why this answer is correct
The correct answer is A. (2). ((f-g)(x)=\(x^2+1\)-\(x^2-1\)=2), so at any (x) the value is (2). Simplifying first reduces calculation.
Step 3
Exam Tip
((f-g)(x)=\(x^2+1\)-\(x^2-1\)=2), इसलिए किसी भी (x) पर मान (2) है। पहले सरल करने से गणना कम होती है।
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यदि (f(x)=2x) और (g(x)=\frac{1}{2}x) हैं, तो ((fg)(x)) क्या है?
If (f(x)=2x) and (g(x)=\frac{1}{2}x), what is ((fg)(x))?
#functions
#product
#linear
#medium
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A \(x^2\)
B (x)
C \(2x^2\)
D \(\frac{x^2}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x^2\)
Step 1
Concept
((fg)(x)=2x\cdot \frac{1}{2}x=x-2 ). Multiply numerical coefficients first.
Step 2
Why this answer is correct
The correct answer is A. \(x^2\). ((fg)(x)=2x\cdot \frac{1}{2}x=x-2 ). Multiply numerical coefficients first.
Step 3
Exam Tip
((fg)(x)=2x\cdot \frac{1}{2}x=x-2 )। संख्यात्मक गुणांकों को पहले गुणा करें।
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यदि (f(x)=x-2 +x+1) और (g(x)=x-2 -x+1) हैं, तो ((f-g)(x)) क्या है?
If (f(x)=x-2 +x+1) and (g(x)=x-2 -x+1), what is ((f-g)(x))?
#functions
#subtraction
#polynomial
#medium
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A (2x)
B (-2x)
C \(2x^2+2\)
D (0)
Explanation opens after your attempt
Step 1
Concept
((f-g)(x)=x-2 +x+1-\(x^2-x+1\)=2x). Open brackets and change signs carefully.
Step 2
Why this answer is correct
The correct answer is A. (2x). ((f-g)(x)=x-2 +x+1-\(x^2-x+1\)=2x). Open brackets and change signs carefully.
Step 3
Exam Tip
((f-g)(x)=x-2 +x+1-\(x^2-x+1\)=2x)। कोष्ठक खोलकर चिह्न सावधानी से बदलें।
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यदि (f(x)=\frac{x+2}{x-3}) और (g(x)=x-3) हैं, तो ((fg)(4)) क्या होगा?
If (f(x)=\frac{x+2}{x-3}) and (g(x)=x-3), what is ((fg)(4))?
#functions
#product
#rational
#value
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A (6)
B (1)
C (4)
D परिभाषित नहीं / Not defined
Explanation opens after your attempt
Step 1
Concept
((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2), where \(x \neq 3\). At (x=4), the value is (6).
Step 2
Why this answer is correct
The correct answer is A. (6). ((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2), where \(x \neq 3\). At (x=4), the value is (6).
Step 3
Exam Tip
((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2), जहाँ \(x \neq 3\)। (x=4) पर मान (6) है।
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यदि (f(x)=\sqrt{x}) और (g(x)=\sqrt{x}) हैं, तो ((fg)(x)) और उसका प्रांत क्या है?
If (f(x)=\sqrt{x}) and (g(x)=\sqrt{x}), what are ((fg)(x)) and its domain?
#functions
#product
#square-root
#medium
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A \(x, x \geq 0\)
B \(x, x \in \mathbb{R}\)
C \(2\sqrt{x}, x \geq 0\)
D \(x^2, x \geq 0\)
Explanation opens after your attempt
Correct Answer
A. \(x, x \geq 0\)
Step 1
Concept
((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), but both square roots require \(x \geq 0\). The simplified form does not change the original domain.
Step 2
Why this answer is correct
The correct answer is A. \(x, x \geq 0\). ((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), but both square roots require \(x \geq 0\). The simplified form does not change the original domain.
Step 3
Exam Tip
((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), लेकिन दोनों वर्गमूलों के लिए \(x \geq 0\)। सरल रूप से प्रांत नहीं बदलता।
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यदि (f(x)=x-2 -3x+2) और (g(x)=x-1) हैं, तो \(x \neq 1\) के लिए (\left\(\frac{f}{g}\right\)(3)) क्या है?
If (f(x)=x-2 -3x+2) and (g(x)=x-1), what is (\left\(\frac{f}{g}\right\)(3)) for \(x \neq 1\)?
#functions
#quotient
#factorisation
#medium
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A (1)
B (2)
C (0)
D (3)
Explanation opens after your attempt
Step 1
Concept
(\frac{x-2 -3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2), so at (x=3) the value is (1). Factorisation makes division simple.
Step 2
Why this answer is correct
The correct answer is A. (1). (\frac{x-2 -3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2), so at (x=3) the value is (1). Factorisation makes division simple.
Step 3
Exam Tip
(\frac{x-2 -3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2), अतः (x=3) पर मान (1) है। गुणनखंडन से भाग सरल होता है।
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यदि (f(x)=x+4) और (g(x)=4-x) हैं, तो ((fg)(x)) किसके बराबर है?
If (f(x)=x+4) and (g(x)=4-x), what is ((fg)(x)) equal to?
#functions
#product
#identity
#medium
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A \(16-x^2\)
B \(x^2-16\)
C (8x)
D \(x^2+16\)
Explanation opens after your attempt
Correct Answer
A. \(16-x^2\)
Step 1
Concept
((fg)(x)=(x+4)(4-x)=16-x-2 ). View it as ((4+x)(4-x)).
Step 2
Why this answer is correct
The correct answer is A. \(16-x^2\). ((fg)(x)=(x+4)(4-x)=16-x-2 ). View it as ((4+x)(4-x)).
Step 3
Exam Tip
((fg)(x)=(x+4)(4-x)=16-x-2 )। इसे ((4+x)(4-x)) के रूप में देखें।
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