रेखा (y=2x-5) का (x)-अक्ष से प्रतिच्छेद बिंदु किसका निर्देशांक होगा?
What will be the intercept point of the line (y=2x-5) on the (x)-axis?
#graphs
#linear
#intercept
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A ( \left\(\frac{5}{2},0\right\) )
B ( \left\(0,-5\right\) )
C ( \left\(-\frac{5}{2},0\right\) )
D ( \left\(0,\frac{5}{2}\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(\frac{5}{2},0\right\) )
Step 1
Concept
On the (x)-axis (y=0) so (2x-5=0) gives \(x=\frac{5}{2}\). For an intercept set the other coordinate equal to zero.
Step 2
Why this answer is correct
The correct answer is A. ( \left\(\frac{5}{2},0\right\) ). On the (x)-axis (y=0) so (2x-5=0) gives \(x=\frac{5}{2}\). For an intercept set the other coordinate equal to zero.
Step 3
Exam Tip
(x)-अक्ष पर (y=0) होता है इसलिए (2x-5=0) से \(x=\frac{5}{2}\) मिलता है। ग्राफ में प्रतिच्छेद के लिए दूसरे निर्देशांक को शून्य रखें।
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फलन (y=|x+3|+|x-1|) का ग्राफ किस अंतराल पर क्षैतिज रहता है?
On which interval is the graph of (y=|x+3|+|x-1|) horizontal?
#graphs
#modulus
#piecewise
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A \(-3\le x\le 1\)
B (x<-3)
C (x>1)
D \(-1\le x\le 3\)
Explanation opens after your attempt
Correct Answer
A. \(-3\le x\le 1\)
Step 1
Concept
On \(-3\le x\le 1\), the sum ((x+3)+(1-x)=4) is constant. In such modulus sums the graph can be horizontal between the two points.
Step 2
Why this answer is correct
The correct answer is A. \(-3\le x\le 1\). On \(-3\le x\le 1\), the sum ((x+3)+(1-x)=4) is constant. In such modulus sums the graph can be horizontal between the two points.
Step 3
Exam Tip
\(-3\le x\le 1\) पर योग ((x+3)+(1-x)=4) स्थिर रहता है। ऐसे मापांक योग में दोनों बिंदुओं के बीच ग्राफ क्षैतिज हो सकता है।
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फलन (f(x)=|x-3|) के ग्राफ का शीर्ष बिंदु कौन सा है?
Which point is the vertex of the graph of (f(x)=|x-3|)?
#graphs
#modulus
#vertex
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A ( \left\(3,0\right\) )
B ( \left\(0,3\right\) )
C ( \left\(-3,0\right\) )
D ( \left\(0,-3\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(3,0\right\) )
Step 1
Concept
( |x-3| ) is minimum (0) when (x=3). For a modulus graph set the inside expression equal to zero.
Step 2
Why this answer is correct
The correct answer is A. ( \left\(3,0\right\) ). ( |x-3| ) is minimum (0) when (x=3). For a modulus graph set the inside expression equal to zero.
Step 3
Exam Tip
( |x-3| ) न्यूनतम (0) तब होता है जब (x=3)। मापांक ग्राफ का शीर्ष अंदर वाले भाग को शून्य करने से मिलता है।
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फलन \(y=\sqrt{x^2-9}\) के ग्राफ का डोमेन क्या है?
What is the domain of the graph of \(y=\sqrt{x^2-9}\)?
#graphs
#square-root
#domain
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A (\(-\infty,-3]\cup[3,\infty\))
B ([-3,3])
C (\(-\infty,3]\)
D \([0,\infty\))
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-3]\cup[3,\infty\))
Step 1
Concept
For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-3]\cup[3,\infty\)). For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-9\ge 0\) चाहिए। इससे \(x\le -3\) या \(x\ge 3\) मिलता है।
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परवलय (y=(x+2)2 -7) की सममिति अक्ष कौन सी है?
What is the axis of symmetry of the parabola (y=(x+2)2 -7)?
#graphs
#quadratic
#axis
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A (x=-2)
B (x=2)
C (y=-7)
D (y=7)
Explanation opens after your attempt
Step 1
Concept
In (y=(x-h)2 +k) the axis of symmetry is (x=h). Here (h=-2).
Step 2
Why this answer is correct
The correct answer is A. (x=-2). In (y=(x-h)2 +k) the axis of symmetry is (x=h). Here (h=-2).
Step 3
Exam Tip
(y=(x-h)2 +k) में सममिति अक्ष (x=h) होती है। यहां (h=-2) है।
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फलन \(y=\frac{1}{x^2-4}\) के ग्राफ के ऊर्ध्वाधर आसम्टोट कौन से हैं?
What are the vertical asymptotes of the graph of \(y=\frac{1}{x^2-4}\)?
#graphs
#rational
#asymptote
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A (x=-2) और (x=2) / (x=-2) and (x=2)
B (x=0) और (x=4) / (x=0) and (x=4)
C (y=-2) और (y=2) / (y=-2) and (y=2)
D (y=0)
Explanation opens after your attempt
Correct Answer
A. (x=-2) और (x=2) / (x=-2) and (x=2)
Step 1
Concept
Asymptotes are found by making the denominator zero. From \(x^2-4=0\), we get (x=-2,2).
Step 2
Why this answer is correct
The correct answer is A. (x=-2) और (x=2) / (x=-2) and (x=2). Asymptotes are found by making the denominator zero. From \(x^2-4=0\), we get (x=-2,2).
Step 3
Exam Tip
आसम्टोट हर को शून्य करने से मिलते हैं। \(x^2-4=0\) से (x=-2,2) मिलते हैं।
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फलन (y=-3(x-1)2 +4) का अधिकतम मान क्या है?
What is the maximum value of the function (y=-3(x-1)2 +4)?
#graphs
#quadratic
#maximum
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A (4)
B (1)
C (-3)
D (-4)
Explanation opens after your attempt
Step 1
Concept
The coefficient (-3) is negative so the parabola opens downward. At the vertex (y=4) is the maximum value.
Step 2
Why this answer is correct
The correct answer is A. (4). The coefficient (-3) is negative so the parabola opens downward. At the vertex (y=4) is the maximum value.
Step 3
Exam Tip
गुणांक (-3) ऋणात्मक है इसलिए परवलय नीचे खुलता है। शीर्ष पर (y=4) अधिकतम मान है।
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फलन (f(x)=\frac{x}{x-2 +1}) के ग्राफ की सममिति कौन सी है?
What is the symmetry of the graph of (f(x)=\frac{x}{x-2 +1})?
#graphs
#rational
#symmetry
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A मूल बिंदु के सापेक्ष सममिति / Symmetry about the origin
B (y)-अक्ष के सापेक्ष सममिति / Symmetry about the (y)-axis
C (x)-अक्ष के सापेक्ष सममिति / Symmetry about the (x)-axis
D कोई डोमेन नहीं / No domain
Explanation opens after your attempt
Correct Answer
A. मूल बिंदु के सापेक्ष सममिति / Symmetry about the origin
Step 1
Concept
(f(-x)=\frac{-x}{x-2 +1}=-f(x)). Therefore it is an odd function and its graph is symmetric about the origin.
Step 2
Why this answer is correct
The correct answer is A. मूल बिंदु के सापेक्ष सममिति / Symmetry about the origin. (f(-x)=\frac{-x}{x-2 +1}=-f(x)). Therefore it is an odd function and its graph is symmetric about the origin.
Step 3
Exam Tip
(f(-x)=\frac{-x}{x-2 +1}=-f(x)) है। इसलिए यह विषम फलन है और ग्राफ मूल बिंदु के सापेक्ष सममित है।
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फलन (f(x)=\sqrt{x+5}) के ग्राफ का डोमेन क्या है?
What is the domain of the graph of (f(x)=\sqrt{x+5})?
#graphs
#square-root
#domain
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A \( [-5,\infty\) )
B ( \(-5,\infty\) )
C ( \(-\infty,-5] \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [-5,\infty\) )
Step 1
Concept
For a square root \(x+5\ge 0\) must hold. Hence \(x\ge -5\).
Step 2
Why this answer is correct
The correct answer is A. \( [-5,\infty\) ). For a square root \(x+5\ge 0\) must hold. Hence \(x\ge -5\).
Step 3
Exam Tip
वर्गमूल के लिए \(x+5\ge 0\) होना चाहिए। इसलिए \(x\ge -5\) है।
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फलन (y=-|x+2|+5) के ग्राफ का शीर्ष बिंदु कौन सा है?
Which point is the vertex of the graph of (y=-|x+2|+5)?
#graphs
#modulus
#vertex
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A ((-2,5))
B ((2,5))
C ((-2,-5))
D ((5,-2))
Explanation opens after your attempt
Correct Answer
A. ((-2,5))
Step 1
Concept
The inside part of the modulus is zero at (x+2=0). Then (y=5), so the vertex is ((-2,5)).
Step 2
Why this answer is correct
The correct answer is A. ((-2,5)). The inside part of the modulus is zero at (x+2=0). Then (y=5), so the vertex is ((-2,5)).
Step 3
Exam Tip
मापांक का अंदरूनी भाग (x+2=0) पर शून्य होता है। तब (y=5) है इसलिए शीर्ष ((-2,5)) है।
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फलन \(y=\frac{1}{x-4}\) के ग्राफ का ऊर्ध्वाधर आसम्टोट कौन सा है?
What is the vertical asymptote of the graph of \(y=\frac{1}{x-4}\)?
#graphs
#rational
#asymptote
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A (x=4)
B (x=-4)
C (y=4)
D (y=0)
Explanation opens after your attempt
Step 1
Concept
The denominator (x-4) makes the function undefined when it is zero. So the vertical asymptote is (x=4).
Step 2
Why this answer is correct
The correct answer is A. (x=4). The denominator (x-4) makes the function undefined when it is zero. So the vertical asymptote is (x=4).
Step 3
Exam Tip
हर (x-4) शून्य होने पर फलन अपरिभाषित होता है। इसलिए ऊर्ध्वाधर आसम्टोट (x=4) है।
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ग्राफ (y=|x|) और (y=3-x) किस बिंदु पर मिलते हैं?
At which point do the graphs (y=|x|) and (y=3-x) meet?
#graphs
#intersection
#modulus
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A (\left\(\frac{3}{2},\frac{3}{2}\right\))
B ((-3,6))
C ((0,3))
D ((3,0))
Explanation opens after your attempt
Correct Answer
A. (\left\(\frac{3}{2},\frac{3}{2}\right\))
Step 1
Concept
For \(x\ge 0\), put (|x|=x), so (x=3-x) gives \(x=\frac{3}{2}\). Hence the intersection is (\left\(\frac{3}{2},\frac{3}{2}\right\)).
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{3}{2},\frac{3}{2}\right\)). For \(x\ge 0\), put (|x|=x), so (x=3-x) gives \(x=\frac{3}{2}\). Hence the intersection is (\left\(\frac{3}{2},\frac{3}{2}\right\)).
Step 3
Exam Tip
\(x\ge 0\) पर (|x|=x) रखकर (x=3-x) से \(x=\frac{3}{2}\) मिलता है। इसलिए प्रतिच्छेद (\left\(\frac{3}{2},\frac{3}{2}\right\)) है।
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फलन \(y=\frac{2}{x+1}-3\) के ग्राफ का क्षैतिज आसम्टोट कौन सा है?
What is the horizontal asymptote of the graph of \(y=\frac{2}{x+1}-3\)?
#graphs
#rational
#horizontal-asymptote
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A (y=-3)
B (x=-1)
C (y=2)
D (x=3)
Explanation opens after your attempt
Step 1
Concept
The term \(\frac{2}{x+1}\) approaches (0) for large (|x|). Hence the graph approaches (y=-3).
Step 2
Why this answer is correct
The correct answer is A. (y=-3). The term \(\frac{2}{x+1}\) approaches (0) for large (|x|). Hence the graph approaches (y=-3).
Step 3
Exam Tip
\(\frac{2}{x+1}\) का मान बड़े (|x|) पर (0) के पास जाता है। इसलिए ग्राफ (y=-3) के पास जाता है।
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फलन (f(x)=\lfloor x \rfloor) के ग्राफ में (x=2.7) पर (y) का मान क्या है?
For the graph of (f(x)=\lfloor x \rfloor), what is the value of (y) at (x=2.7)?
#graphs
#greatest-integer
#value
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A (2)
B (3)
C (2.7)
D (-2)
Explanation opens after your attempt
Step 1
Concept
The greatest integer function gives the largest integer less than or equal to (x). Hence \(\lfloor 2.7 \rfloor=2\).
Step 2
Why this answer is correct
The correct answer is A. (2). The greatest integer function gives the largest integer less than or equal to (x). Hence \(\lfloor 2.7 \rfloor=2\).
Step 3
Exam Tip
महत्तम पूर्णांक फलन (x) से कम या बराबर सबसे बड़ा पूर्णांक देता है। इसलिए \(\lfloor 2.7 \rfloor=2\)।
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फलन (f(x)=\lfloor x \rfloor) के ग्राफ में बिंदु (x=3) पर कौन सा कथन सही है?
Which statement is correct at (x=3) on the graph of (f(x)=\lfloor x \rfloor)?
#graphs
#greatest-integer
#discontinuity
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A ग्राफ में छलांग होती है / The graph has a jump
B ग्राफ हर जगह चिकना है / The graph is smooth everywhere
C ग्राफ (y=x) है / The graph is (y=x)
D ग्राफ केवल (x<0) के लिए है / The graph is only for (x<0)
Explanation opens after your attempt
Correct Answer
A. ग्राफ में छलांग होती है / The graph has a jump
Step 1
Concept
The greatest integer function changes suddenly at every integer. Therefore a jump occurs at (x=3).
Step 2
Why this answer is correct
The correct answer is A. ग्राफ में छलांग होती है / The graph has a jump. The greatest integer function changes suddenly at every integer. Therefore a jump occurs at (x=3).
Step 3
Exam Tip
महत्तम पूर्णांक फलन हर पूर्णांक पर अचानक बदलता है। इसलिए (x=3) पर छलांग मिलती है।
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फलन (f(x)=\operatorname{sgn}(x)) के ग्राफ में (x=-8) पर मान क्या है?
What is the value of the graph of (f(x)=\operatorname{sgn}(x)) at (x=-8)?
#graphs
#signum
#value
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A (-1)
B (0)
C (1)
D (8)
Explanation opens after your attempt
Step 1
Concept
The signum function gives (-1) for negative (x). Therefore (f(-8)=-1).
Step 2
Why this answer is correct
The correct answer is A. (-1). The signum function gives (-1) for negative (x). Therefore (f(-8)=-1).
Step 3
Exam Tip
साइनम फलन ऋणात्मक (x) पर (-1) देता है। इसलिए (f(-8)=-1)।
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फलन (f(x)=\operatorname{sgn}(x-2)) के ग्राफ में छलांग किस (x) पर होगी?
At which (x) will the graph of (f(x)=\operatorname{sgn}(x-2)) have a jump?
#graphs
#signum
#jump
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A (x=2)
B (x=0)
C (x=-2)
D (x=1)
Explanation opens after your attempt
Step 1
Concept
A signum graph changes when the inside expression becomes zero. From (x-2=0), we get (x=2).
Step 2
Why this answer is correct
The correct answer is A. (x=2). A signum graph changes when the inside expression becomes zero. From (x-2=0), we get (x=2).
Step 3
Exam Tip
साइनम ग्राफ अंदर की अभिव्यक्ति शून्य होने पर बदलता है। (x-2=0) से (x=2) मिलता है।
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ग्राफ \(y=x^3\) के बारे में कौन सा कथन सही है?
Which statement is correct about the graph of \(y=x^3\)?
#graphs
#cubic
#symmetry
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A यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin
B यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
C यह नीचे खुलने वाला परवलय है / It is a downward opening parabola
D इसका डोमेन \( [0,\infty\) ) है / Its domain is \( [0,\infty\) )
Explanation opens after your attempt
Correct Answer
A. यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin
Step 1
Concept
\(y=x^3\) is an odd function because (f(-x)=-f(x)). The graph of an odd function is symmetric about the origin.
Step 2
Why this answer is correct
The correct answer is A. यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin. \(y=x^3\) is an odd function because (f(-x)=-f(x)). The graph of an odd function is symmetric about the origin.
Step 3
Exam Tip
\(y=x^3\) विषम फलन है क्योंकि (f(-x)=-f(x))। विषम फलन का ग्राफ मूल बिंदु के सापेक्ष सममित होता है।
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ग्राफ \(y=x^2-6x+11\) का शीर्ष बिंदु कौन सा है?
What is the vertex of the graph \(y=x^2-6x+11\)?
#graphs
#quadratic
#vertex
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A ( \left\(3,2\right\) )
B ( \left\(-3,2\right\) )
C ( \left\(6,11\right\) )
D ( \left\(2,3\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(3,2\right\) )
Step 1
Concept
Completing the square gives (y=(x-3)2 +2). Therefore the vertex is (\left\(3,2\right\)).
Step 2
Why this answer is correct
The correct answer is A. ( \left\(3,2\right\) ). Completing the square gives (y=(x-3)2 +2). Therefore the vertex is (\left\(3,2\right\)).
Step 3
Exam Tip
पूर्ण वर्ग बनाने पर (y=(x-3)2 +2) मिलता है। इसलिए शीर्ष (\left\(3,2\right\)) है।
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फलन (y=|x+1|+|x-3|) का न्यूनतम मान क्या है?
What is the minimum value of (y=|x+1|+|x-3|)?
#graphs
#modulus
#minimum
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A (4)
B (2)
C (0)
D (8)
Explanation opens after your attempt
Step 1
Concept
It is the sum of distances from (-1) and (3). Between them the sum remains (4).
Step 2
Why this answer is correct
The correct answer is A. (4). It is the sum of distances from (-1) and (3). Between them the sum remains (4).
Step 3
Exam Tip
यह (-1) और (3) से दूरीयों का योग है। इनके बीच योग हमेशा (4) रहता है।
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फलन (y=|2x-6|) के ग्राफ का (x)-अक्ष से प्रतिच्छेद क्या है?
What is the (x)-intercept of the graph of (y=|2x-6|)?
#graphs
#modulus
#intercept
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A ( \left\(3,0\right\) )
B ( \left\(-3,0\right\) )
C ( \left\(0,6\right\) )
D ( \left\(6,0\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(3,0\right\) )
Step 1
Concept
( |2x-6|=0 ) only when (2x-6=0). Hence (x=3).
Step 2
Why this answer is correct
The correct answer is A. ( \left\(3,0\right\) ). ( |2x-6|=0 ) only when (2x-6=0). Hence (x=3).
Step 3
Exam Tip
( |2x-6|=0 ) तभी होगा जब (2x-6=0)। इसलिए (x=3) है।
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रेखा (3x+2y=12) का ढाल क्या है?
What is the slope of the line (3x+2y=12)?
#graphs
#linear
#slope
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A \(-\frac{3}{2}\)
B \(\frac{3}{2}\)
C (2)
D (3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{3}{2}\)
Step 1
Concept
Write the equation as \(y=-\frac{3}{2}x+6\). The coefficient of (x) is the slope.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). Write the equation as \(y=-\frac{3}{2}x+6\). The coefficient of (x) is the slope.
Step 3
Exam Tip
समीकरण को \(y=-\frac{3}{2}x+6\) लिखें। (x) का गुणांक ढाल होता है।
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फलन (y=5) के ग्राफ के लिए कौन सा कथन सही है?
Which statement is correct for the graph of (y=5)?
#graphs
#constant
#function
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A यह (x)-अक्ष के समानांतर रेखा है / It is a line parallel to the (x)-axis
B यह (y)-अक्ष के समानांतर रेखा है / It is a line parallel to the (y)-axis
C यह मूल बिंदु से गुजरता है / It passes through the origin
D यह परवलय है / It is a parabola
Explanation opens after your attempt
Correct Answer
A. यह (x)-अक्ष के समानांतर रेखा है / It is a line parallel to the (x)-axis
Step 1
Concept
The value of (y) is (5) for every (x). The graph of a constant function is a horizontal line.
Step 2
Why this answer is correct
The correct answer is A. यह (x)-अक्ष के समानांतर रेखा है / It is a line parallel to the (x)-axis. The value of (y) is (5) for every (x). The graph of a constant function is a horizontal line.
Step 3
Exam Tip
(y) का मान हर (x) के लिए (5) है। स्थिर फलन का ग्राफ क्षैतिज रेखा होता है।
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पहचान फलन (f(x)=x) के ग्राफ पर कौन सा बिंदु स्थित है?
Which point lies on the graph of the identity function (f(x)=x)?
#graphs
#identity
#point
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A ( \left\(-4,-4\right\) )
B ( \left\(-4,4\right\) )
C ( \left\(4,-4\right\) )
D ( \left\(0,4\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(-4,-4\right\) )
Step 1
Concept
In the identity function (y=x). Therefore both coordinates must be equal.
Step 2
Why this answer is correct
The correct answer is A. ( \left\(-4,-4\right\) ). In the identity function (y=x). Therefore both coordinates must be equal.
Step 3
Exam Tip
पहचान फलन में (y=x) होता है। इसलिए दोनों निर्देशांक बराबर होने चाहिए।
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फलन \(y=x^2\) और (y=2x+3) के ग्राफ किस (x) मान पर प्रतिच्छेद करते हैं?
For which (x) values do the graphs \(y=x^2\) and (y=2x+3) intersect?
#graphs
#intersection
#quadratic
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A (x=-1,3)
B (x=1,3)
C (x=-3,1)
D (x=0,3)
Explanation opens after your attempt
Correct Answer
A. (x=-1,3)
Step 1
Concept
For intersection set \(x^2=2x+3\). From \(x^2-2x-3=0\), (x=-1,3).
Step 2
Why this answer is correct
The correct answer is A. (x=-1,3). For intersection set \(x^2=2x+3\). From \(x^2-2x-3=0\), (x=-1,3).
Step 3
Exam Tip
प्रतिच्छेद के लिए \(x^2=2x+3\) रखें। \(x^2-2x-3=0\) से (x=-1,3) मिलते हैं।
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फलन \(y=x^2\) और (y=|x|) के ग्राफ किन (x) मानों पर मिलते हैं?
At which (x) values do the graphs \(y=x^2\) and (y=|x|) meet?
#graphs
#intersection
#modulus
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A (x=-1,0,1)
B (x=0,1)
C (x=-1,1)
D (x=-2,0,2)
Explanation opens after your attempt
Correct Answer
A. (x=-1,0,1)
Step 1
Concept
Putting \(x^2=|x|\) gives (x=0) or (|x|=1). Hence (x=-1,0,1).
Step 2
Why this answer is correct
The correct answer is A. (x=-1,0,1). Putting \(x^2=|x|\) gives (x=0) or (|x|=1). Hence (x=-1,0,1).
Step 3
Exam Tip
\(x^2=|x|\) रखने पर (x=0) या (|x|=1) मिलता है। इसलिए (x=-1,0,1) हैं।
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फलन (f(x)=\frac{x}{|x|}) का ग्राफ (x>0) के लिए किस रेखा पर है?
For (x>0), the graph of (f(x)=\frac{x}{|x|}) lies on which line?
#graphs
#piecewise
#signum
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A (y=1)
B (y=-1)
C (y=0)
D (y=x)
Explanation opens after your attempt
Step 1
Concept
For (x>0), (|x|=x). Hence \(\frac{x}{|x|}=1\).
Step 2
Why this answer is correct
The correct answer is A. (y=1). For (x>0), (|x|=x). Hence \(\frac{x}{|x|}=1\).
Step 3
Exam Tip
(x>0) पर (|x|=x) होता है। इसलिए \(\frac{x}{|x|}=1\)।
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फलन (y=\min(x,-1)) के ग्राफ में (x>-1) के लिए (y) क्या है?
For (x>-1), what is (y) in the graph of (y=\min(x,-1))?
#graphs
#piecewise
#minimum
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A (y=-1)
B (y=x)
C (y=1)
D (y=x-1)
Explanation opens after your attempt
Step 1
Concept
When (x>-1), the smaller value between (x) and (-1) is (-1). Hence the graph is (y=-1).
Step 2
Why this answer is correct
The correct answer is A. (y=-1). When (x>-1), the smaller value between (x) and (-1) is (-1). Hence the graph is (y=-1).
Step 3
Exam Tip
जब (x>-1) हो तो (x) और (-1) में छोटा मान (-1) है। इसलिए ग्राफ (y=-1) है।
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ग्राफ (y=|x-2|+1) की रेंज क्या है?
What is the range of the graph (y=|x-2|+1)?
#graphs
#modulus
#range
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A \( [1,\infty\) )
B \( [2,\infty\) )
C ( \(-\infty,1] \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [1,\infty\) )
Step 1
Concept
The minimum value of (|x-2|) is (0). Therefore the minimum value of (y) is (1).
Step 2
Why this answer is correct
The correct answer is A. \( [1,\infty\) ). The minimum value of (|x-2|) is (0). Therefore the minimum value of (y) is (1).
Step 3
Exam Tip
(|x-2|) का न्यूनतम मान (0) है। इसलिए (y) का न्यूनतम मान (1) है।
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ग्राफ (y=-(x+4)2 +9) की रेंज क्या है?
What is the range of the graph (y=-(x+4)2 +9)?
#graphs
#quadratic
#range
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A ( \(-\infty,9] \)
B \( [9,\infty\) )
C ( \(-\infty,-4] \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,9] \)
Step 1
Concept
The parabola opens downward and its vertex has (y=9). Hence the range is ( \(-\infty,9] \).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,9] \). The parabola opens downward and its vertex has (y=9). Hence the range is ( \(-\infty,9] \).
Step 3
Exam Tip
परवलय नीचे खुलता है और शीर्ष पर (y=9) है। इसलिए रेंज ( \(-\infty,9] \) है।
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फलन \(y=\sqrt{4-x}\) के ग्राफ का डोमेन क्या है?
What is the domain of the graph \(y=\sqrt{4-x}\)?
#graphs
#square-root
#domain
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A ( \(-\infty,4] \)
B \( [4,\infty\) )
C ( [0,4] )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,4] \)
Step 1
Concept
For the square root \(4-x\ge 0\) is required. Therefore \(x\le 4\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,4] \). For the square root \(4-x\ge 0\) is required. Therefore \(x\le 4\).
Step 3
Exam Tip
वर्गमूल के लिए \(4-x\ge 0\) चाहिए। इसलिए \(x\le 4\) है।
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ग्राफ \(y=\frac{x+2}{x-1}\) में कौन सा (x) मान डोमेन में नहीं है?
Which (x) value is not in the domain of the graph \(y=\frac{x+2}{x-1}\)?
#graphs
#rational
#domain
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A (1)
B (-2)
C (0)
D (2)
Explanation opens after your attempt
Step 1
Concept
The denominator (x-1) cannot be zero. From (x-1=0), (x=1) is excluded.
Step 2
Why this answer is correct
The correct answer is A. (1). The denominator (x-1) cannot be zero. From (x-1=0), (x=1) is excluded.
Step 3
Exam Tip
हर (x-1) शून्य नहीं हो सकता। (x-1=0) से (x=1) निषिद्ध है।
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फलन \(y=x^4\) के ग्राफ के बारे में कौन सा कथन सही है?
Which statement is correct about the graph of \(y=x^4\)?
#graphs
#power-function
#symmetry
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A यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
B यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin
C यह केवल \(x\ge 0\) पर परिभाषित है / It is defined only for \(x\ge 0\)
D यह सदैव घटता है / It is always decreasing
Explanation opens after your attempt
Correct Answer
A. यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
Step 1
Concept
(f(-x)=(-x)4 =x-4 =f(x)). Therefore it is an even function and is symmetric about the (y)-axis.
Step 2
Why this answer is correct
The correct answer is A. यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis. (f(-x)=(-x)4 =x-4 =f(x)). Therefore it is an even function and is symmetric about the (y)-axis.
Step 3
Exam Tip
(f(-x)=(-x)4 =x-4 =f(x)) है। इसलिए यह सम फलन है और (y)-अक्ष के सापेक्ष सममित है।
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फलन (y=-|x|) के ग्राफ की रेंज क्या है?
What is the range of the graph (y=-|x|)?
#graphs
#modulus
#range
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A ( \(-\infty,0] \)
B \( [0,\infty\) )
C \( \mathbb{R} \)
D ( [-1,1] )
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,0] \)
Step 1
Concept
Since \(|x|\ge 0\), we have \(-|x|\le 0\). The maximum value is (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,0] \). Since \(|x|\ge 0\), we have \(-|x|\le 0\). The maximum value is (0).
Step 3
Exam Tip
\(|x|\ge 0\) होने से \(-|x|\le 0\) होता है। अधिकतम मान (0) है।
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फलन (y=|x+2|-|x-2|) का (x>2) के लिए मान क्या होगा?
For (x>2), what is the value of (y=|x+2|-|x-2|)?
#graphs
#modulus
#piecewise
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A (4)
B (2x)
C (-4)
D (0)
Explanation opens after your attempt
Step 1
Concept
For (x>2), (|x+2|=x+2) and (|x-2|=x-2). Their difference is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). For (x>2), (|x+2|=x+2) and (|x-2|=x-2). Their difference is (4).
Step 3
Exam Tip
(x>2) पर (|x+2|=x+2) और (|x-2|=x-2) है। अंतर (4) मिलता है।
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फलन \(y=\frac{1}{x^2}\) के ग्राफ के बारे में कौन सा कथन सही है?
Which statement is correct about the graph of \(y=\frac{1}{x^2}\)?
#graphs
#rational
#symmetry
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A यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
B यह मूल बिंदु से गुजरता है / It passes through the origin
C यह (x=0) पर परिभाषित है / It is defined at (x=0)
D इसकी रेंज \( \mathbb{R} \) है / Its range is \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
Step 1
Concept
(f(-x)=\frac{1}{(-x)2 }=\frac{1}{x-2 }). Hence the graph is symmetric about the (y)-axis.
Step 2
Why this answer is correct
The correct answer is A. यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis. (f(-x)=\frac{1}{(-x)2 }=\frac{1}{x-2 }). Hence the graph is symmetric about the (y)-axis.
Step 3
Exam Tip
(f(-x)=\frac{1}{(-x)2 }=\frac{1}{x-2 }) है। इसलिए ग्राफ (y)-अक्ष के सापेक्ष सममित है।
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फलन \(y=\frac{1}{x}\) के ग्राफ के बारे में कौन सा कथन सही है?
Which statement is correct about the graph of \(y=\frac{1}{x}\)?
#graphs
#rational
#symmetry
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A यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin
B यह (y)-अक्ष के सापेक्ष सममित है / It is symmetric about the (y)-axis
C यह (x=0) पर मान (0) लेता है / It has value (0) at (x=0)
D यह केवल प्रथम चतुर्थांश में है / It lies only in the first quadrant
Explanation opens after your attempt
Correct Answer
A. यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin
Step 1
Concept
(f(-x)=-\frac{1}{x}=-f(x)). Therefore it is an odd function and symmetric about the origin.
Step 2
Why this answer is correct
The correct answer is A. यह मूल बिंदु के सापेक्ष सममित है / It is symmetric about the origin. (f(-x)=-\frac{1}{x}=-f(x)). Therefore it is an odd function and symmetric about the origin.
Step 3
Exam Tip
(f(-x)=-\frac{1}{x}=-f(x)) है। इसलिए यह विषम फलन है और मूल बिंदु के सापेक्ष सममित है।
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फलन \(y=x^2-4\) का ग्राफ (x)-अक्ष को किन बिंदुओं पर काटता है?
At which points does the graph of \(y=x^2-4\) cut the (x)-axis?
#graphs
#quadratic
#zeros
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A ( \left\(-2,0\right\),\left\(2,0\right\) )
B ( \left\(0,-2\right\),\left\(0,2\right\) )
C ( \left\(4,0\right\),\left\(-4,0\right\) )
D ( \left\(0,-4\right\) )
Explanation opens after your attempt
Correct Answer
A. ( \left\(-2,0\right\),\left\(2,0\right\) )
Step 1
Concept
Set (y=0) on the (x)-axis. From \(x^2-4=0\), \(x=\pm 2\).
Step 2
Why this answer is correct
The correct answer is A. ( \left\(-2,0\right\),\left\(2,0\right\) ). Set (y=0) on the (x)-axis. From \(x^2-4=0\), \(x=\pm 2\).
Step 3
Exam Tip
(x)-अक्ष पर (y=0) रखें। \(x^2-4=0\) से \(x=\pm 2\) मिलता है।
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फलन (y=(x-5)2 ) का ग्राफ \(y=x^2\) से कैसे प्राप्त होगा?
How is the graph of (y=(x-5)2 ) obtained from \(y=x^2\)?
#graphs
#transformation
#quadratic
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A (5) इकाई दाईं ओर खिसकाकर / By shifting (5) units right
B (5) इकाई बाईं ओर खिसकाकर / By shifting (5) units left
C (5) इकाई ऊपर खिसकाकर / By shifting (5) units up
D (5) इकाई नीचे खिसकाकर / By shifting (5) units down
Explanation opens after your attempt
Correct Answer
A. (5) इकाई दाईं ओर खिसकाकर / By shifting (5) units right
Step 1
Concept
Replacing (x) by (x-5) shifts the graph (5) units right. In horizontal shift read the sign oppositely.
Step 2
Why this answer is correct
The correct answer is A. (5) इकाई दाईं ओर खिसकाकर / By shifting (5) units right. Replacing (x) by (x-5) shifts the graph (5) units right. In horizontal shift read the sign oppositely.
Step 3
Exam Tip
(x) की जगह (x-5) आने पर ग्राफ दाईं ओर (5) इकाई खिसकता है। क्षैतिज खिसकाव में संकेत उल्टा पढ़ें।
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फलन (y=(x+3)2 ) का ग्राफ \(y=x^2\) से कैसे प्राप्त होगा?
How is the graph of (y=(x+3)2 ) obtained from \(y=x^2\)?
#graphs
#transformation
#quadratic
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A (3) इकाई बाईं ओर खिसकाकर / By shifting (3) units left
B (3) इकाई दाईं ओर खिसकाकर / By shifting (3) units right
C (3) इकाई ऊपर खिसकाकर / By shifting (3) units up
D (3) इकाई नीचे खिसकाकर / By shifting (3) units down
Explanation opens after your attempt
Correct Answer
A. (3) इकाई बाईं ओर खिसकाकर / By shifting (3) units left
Step 1
Concept
(x+3) means (x-(-3)). Therefore the graph shifts (3) units left.
Step 2
Why this answer is correct
The correct answer is A. (3) इकाई बाईं ओर खिसकाकर / By shifting (3) units left. (x+3) means (x-(-3)). Therefore the graph shifts (3) units left.
Step 3
Exam Tip
(x+3) का अर्थ (x-(-3)) है। इसलिए ग्राफ बाईं ओर (3) इकाई खिसकता है।
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फलन \(y=x^2+6\) का ग्राफ \(y=x^2\) से कैसे प्राप्त होगा?
How is the graph of \(y=x^2+6\) obtained from \(y=x^2\)?
#graphs
#transformation
#vertical-shift
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A (6) इकाई ऊपर खिसकाकर / By shifting (6) units up
B (6) इकाई नीचे खिसकाकर / By shifting (6) units down
C (6) इकाई दाईं ओर खिसकाकर / By shifting (6) units right
D (6) इकाई बाईं ओर खिसकाकर / By shifting (6) units left
Explanation opens after your attempt
Correct Answer
A. (6) इकाई ऊपर खिसकाकर / By shifting (6) units up
Step 1
Concept
Adding (+6) outside the function increases every (y)-value by (6). Hence the graph moves up.
Step 2
Why this answer is correct
The correct answer is A. (6) इकाई ऊपर खिसकाकर / By shifting (6) units up. Adding (+6) outside the function increases every (y)-value by (6). Hence the graph moves up.
Step 3
Exam Tip
फलन के बाहर (+6) जुड़ने से सभी (y)-मान (6) बढ़ते हैं। इसलिए ग्राफ ऊपर जाता है।
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फलन \(y=-\sqrt{x}\) का ग्राफ \(y=\sqrt{x}\) से कैसे संबंधित है?
How is the graph of \(y=-\sqrt{x}\) related to \(y=\sqrt{x}\)?
#graphs
#transformation
#reflection
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A (x)-अक्ष में परावर्तन / Reflection in the (x)-axis
B (y)-अक्ष में परावर्तन / Reflection in the (y)-axis
C (1) इकाई ऊपर खिसकाव / Shift (1) unit up
D (1) इकाई दाईं ओर खिसकाव / Shift (1) unit right
Explanation opens after your attempt
Correct Answer
A. (x)-अक्ष में परावर्तन / Reflection in the (x)-axis
Step 1
Concept
A negative sign outside the function makes every (y)-value negative. So it is a reflection in the (x)-axis.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष में परावर्तन / Reflection in the (x)-axis. A negative sign outside the function makes every (y)-value negative. So it is a reflection in the (x)-axis.
Step 3
Exam Tip
फलन के बाहर ऋण चिह्न सभी (y)-मानों को ऋणात्मक कर देता है। इसलिए (x)-अक्ष में परावर्तन होता है।
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फलन (y=2|x|) का ग्राफ (y=|x|) से कैसे बदलेगा?
How does the graph of (y=2|x|) change from (y=|x|)?
#graphs
#transformation
#modulus
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A ऊर्ध्वाधर खिंचाव (2) गुना / Vertical stretch by factor (2)
B क्षैतिज खिंचाव (2) गुना / Horizontal stretch by factor (2)
C दाईं ओर (2) इकाई खिसकाव / Shift (2) units right
D नीचे (2) इकाई खिसकाव / Shift (2) units down
Explanation opens after your attempt
Correct Answer
A. ऊर्ध्वाधर खिंचाव (2) गुना / Vertical stretch by factor (2)
Step 1
Concept
Multiplying outside by (2) doubles all (y)-values. So it is a vertical stretch.
Step 2
Why this answer is correct
The correct answer is A. ऊर्ध्वाधर खिंचाव (2) गुना / Vertical stretch by factor (2). Multiplying outside by (2) doubles all (y)-values. So it is a vertical stretch.
Step 3
Exam Tip
बाहर (2) से गुणा होने पर सभी (y)-मान दुगुने होते हैं। इसलिए ऊर्ध्वाधर खिंचाव होता है।
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फलन (y=|2x|) का ग्राफ (y=|x|) से कैसे बदलेगा?
How does the graph of (y=|2x|) change from (y=|x|)?
#graphs
#transformation
#compression
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A क्षैतिज संपीड़न (2) गुना / Horizontal compression by factor (2)
B ऊर्ध्वाधर संपीड़न (2) गुना / Vertical compression by factor (2)
C ऊपर (2) इकाई खिसकाव / Shift (2) units up
D बाईं ओर (2) इकाई खिसकाव / Shift (2) units left
Explanation opens after your attempt
Correct Answer
A. क्षैतिज संपीड़न (2) गुना / Horizontal compression by factor (2)
Step 1
Concept
Having (2x) inside makes the graph shrink toward the (y)-axis. This is horizontal compression.
Step 2
Why this answer is correct
The correct answer is A. क्षैतिज संपीड़न (2) गुना / Horizontal compression by factor (2). Having (2x) inside makes the graph shrink toward the (y)-axis. This is horizontal compression.
Step 3
Exam Tip
अंदर (2x) होने से ग्राफ (y)-अक्ष की ओर सिमटता है। यह क्षैतिज संपीड़न है।
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ग्राफ (y=|x-1|+|x+1|) की रेंज क्या है?
What is the range of the graph (y=|x-1|+|x+1|)?
#graphs
#modulus
#range
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A \( [2,\infty\) )
B \( [0,\infty\) )
C ( \(-\infty,2] \)
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [2,\infty\) )
Step 1
Concept
It is the sum of distances from (1) and (-1). The minimum sum of distances is (2).
Step 2
Why this answer is correct
The correct answer is A. \( [2,\infty\) ). It is the sum of distances from (1) and (-1). The minimum sum of distances is (2).
Step 3
Exam Tip
यह (1) और (-1) से दूरीयों का योग है। न्यूनतम दूरी योग (2) है।
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फलन \(y=x^3-1\) का ग्राफ \(y=x^3\) से कैसे प्राप्त होगा?
How is the graph of \(y=x^3-1\) obtained from \(y=x^3\)?
#graphs
#transformation
#cubic
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A (1) इकाई नीचे खिसकाकर / By shifting (1) unit down
B (1) इकाई ऊपर खिसकाकर / By shifting (1) unit up
C (1) इकाई दाईं ओर खिसकाकर / By shifting (1) unit right
D (1) इकाई बाईं ओर खिसकाकर / By shifting (1) unit left
Explanation opens after your attempt
Correct Answer
A. (1) इकाई नीचे खिसकाकर / By shifting (1) unit down
Step 1
Concept
The outside (-1) decreases every (y)-value by (1). Hence the graph shifts down.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई नीचे खिसकाकर / By shifting (1) unit down. The outside (-1) decreases every (y)-value by (1). Hence the graph shifts down.
Step 3
Exam Tip
फलन के बाहर (-1) सभी (y)-मानों को (1) घटाता है। इसलिए ग्राफ नीचे खिसकता है।
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फलन (y=(x-2)3 ) का ग्राफ \(y=x^3\) से कैसे प्राप्त होगा?
How is the graph of (y=(x-2)3 ) obtained from \(y=x^3\)?
#graphs
#transformation
#cubic
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A (2) इकाई दाईं ओर खिसकाकर / By shifting (2) units right
B (2) इकाई बाईं ओर खिसकाकर / By shifting (2) units left
C (2) इकाई ऊपर खिसकाकर / By shifting (2) units up
D (2) इकाई नीचे खिसकाकर / By shifting (2) units down
Explanation opens after your attempt
Correct Answer
A. (2) इकाई दाईं ओर खिसकाकर / By shifting (2) units right
Step 1
Concept
Replacing (x) by (x-2) moves the graph (2) units right. The same rule applies to the cubic function.
Step 2
Why this answer is correct
The correct answer is A. (2) इकाई दाईं ओर खिसकाकर / By shifting (2) units right. Replacing (x) by (x-2) moves the graph (2) units right. The same rule applies to the cubic function.
Step 3
Exam Tip
(x) की जगह (x-2) आने से ग्राफ दाईं ओर (2) इकाई जाता है। घन फलन में भी यही नियम लागू होता है।
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फलन \(y=\frac{1}{|x|}\) के ग्राफ की रेंज क्या है?
What is the range of the graph \(y=\frac{1}{|x|}\)?
#graphs
#rational
#range
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A ( \(0,\infty\) )
B \( [0,\infty\) )
C \( \mathbb{R} \setminus {0} \)
D ( \(-\infty,0\) )
Explanation opens after your attempt
Correct Answer
A. ( \(0,\infty\) )
Step 1
Concept
(|x|>0) when \(x\ne 0\), and \(\frac{1}{|x|}\) is always positive. It never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\infty\) ). (|x|>0) when \(x\ne 0\), and \(\frac{1}{|x|}\) is always positive. It never becomes (0).
Step 3
Exam Tip
(|x|>0) जब \(x\ne 0\) होता है और \(\frac{1}{|x|}\) हमेशा धनात्मक है। यह (0) को कभी नहीं लेता।
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ग्राफ \(y=x^2\) और रेखा (y=4) से घिरा क्षेत्र किन (x) मानों के बीच है?
Between which (x) values is the region bounded by \(y=x^2\) and the line (y=4)?
#graphs
#intersection
#bounded-region
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A \(-2\le x\le 2\)
B \(0\le x\le 4\)
C \(x\le -2\)
D \(x\ge 2\)
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Correct Answer
A. \(-2\le x\le 2\)
Step 1
Concept
The boundaries come from intersections. From \(x^2=4\), (x=-2,2).
Step 2
Why this answer is correct
The correct answer is A. \(-2\le x\le 2\). The boundaries come from intersections. From \(x^2=4\), (x=-2,2).
Step 3
Exam Tip
सीमाएं प्रतिच्छेद से मिलती हैं। \(x^2=4\) से (x=-2,2) मिलते हैं।
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फलन (y=|x|+|x-4|) का ग्राफ किस अंतराल पर क्षैतिज रहता है?
On which interval is the graph of (y=|x|+|x-4|) horizontal?
#graphs
#modulus
#piecewise
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A \(0\le x\le 4\)
B (x<0)
C (x>4)
D \(-4\le x\le 0\)
Explanation opens after your attempt
Correct Answer
A. \(0\le x\le 4\)
Step 1
Concept
For \(0\le x\le 4\), the sum is (x+(4-x)=4), which is constant. Hence the graph is horizontal.
Step 2
Why this answer is correct
The correct answer is A. \(0\le x\le 4\). For \(0\le x\le 4\), the sum is (x+(4-x)=4), which is constant. Hence the graph is horizontal.
Step 3
Exam Tip
\(0\le x\le 4\) पर योग (x+(4-x)=4) स्थिर है। इसलिए ग्राफ क्षैतिज रहता है।
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