फलन (y=|x+2|+|x-4|) का न्यूनतम मान क्या है?
What is the minimum value of (y=|x+2|+|x-4|)?
#graphs
#modulus
#minimum
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A (6)
B (2)
C (4)
D (0)
Explanation opens after your attempt
Step 1
Concept
It is the sum of distances from (-2) and (4). Between the two points the minimum sum remains (6).
Step 2
Why this answer is correct
The correct answer is A. (6). It is the sum of distances from (-2) and (4). Between the two points the minimum sum remains (6).
Step 3
Exam Tip
यह (-2) और (4) से दूरियों का योग है। दोनों बिंदुओं के बीच न्यूनतम योग (6) रहता है।
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फलन \(y=\frac{x^2-16}{x+4}\) के ग्राफ में (x=-4) पर क्या होता है?
What happens at (x=-4) in the graph of \(y=\frac{x^2-16}{x+4}\)?
#graphs
#rational
#hole
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A छिद्र बनता है / A hole occurs
B ऊर्ध्वाधर आसम्टोट बनता है / A vertical asymptote occurs
C शीर्ष बनता है / A vertex occurs
D ग्राफ (x)-अक्ष बन जाता है / The graph becomes the (x)-axis
Explanation opens after your attempt
Correct Answer
A. छिद्र बनता है / A hole occurs
Step 1
Concept
\(\frac{x^2-16}{x+4}=x-4\), but (x=-4) is not in the domain. Therefore the line-like graph has a hole at (x=-4).
Step 2
Why this answer is correct
The correct answer is A. छिद्र बनता है / A hole occurs. \(\frac{x^2-16}{x+4}=x-4\), but (x=-4) is not in the domain. Therefore the line-like graph has a hole at (x=-4).
Step 3
Exam Tip
\(\frac{x^2-16}{x+4}=x-4\) है लेकिन (x=-4) डोमेन में नहीं है। इसलिए रेखा जैसे ग्राफ में (x=-4) पर छिद्र बनता है।
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फलन (f(x)=x-\lfloor x \rfloor) के ग्राफ की रेंज क्या है?
What is the range of the graph of (f(x)=x-\lfloor x \rfloor)?
#graphs
#fractional-part
#range
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A ([0,1))
B ((0,1])
C ([0,1])
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. ([0,1))
Step 1
Concept
It gives the fractional part of (x), which starts at (0) and stays less than (1). Hence the range is ([0,1)).
Step 2
Why this answer is correct
The correct answer is A. ([0,1)). It gives the fractional part of (x), which starts at (0) and stays less than (1). Hence the range is ([0,1)).
Step 3
Exam Tip
यह (x) का भिन्नात्मक भाग देता है जो (0) से शुरू होकर (1) से कम रहता है। इसलिए रेंज ([0,1)) है।
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फलन (y=|x-4|+|x-10|) किस अंतराल पर स्थिर रहता है?
On which interval is (y=|x-4|+|x-10|) constant?
#graphs
#modulus
#piecewise
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A \(4\le x\le 10\)
B (x<4)
C (x>10)
D \(0\le x\le 4\)
Explanation opens after your attempt
Correct Answer
A. \(4\le x\le 10\)
Step 1
Concept
On \(4\le x\le 10\), the sum is ((x-4)+(10-x)=6). In a modulus distance sum the graph is horizontal between the two points.
Step 2
Why this answer is correct
The correct answer is A. \(4\le x\le 10\). On \(4\le x\le 10\), the sum is ((x-4)+(10-x)=6). In a modulus distance sum the graph is horizontal between the two points.
Step 3
Exam Tip
\(4\le x\le 10\) पर योग ((x-4)+(10-x)=6) है। मापांक दूरी योग में दोनों बिंदुओं के बीच ग्राफ क्षैतिज होता है।
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फलन (y=2-|x+5|) की रेंज क्या है?
What is the range of (y=2-|x+5|)?
#graphs
#modulus
#range
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A (\(-\infty,2]\)
B \([2,\infty\))
C \([-5,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,2]\)
Step 1
Concept
Since \(|x+5|\ge 0\), \(2-|x+5|\le 2\). An inverted modulus graph gives a maximum at the vertex.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2]\). Since \(|x+5|\ge 0\), \(2-|x+5|\le 2\). An inverted modulus graph gives a maximum at the vertex.
Step 3
Exam Tip
\(|x+5|\ge 0\) है इसलिए \(2-|x+5|\le 2\) होगा। उल्टा मापांक ग्राफ शीर्ष पर अधिकतम देता है।
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ग्राफ (y=3|x-2|+1) का शीर्ष बिंदु कौन सा है?
Which point is the vertex of (y=3|x-2|+1)?
#graphs
#modulus
#vertex
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A ((2,1))
B ((-2,1))
C ((2,3))
D ((1,2))
Explanation opens after your attempt
Correct Answer
A. ((2,1))
Step 1
Concept
The inside of the modulus is zero when (x-2=0). Then (y=1), so the vertex is ((2,1)).
Step 2
Why this answer is correct
The correct answer is A. ((2,1)). The inside of the modulus is zero when (x-2=0). Then (y=1), so the vertex is ((2,1)).
Step 3
Exam Tip
मापांक का अंदरूनी भाग (x-2=0) पर शून्य होता है। तब (y=1) है इसलिए शीर्ष ((2,1)) है।
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फलन (y=|2x+6|-4) का (x)-अक्ष से प्रतिच्छेद किन (x) मानों पर है?
At which (x) values does (y=|2x+6|-4) intersect the (x)-axis?
#graphs
#modulus
#intercepts
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A (x=-5,-1)
B (x=-3,3)
C (x=1,5)
D (x=-4,4)
Explanation opens after your attempt
Correct Answer
A. (x=-5,-1)
Step 1
Concept
From ( |2x+6|-4=0 ), we get (|2x+6|=4). Thus \(2x+6=\pm4\) gives (x=-5,-1).
Step 2
Why this answer is correct
The correct answer is A. (x=-5,-1). From ( |2x+6|-4=0 ), we get (|2x+6|=4). Thus \(2x+6=\pm4\) gives (x=-5,-1).
Step 3
Exam Tip
( |2x+6|-4=0 ) से (|2x+6|=4) मिलता है। इसलिए \(2x+6=\pm4\) से (x=-5,-1) है।
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फलन \(y=|x^2-1|\) (x)-अक्ष को किन बिंदुओं पर काटता है?
At which points does \(y=|x^2-1|\) cut the (x)-axis?
#graphs
#modulus
#zeros
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A ((-1,0)) और ((1,0)) / ((-1,0)) and ((1,0))
B ((0,-1)) और ((0,1)) / ((0,-1)) and ((0,1))
C ((-1,1)) और ((1,1)) / ((-1,1)) and ((1,1))
D ((0,0))
Explanation opens after your attempt
Correct Answer
A. ((-1,0)) और ((1,0)) / ((-1,0)) and ((1,0))
Step 1
Concept
A modulus is zero only when \(x^2-1=0\). This gives \(x=\pm1\).
Step 2
Why this answer is correct
The correct answer is A. ((-1,0)) और ((1,0)) / ((-1,0)) and ((1,0)). A modulus is zero only when \(x^2-1=0\). This gives \(x=\pm1\).
Step 3
Exam Tip
मापांक शून्य तभी होता है जब \(x^2-1=0\) हो। इससे \(x=\pm1\) मिलता है।
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फलन (y=|x+1|-|x-5|) का (x>5) के लिए मान क्या है?
For (x>5), what is the value of (y=|x+1|-|x-5|)?
#graphs
#modulus
#piecewise
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A (6)
B (-6)
C (2x-4)
D (4-2x)
Explanation opens after your attempt
Step 1
Concept
For (x>5), (|x+1|=x+1) and (|x-5|=x-5). Hence (y=(x+1)-(x-5)=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For (x>5), (|x+1|=x+1) and (|x-5|=x-5). Hence (y=(x+1)-(x-5)=6).
Step 3
Exam Tip
(x>5) पर (|x+1|=x+1) और (|x-5|=x-5) होता है। इसलिए (y=(x+1)-(x-5)=6) है।
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परवलय \(y=2x^2-12x+13\) का शीर्ष बिंदु कौन सा है?
What is the vertex of the parabola \(y=2x^2-12x+13\)?
#graphs
#quadratic
#vertex
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A ((3,-5))
B ((-3,-5))
C ((6,13))
D ((3,5))
Explanation opens after your attempt
Correct Answer
A. ((3,-5))
Step 1
Concept
Here \(x=\frac{-b}{2a}=\frac{12}{4}=3\). Substituting (x=3) gives (y=-5).
Step 2
Why this answer is correct
The correct answer is A. ((3,-5)). Here \(x=\frac{-b}{2a}=\frac{12}{4}=3\). Substituting (x=3) gives (y=-5).
Step 3
Exam Tip
\(x=\frac{-b}{2a}=\frac{12}{4}=3\) है। (x=3) रखने पर (y=-5) मिलता है।
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फलन \(y=-x^2+4x+5\) का अधिकतम मान क्या है?
What is the maximum value of \(y=-x^2+4x+5\)?
#graphs
#quadratic
#maximum
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A (9)
B (5)
C (4)
D (1)
Explanation opens after your attempt
Step 1
Concept
Completing the square gives (y=-(x-2)2 +9). Hence the maximum value is (9).
Step 2
Why this answer is correct
The correct answer is A. (9). Completing the square gives (y=-(x-2)2 +9). Hence the maximum value is (9).
Step 3
Exam Tip
पूर्ण वर्ग से (y=-(x-2)2 +9) मिलता है। इसलिए अधिकतम मान (9) है।
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परवलय (y=3(x+2)2 -12) की रेंज क्या है?
What is the range of (y=3(x+2)2 -12)?
#graphs
#quadratic
#range
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A \([-12,\infty\))
B (\(-\infty,-12]\)
C \([3,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([-12,\infty\))
Step 1
Concept
Since (3(x+2)2 \ge 0), we have \(y\ge -12\).
Step 2
Why this answer is correct
The correct answer is A. \([-12,\infty\)). Since (3(x+2)2 \ge 0), we have \(y\ge -12\).
Step 3
Exam Tip
(3(x+2)2 \ge 0) होता है। इसलिए \(y\ge -12\) है।
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ग्राफ \(y=x^2+2x-8\) (x)-अक्ष को किन (x) मानों पर काटता है?
At which (x) values does \(y=x^2+2x-8\) cut the (x)-axis?
#graphs
#quadratic
#zeros
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A (x=-4,2)
B (x=4,-2)
C (x=-1,8)
D (x=0,-8)
Explanation opens after your attempt
Correct Answer
A. (x=-4,2)
Step 1
Concept
Set (y=0) on the (x)-axis. From \(x^2+2x-8=0\), (x=-4,2).
Step 2
Why this answer is correct
The correct answer is A. (x=-4,2). Set (y=0) on the (x)-axis. From \(x^2+2x-8=0\), (x=-4,2).
Step 3
Exam Tip
(x)-अक्ष पर (y=0) रखें। \(x^2+2x-8=0\) से (x=-4,2) मिलता है।
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ग्राफ (y=(x-3)2 +5) (x)-अक्ष को कितनी बार काटता है?
How many times does (y=(x-3)2 +5) cut the (x)-axis?
#graphs
#quadratic
#intercepts
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A (0) बार / (0) times
B (1) बार / (1) time
C (2) बार / (2) times
D (3) बार / (3) times
Explanation opens after your attempt
Correct Answer
A. (0) बार / (0) times
Step 1
Concept
Since ((x-3)2 \ge 0), \(y\ge 5\). The graph does not reach the (x)-axis.
Step 2
Why this answer is correct
The correct answer is A. (0) बार / (0) times. Since ((x-3)2 \ge 0), \(y\ge 5\). The graph does not reach the (x)-axis.
Step 3
Exam Tip
((x-3)2 \ge 0) इसलिए \(y\ge 5\) है। ग्राफ (x)-अक्ष तक नहीं पहुंचता।
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रेखा (4x-3y=15) की ढाल क्या है?
What is the slope of the line (4x-3y=15)?
#graphs
#linear
#slope
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A \(\frac{4}{3}\)
B \(-\frac{4}{3}\)
C \(\frac{3}{4}\)
D \(-\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4}{3}\)
Step 1
Concept
Write the equation as \(y=\frac{4}{3}x-5\). The coefficient of (x) is the slope.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{3}\). Write the equation as \(y=\frac{4}{3}x-5\). The coefficient of (x) is the slope.
Step 3
Exam Tip
समीकरण को \(y=\frac{4}{3}x-5\) लिखें। (x) का गुणांक ढाल होता है।
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रेखा (5x+2y=20) का (x)-अक्ष से प्रतिच्छेद क्या है?
What is the (x)-intercept of (5x+2y=20)?
#graphs
#linear
#intercept
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A ((4,0))
B ((0,10))
C ((-4,0))
D ((0,4))
Explanation opens after your attempt
Correct Answer
A. ((4,0))
Step 1
Concept
On the (x)-axis set (y=0). From (5x=20), we get (x=4).
Step 2
Why this answer is correct
The correct answer is A. ((4,0)). On the (x)-axis set (y=0). From (5x=20), we get (x=4).
Step 3
Exam Tip
(x)-अक्ष पर (y=0) रखें। (5x=20) से (x=4) मिलता है।
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रेखा (y=-2) के ग्राफ के लिए सही कथन कौन सा है?
Which statement is correct for the graph of the line (y=-2)?
#graphs
#linear
#horizontal-line
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A यह (x)-अक्ष के समानांतर है / It is parallel to the (x)-axis
B यह (y)-अक्ष के समानांतर है / It is parallel to the (y)-axis
C यह मूल बिंदु से गुजरती है / It passes through the origin
D इसकी ढाल अपरिभाषित है / Its slope is undefined
Explanation opens after your attempt
Correct Answer
A. यह (x)-अक्ष के समानांतर है / It is parallel to the (x)-axis
Step 1
Concept
In (y=-2), (y) is fixed and (x) can be any value. Hence the graph is a horizontal line.
Step 2
Why this answer is correct
The correct answer is A. यह (x)-अक्ष के समानांतर है / It is parallel to the (x)-axis. In (y=-2), (y) is fixed and (x) can be any value. Hence the graph is a horizontal line.
Step 3
Exam Tip
(y=-2) में (y) स्थिर है और (x) कोई भी हो सकता है। इसलिए ग्राफ क्षैतिज रेखा है।
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रेखाएं (y=2x+1) और (y=2x-7) कैसी हैं?
How are the lines (y=2x+1) and (y=2x-7) related?
#graphs
#linear
#parallel
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A समांतर / Parallel
B लंबवत / Perpendicular
C एक ही रेखा / Same line
D प्रतिच्छेदी लेकिन समांतर नहीं / Intersecting but not parallel
Explanation opens after your attempt
Correct Answer
A. समांतर / Parallel
Step 1
Concept
Both lines have slope (2), but different intercepts. Distinct lines with equal slope are parallel.
Step 2
Why this answer is correct
The correct answer is A. समांतर / Parallel. Both lines have slope (2), but different intercepts. Distinct lines with equal slope are parallel.
Step 3
Exam Tip
दोनों रेखाओं की ढाल (2) है लेकिन अवरोध अलग हैं। समान ढाल वाली अलग रेखाएं समांतर होती हैं।
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फलन \(y=\sqrt{2x-8}\) का डोमेन क्या है?
What is the domain of \(y=\sqrt{2x-8}\)?
#graphs
#square-root
#domain
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A \([4,\infty\))
B (\(4,\infty\))
C (\(-\infty,4]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([4,\infty\))
Step 1
Concept
For the square root \(2x-8\ge 0\) is required. This gives \(x\ge 4\).
Step 2
Why this answer is correct
The correct answer is A. \([4,\infty\)). For the square root \(2x-8\ge 0\) is required. This gives \(x\ge 4\).
Step 3
Exam Tip
वर्गमूल के लिए \(2x-8\ge 0\) चाहिए। इससे \(x\ge 4\) मिलता है।
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फलन \(y=7-\sqrt{x-1}\) की रेंज क्या है?
What is the range of \(y=7-\sqrt{x-1}\)?
#graphs
#square-root
#range
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A (\(-\infty,7]\)
B \([7,\infty\))
C \([1,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,7]\)
Step 1
Concept
Since \(\sqrt{x-1}\ge 0\), \(7-\sqrt{x-1}\le 7\). The values can decrease without bound.
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,7]\). Since \(\sqrt{x-1}\ge 0\), \(7-\sqrt{x-1}\le 7\). The values can decrease without bound.
Step 3
Exam Tip
\(\sqrt{x-1}\ge 0\) है इसलिए \(7-\sqrt{x-1}\le 7\) होता है। मान नीचे की ओर असीमित जा सकते हैं।
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ग्राफ \(y=\sqrt{x+9}-4\) का प्रारंभिक बिंदु कौन सा है?
What is the starting point of \(y=\sqrt{x+9}-4\)?
#graphs
#square-root
#start-point
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A ((-9,-4))
B ((9,-4))
C ((-4,-9))
D ((0,-4))
Explanation opens after your attempt
Correct Answer
A. ((-9,-4))
Step 1
Concept
From the inside (x+9=0), (x=-9). Then (y=-4).
Step 2
Why this answer is correct
The correct answer is A. ((-9,-4)). From the inside (x+9=0), (x=-9). Then (y=-4).
Step 3
Exam Tip
अंदर (x+9=0) से (x=-9) मिलता है। तब (y=-4) है।
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फलन \(y=\sqrt{25-x^2}\) किस वृत्त का ऊपरी अर्धभाग है?
The graph \(y=\sqrt{25-x^2}\) is the upper semicircle of which circle?
#graphs
#circle
#square-root
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A \(x^2+y^2=25\)
B \(x^2+y^2=5\)
C (x+y=25)
D \(y=x^2-25\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+y^2=25\)
Step 1
Concept
Squaring gives \(y^2=25-x^2\), so \(x^2+y^2=25\). Because of the square root only the upper part appears.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+y^2=25\). Squaring gives \(y^2=25-x^2\), so \(x^2+y^2=25\). Because of the square root only the upper part appears.
Step 3
Exam Tip
वर्ग करने पर \(y^2=25-x^2\) से \(x^2+y^2=25\) मिलता है। वर्गमूल के कारण केवल ऊपरी भाग आता है।
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फलन \(y=\sqrt{x^2-16}\) का डोमेन क्या है?
What is the domain of \(y=\sqrt{x^2-16}\)?
#graphs
#square-root
#domain
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A (\(-\infty,-4]\cup[4,\infty\))
B ([-4,4])
C ([0,4])
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-4]\cup[4,\infty\))
Step 1
Concept
For the square root \(x^2-16\ge 0\) is required. Hence \(x\le -4\) or \(x\ge 4\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-4]\cup[4,\infty\)). For the square root \(x^2-16\ge 0\) is required. Hence \(x\le -4\) or \(x\ge 4\).
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-16\ge 0\) चाहिए। इसलिए \(x\le -4\) या \(x\ge 4\) है।
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फलन \(y=\frac{1}{x-6}\) के ग्राफ का ऊर्ध्वाधर आसम्टोट क्या है?
What is the vertical asymptote of \(y=\frac{1}{x-6}\)?
#graphs
#rational
#asymptote
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A (x=6)
B (x=-6)
C (y=0)
D (y=6)
Explanation opens after your attempt
Step 1
Concept
The denominator (x-6) makes the function undefined when it is zero. Therefore (x=6) is the vertical asymptote.
Step 2
Why this answer is correct
The correct answer is A. (x=6). The denominator (x-6) makes the function undefined when it is zero. Therefore (x=6) is the vertical asymptote.
Step 3
Exam Tip
हर (x-6) शून्य होने पर फलन अपरिभाषित होता है। इसलिए (x=6) ऊर्ध्वाधर आसम्टोट है।
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फलन \(y=\frac{4}{x+2}-5\) का क्षैतिज आसम्टोट क्या है?
What is the horizontal asymptote of \(y=\frac{4}{x+2}-5\)?
#graphs
#rational
#horizontal-asymptote
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A (y=-5)
B (x=-2)
C (y=4)
D (x=5)
Explanation opens after your attempt
Step 1
Concept
The term \(\frac{4}{x+2}\) approaches (0) for large (|x|). Hence the graph approaches (y=-5).
Step 2
Why this answer is correct
The correct answer is A. (y=-5). The term \(\frac{4}{x+2}\) approaches (0) for large (|x|). Hence the graph approaches (y=-5).
Step 3
Exam Tip
\(\frac{4}{x+2}\) बड़े (|x|) पर (0) के पास जाता है। इसलिए ग्राफ (y=-5) के पास जाता है।
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फलन \(y=\frac{x+4}{x-2}\) के ग्राफ का क्षैतिज आसम्टोट क्या है?
What is the horizontal asymptote of \(y=\frac{x+4}{x-2}\)?
#graphs
#rational
#asymptote
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A (y=1)
B (x=2)
C (y=4)
D (x=-4)
Explanation opens after your attempt
Step 1
Concept
The numerator and denominator have equal degree. The ratio of leading coefficients is \(\frac{1}{1}=1\).
Step 2
Why this answer is correct
The correct answer is A. (y=1). The numerator and denominator have equal degree. The ratio of leading coefficients is \(\frac{1}{1}=1\).
Step 3
Exam Tip
अंश और हर की डिग्री समान है। प्रमुख गुणांकों का अनुपात \(\frac{1}{1}=1\) है।
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फलन \(y=\frac{x^2-9}{x-3}\) में (x=3) पर ग्राफ में क्या दिखता है?
What appears in the graph of \(y=\frac{x^2-9}{x-3}\) at (x=3)?
#graphs
#rational
#hole
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A छिद्र / A hole
B ऊर्ध्वाधर आसम्टोट / A vertical asymptote
C शीर्ष / A vertex
D अधिकतम बिंदु / A maximum point
Explanation opens after your attempt
Correct Answer
A. छिद्र / A hole
Step 1
Concept
\(\frac{x^2-9}{x-3}=x+3\), but (x=3) remains excluded. Therefore the line-like graph has a hole.
Step 2
Why this answer is correct
The correct answer is A. छिद्र / A hole. \(\frac{x^2-9}{x-3}=x+3\), but (x=3) remains excluded. Therefore the line-like graph has a hole.
Step 3
Exam Tip
\(\frac{x^2-9}{x-3}=x+3\) है लेकिन (x=3) निषिद्ध रहता है। इसलिए रेखा जैसे ग्राफ में छिद्र होता है।
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फलन (y=\frac{1}{(x+1)2 }) की सममिति किस रेखा के सापेक्ष है?
The graph of (y=\frac{1}{(x+1)2 }) is symmetric about which line?
#graphs
#rational
#symmetry
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A (x=-1)
B (x=1)
C (y=0)
D (y=-1)
Explanation opens after your attempt
Step 1
Concept
It is the graph of \(y=\frac{1}{x^2}\) shifted (1) unit left. Therefore the axis of symmetry is (x=-1).
Step 2
Why this answer is correct
The correct answer is A. (x=-1). It is the graph of \(y=\frac{1}{x^2}\) shifted (1) unit left. Therefore the axis of symmetry is (x=-1).
Step 3
Exam Tip
यह \(y=\frac{1}{x^2}\) का (1) इकाई बाईं ओर खिसका हुआ ग्राफ है। इसलिए सममिति अक्ष (x=-1) है।
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फलन \(y=\frac{1}{x^2+9}\) का अधिकतम मान क्या है?
What is the maximum value of \(y=\frac{1}{x^2+9}\)?
#graphs
#rational
#maximum
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A \(\frac{1}{9}\)
B (9)
C (0)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{9}\)
Step 1
Concept
The minimum value of \(x^2+9\) is (9). Hence the maximum value of \(\frac{1}{x^2+9}\) is \(\frac{1}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{9}\). The minimum value of \(x^2+9\) is (9). Hence the maximum value of \(\frac{1}{x^2+9}\) is \(\frac{1}{9}\).
Step 3
Exam Tip
\(x^2+9\) का न्यूनतम मान (9) है। इसलिए \(\frac{1}{x^2+9}\) का अधिकतम मान \(\frac{1}{9}\) है।
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फलन (f(x)=\frac{x-3 }{x-2 +1}) के ग्राफ की सममिति कौन सी है?
What is the symmetry of the graph of (f(x)=\frac{x-3 }{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 सममिति नहीं / No symmetry
Explanation opens after your attempt
Correct Answer
A. मूल बिंदु के सापेक्ष सममिति / Symmetry about the origin
Step 1
Concept
(f(-x)=\frac{-x-3 }{x-2 +1}=-f(x)). Therefore it is an odd function.
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-3 }{x-2 +1}=-f(x)). Therefore it is an odd function.
Step 3
Exam Tip
(f(-x)=\frac{-x-3 }{x-2 +1}=-f(x)) है। इसलिए यह विषम फलन है।
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महत्तम पूर्णांक फलन (f(x)=\lfloor x \rfloor) में (f(-4)) का मान क्या है?
For (f(x)=\lfloor x \rfloor), what is (f(-4))?
#graphs
#greatest-integer
#value
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A (-4)
B (-5)
C (4)
D (0)
Explanation opens after your attempt
Step 1
Concept
When (x) itself is an integer, \(\lfloor x \rfloor=x\). Hence \(\lfloor -4 \rfloor=-4\).
Step 2
Why this answer is correct
The correct answer is A. (-4). When (x) itself is an integer, \(\lfloor x \rfloor=x\). Hence \(\lfloor -4 \rfloor=-4\).
Step 3
Exam Tip
जब (x) स्वयं पूर्णांक हो तो \(\lfloor x \rfloor=x\) होता है। इसलिए \(\lfloor -4 \rfloor=-4\) है।
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फलन (f(x)=\lfloor x+0.5 \rfloor) में (f(2.6)) का मान क्या है?
What is (f(2.6)) for (f(x)=\lfloor x+0.5 \rfloor)?
#graphs
#greatest-integer
#value
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A (3)
B (2)
C (2.6)
D (4)
Explanation opens after your attempt
Step 1
Concept
(2.6+0.5=3.1), and \(\lfloor 3.1 \rfloor=3\). First calculate the inside value.
Step 2
Why this answer is correct
The correct answer is A. (3). (2.6+0.5=3.1), and \(\lfloor 3.1 \rfloor=3\). First calculate the inside value.
Step 3
Exam Tip
(2.6+0.5=3.1) है और \(\lfloor 3.1 \rfloor=3\)। पहले अंदर का मान निकालें।
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फलन (f(x)=\lfloor 2x \rfloor) में \(x\in[1,1.5\)) पर मानों का समूह क्या है?
What is the set of values of (f(x)=\lfloor 2x \rfloor) for \(x\in[1,1.5\))?
#graphs
#greatest-integer
#range
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A ({2})
B ({1,2})
C ({2,3})
D ([2,3))
Explanation opens after your attempt
Step 1
Concept
From \(x\in[1,1.5\)), \(2x\in[2,3\)). On this whole interval \(\lfloor 2x \rfloor=2\).
Step 2
Why this answer is correct
The correct answer is A. ({2}). From \(x\in[1,1.5\)), \(2x\in[2,3\)). On this whole interval \(\lfloor 2x \rfloor=2\).
Step 3
Exam Tip
\(x\in[1,1.5\)) से \(2x\in[2,3\)) है। इस पूरे अंतराल पर \(\lfloor 2x \rfloor=2\) होता है।
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सीलिंग फलन (g(x)=\lceil x \rceil) में (g(-2.2)) का मान क्या है?
What is (g(-2.2)) for the ceiling function (g(x)=\lceil x \rceil)?
#graphs
#ceiling-function
#value
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A (-2)
B (-3)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
\(\lceil x \rceil\) gives the smallest integer greater than or equal to (x). For (-2.2), it is (-2).
Step 2
Why this answer is correct
The correct answer is A. (-2). \(\lceil x \rceil\) gives the smallest integer greater than or equal to (x). For (-2.2), it is (-2).
Step 3
Exam Tip
\(\lceil x \rceil\) (x) से बड़ा या बराबर सबसे छोटा पूर्णांक देता है। (-2.2) के लिए वह (-2) है।
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साइनम फलन (f(x)=\operatorname{sgn}(5-x)) में (x=7) पर मान क्या है?
What is the value of (f(x)=\operatorname{sgn}(5-x)) at (x=7)?
#graphs
#signum
#value
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A (-1)
B (0)
C (1)
D (7)
Explanation opens after your attempt
Step 1
Concept
At (x=7), (5-x=-2), which is negative. The signum function gives (-1) for a negative input.
Step 2
Why this answer is correct
The correct answer is A. (-1). At (x=7), (5-x=-2), which is negative. The signum function gives (-1) for a negative input.
Step 3
Exam Tip
(x=7) पर (5-x=-2) ऋणात्मक है। साइनम फलन ऋणात्मक इनपुट पर (-1) देता है।
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फलन (f(x)=\operatorname{sgn}\(x^2-4\)) में किन (x) मानों पर मान (0) है?
For which (x) values is (f(x)=\operatorname{sgn}\(x^2-4\)) equal to (0)?
#graphs
#signum
#zeros
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A (x=-2,2)
B (x=0,4)
C (x=-4,4)
D (x=2,4)
Explanation opens after your attempt
Correct Answer
A. (x=-2,2)
Step 1
Concept
The signum function gives (0) when the inside expression is (0). From \(x^2-4=0\), \(x=\pm2\).
Step 2
Why this answer is correct
The correct answer is A. (x=-2,2). The signum function gives (0) when the inside expression is (0). From \(x^2-4=0\), \(x=\pm2\).
Step 3
Exam Tip
साइनम फलन (0) तब देता है जब अंदर की अभिव्यक्ति (0) हो। \(x^2-4=0\) से \(x=\pm2\) है।
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फलन (y=\max(x,3)) के ग्राफ में कोना किस बिंदु पर है?
At which point is the corner of the graph (y=\max(x,3))?
#graphs
#piecewise
#maximum
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A ((3,3))
B ((0,3))
C ((3,0))
D ((-3,3))
Explanation opens after your attempt
Correct Answer
A. ((3,3))
Step 1
Concept
The two rules (y=x) and (y=3) meet when (x=3). Hence the corner is at ((3,3)).
Step 2
Why this answer is correct
The correct answer is A. ((3,3)). The two rules (y=x) and (y=3) meet when (x=3). Hence the corner is at ((3,3)).
Step 3
Exam Tip
दोनों नियम (y=x) और (y=3) तब मिलते हैं जब (x=3)। इसलिए कोना ((3,3)) पर है।
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फलन (y=\min(x,-3)) के ग्राफ में कोना किस बिंदु पर है?
At which point is the corner of (y=\min(x,-3))?
#graphs
#piecewise
#minimum
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A ((-3,-3))
B ((3,-3))
C ((-3,3))
D ((0,-3))
Explanation opens after your attempt
Correct Answer
A. ((-3,-3))
Step 1
Concept
The rules (y=x) and (y=-3) meet at (x=-3). Therefore the corner is ((-3,-3)).
Step 2
Why this answer is correct
The correct answer is A. ((-3,-3)). The rules (y=x) and (y=-3) meet at (x=-3). Therefore the corner is ((-3,-3)).
Step 3
Exam Tip
नियम (y=x) और (y=-3) (x=-3) पर मिलते हैं। इसलिए कोना ((-3,-3)) है।
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फलन (y=|x-2|+x) में \(x\ge 2\) के लिए ग्राफ कौन सी रेखा है?
For \(x\ge 2\), which line represents (y=|x-2|+x)?
#graphs
#modulus
#piecewise
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A (y=2x-2)
B (y=2)
C (y=-2)
D (y=x-2)
Explanation opens after your attempt
Correct Answer
A. (y=2x-2)
Step 1
Concept
For \(x\ge 2\), (|x-2|=x-2). Hence (y=x-2+x=2x-2).
Step 2
Why this answer is correct
The correct answer is A. (y=2x-2). For \(x\ge 2\), (|x-2|=x-2). Hence (y=x-2+x=2x-2).
Step 3
Exam Tip
\(x\ge 2\) पर (|x-2|=x-2) होता है। इसलिए (y=x-2+x=2x-2) है।
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ग्राफ \(y=x^2\) और (y=3x+4) किन (x) मानों पर मिलते हैं?
At which (x) values do \(y=x^2\) and (y=3x+4) meet?
#graphs
#intersection
#quadratic
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A (x=-1,4)
B (x=1,4)
C (x=-4,1)
D (x=0,4)
Explanation opens after your attempt
Correct Answer
A. (x=-1,4)
Step 1
Concept
For intersection set \(x^2=3x+4\). From \(x^2-3x-4=0\), (x=-1,4).
Step 2
Why this answer is correct
The correct answer is A. (x=-1,4). For intersection set \(x^2=3x+4\). From \(x^2-3x-4=0\), (x=-1,4).
Step 3
Exam Tip
प्रतिच्छेद के लिए \(x^2=3x+4\) रखें। \(x^2-3x-4=0\) से (x=-1,4) है।
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ग्राफ (y=|x-2|) और (y=4) किन (x) मानों पर मिलते हैं?
At which (x) values do (y=|x-2|) and (y=4) meet?
#graphs
#intersection
#modulus
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A (x=-2,6)
B (x=2,4)
C (x=-4,4)
D (x=0,6)
Explanation opens after your attempt
Correct Answer
A. (x=-2,6)
Step 1
Concept
From (|x-2|=4), \(x-2=\pm4\). Therefore (x=-2,6).
Step 2
Why this answer is correct
The correct answer is A. (x=-2,6). From (|x-2|=4), \(x-2=\pm4\). Therefore (x=-2,6).
Step 3
Exam Tip
(|x-2|=4) से \(x-2=\pm4\) मिलता है। इसलिए (x=-2,6) हैं।
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ग्राफ \(y=\sqrt{x+1}\) और (y=3) किस (x) मान पर मिलते हैं?
At which (x) value do \(y=\sqrt{x+1}\) and (y=3) meet?
#graphs
#intersection
#square-root
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A (x=8)
B (x=9)
C (x=2)
D (x=-1)
Explanation opens after your attempt
Step 1
Concept
From \(\sqrt{x+1}=3\), we get (x+1=9). Hence (x=8).
Step 2
Why this answer is correct
The correct answer is A. (x=8). From \(\sqrt{x+1}=3\), we get (x+1=9). Hence (x=8).
Step 3
Exam Tip
\(\sqrt{x+1}=3\) से (x+1=9) मिलता है। इसलिए (x=8) है।
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ग्राफ \(y=\frac{2}{x}\) और (y=x) किन बिंदुओं पर मिलते हैं?
At which points do \(y=\frac{2}{x}\) and (y=x) meet?
#graphs
#intersection
#rational
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A (\(\sqrt{2},\sqrt{2}\)) और (\(-\sqrt{2},-\sqrt{2}\)) / (\(\sqrt{2},\sqrt{2}\)) and (\(-\sqrt{2},-\sqrt{2}\))
B ((2,2)) और ((-2,-2)) / ((2,2)) and ((-2,-2))
C ((1,2)) और ((-1,-2)) / ((1,2)) and ((-1,-2))
D ((0,0))
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{2},\sqrt{2}\)) और (\(-\sqrt{2},-\sqrt{2}\)) / (\(\sqrt{2},\sqrt{2}\)) and (\(-\sqrt{2},-\sqrt{2}\))
Step 1
Concept
From \(\frac{2}{x}=x\), we get \(x^2=2\). Hence \(x=\pm\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{2},\sqrt{2}\)) और (\(-\sqrt{2},-\sqrt{2}\)) / (\(\sqrt{2},\sqrt{2}\)) and (\(-\sqrt{2},-\sqrt{2}\)). From \(\frac{2}{x}=x\), we get \(x^2=2\). Hence \(x=\pm\sqrt{2}\).
Step 3
Exam Tip
\(\frac{2}{x}=x\) से \(x^2=2\) मिलता है। इसलिए \(x=\pm\sqrt{2}\) है।
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फलन (y=f(x+4)-3) का ग्राफ (y=f(x)) से कैसे प्राप्त होगा?
How is (y=f(x+4)-3) obtained from (y=f(x))?
#graphs
#transformation
#shift
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A (4) इकाई बाईं ओर और (3) इकाई नीचे / (4) units left and (3) units down
B (4) इकाई दाईं ओर और (3) इकाई नीचे / (4) units right and (3) units down
C (4) इकाई बाईं ओर और (3) इकाई ऊपर / (4) units left and (3) units up
D (3) इकाई बाईं ओर और (4) इकाई नीचे / (3) units left and (4) units down
Explanation opens after your attempt
Correct Answer
A. (4) इकाई बाईं ओर और (3) इकाई नीचे / (4) units left and (3) units down
Step 1
Concept
The term (x+4) shifts the graph (4) units left. The outside (-3) shifts it (3) units down.
Step 2
Why this answer is correct
The correct answer is A. (4) इकाई बाईं ओर और (3) इकाई नीचे / (4) units left and (3) units down. The term (x+4) shifts the graph (4) units left. The outside (-3) shifts it (3) units down.
Step 3
Exam Tip
(x+4) ग्राफ को (4) इकाई बाईं ओर ले जाता है। बाहर (-3) ग्राफ को (3) इकाई नीचे ले जाता है।
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फलन (y=2f(x)-1) में (y=f(x)) के ग्राफ पर क्या परिवर्तन होगा?
What transformation occurs from (y=f(x)) to (y=2f(x)-1)?
#graphs
#transformation
#stretch
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A ऊर्ध्वाधर खिंचाव (2) गुना और (1) इकाई नीचे / Vertical stretch by (2) and (1) unit down
B क्षैतिज खिंचाव (2) गुना और (1) इकाई ऊपर / Horizontal stretch by (2) and (1) unit up
C (2) इकाई दाईं ओर और (1) इकाई नीचे / (2) units right and (1) unit down
D केवल (1) इकाई ऊपर / Only (1) unit up
Explanation opens after your attempt
Correct Answer
A. ऊर्ध्वाधर खिंचाव (2) गुना और (1) इकाई नीचे / Vertical stretch by (2) and (1) unit down
Step 1
Concept
Multiplication by (2) outside doubles all (y)-values. Then (-1) moves the graph downward.
Step 2
Why this answer is correct
The correct answer is A. ऊर्ध्वाधर खिंचाव (2) गुना और (1) इकाई नीचे / Vertical stretch by (2) and (1) unit down. Multiplication by (2) outside doubles all (y)-values. Then (-1) moves the graph downward.
Step 3
Exam Tip
बाहर (2) से गुणा सभी (y)-मानों को दुगुना करता है। फिर (-1) ग्राफ को नीचे ले जाता है।
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फलन (y=f(3x)) का ग्राफ (y=f(x)) से कैसे बदलेगा?
How does (y=f(3x)) change from (y=f(x))?
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#transformation
#compression
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A क्षैतिज संपीड़न (3) गुना / Horizontal compression by factor (3)
B ऊर्ध्वाधर खिंचाव (3) गुना / Vertical stretch by factor (3)
C दाईं ओर (3) इकाई खिसकाव / Shift (3) units right
D ऊपर (3) इकाई खिसकाव / Shift (3) units up
Explanation opens after your attempt
Correct Answer
A. क्षैतिज संपीड़न (3) गुना / Horizontal compression by factor (3)
Step 1
Concept
Having (3x) inside shrinks the graph toward the (y)-axis. An inside multiplier gives a horizontal change.
Step 2
Why this answer is correct
The correct answer is A. क्षैतिज संपीड़न (3) गुना / Horizontal compression by factor (3). Having (3x) inside shrinks the graph toward the (y)-axis. An inside multiplier gives a horizontal change.
Step 3
Exam Tip
अंदर (3x) होने से ग्राफ (y)-अक्ष की ओर सिमटता है। अंदर का गुणक क्षैतिज परिवर्तन देता है।
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फलन (y=f\left\(\frac{x}{2}\right\)) का ग्राफ (y=f(x)) से कैसे बदलेगा?
How does (y=f\left\(\frac{x}{2}\right\)) change from (y=f(x))?
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#transformation
#stretch
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A क्षैतिज खिंचाव (2) गुना / Horizontal stretch by factor (2)
B क्षैतिज संपीड़न (2) गुना / Horizontal compression by factor (2)
C ऊर्ध्वाधर खिंचाव (2) गुना / Vertical stretch by factor (2)
D नीचे (2) इकाई खिसकाव / Shift (2) units down
Explanation opens after your attempt
Correct Answer
A. क्षैतिज खिंचाव (2) गुना / Horizontal stretch by factor (2)
Step 1
Concept
With \(\frac{x}{2}\) inside, (x) must double to get the same (f)-value. Hence the graph stretches horizontally.
Step 2
Why this answer is correct
The correct answer is A. क्षैतिज खिंचाव (2) गुना / Horizontal stretch by factor (2). With \(\frac{x}{2}\) inside, (x) must double to get the same (f)-value. Hence the graph stretches horizontally.
Step 3
Exam Tip
अंदर \(\frac{x}{2}\) होने से वही (f)-मान पाने के लिए (x) दुगुना चाहिए। इसलिए ग्राफ क्षैतिज रूप से खिंचता है।
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फलन (y=f(-x)+2) का ग्राफ (y=f(x)) से कैसे प्राप्त होगा?
How is (y=f(-x)+2) obtained from (y=f(x))?
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#transformation
#reflection
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A (y)-अक्ष में परावर्तन और (2) इकाई ऊपर / Reflection in the (y)-axis and (2) units up
B (x)-अक्ष में परावर्तन और (2) इकाई ऊपर / Reflection in the (x)-axis and (2) units up
C (y)-अक्ष में परावर्तन और (2) इकाई नीचे / Reflection in the (y)-axis and (2) units down
D केवल (2) इकाई दाईं ओर / Only (2) units right
Explanation opens after your attempt
Correct Answer
A. (y)-अक्ष में परावर्तन और (2) इकाई ऊपर / Reflection in the (y)-axis and (2) units up
Step 1
Concept
(f(-x)) gives reflection in the (y)-axis. The outside (+2) shifts the graph upward.
Step 2
Why this answer is correct
The correct answer is A. (y)-अक्ष में परावर्तन और (2) इकाई ऊपर / Reflection in the (y)-axis and (2) units up. (f(-x)) gives reflection in the (y)-axis. The outside (+2) shifts the graph upward.
Step 3
Exam Tip
(f(-x)) (y)-अक्ष में परावर्तन देता है। बाहर (+2) ग्राफ को ऊपर खिसकाता है।
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फलन (y=|f(x)-2|) बनाते समय (y=f(x)) के किस भाग का परावर्तन होता है?
While making (y=|f(x)-2|), which part of (y=f(x)) is reflected?
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A रेखा (y=2) के नीचे वाला भाग ऊपर / The part below (y=2) upward
B रेखा (x=2) के बाईं ओर वाला भाग दाईं ओर / The part left of (x=2) to the right
C रेखा (y=0) के ऊपर वाला भाग नीचे / The part above (y=0) downward
D पूरा ग्राफ (y)-अक्ष में / The whole graph in the (y)-axis
Explanation opens after your attempt
Correct Answer
A. रेखा (y=2) के नीचे वाला भाग ऊपर / The part below (y=2) upward
Step 1
Concept
(|f(x)-2|) makes (f(x)-2) non-negative. Therefore the part with (f(x)<2) folds upward about the line (y=2).
Step 2
Why this answer is correct
The correct answer is A. रेखा (y=2) के नीचे वाला भाग ऊपर / The part below (y=2) upward. (|f(x)-2|) makes (f(x)-2) non-negative. Therefore the part with (f(x)<2) folds upward about the line (y=2).
Step 3
Exam Tip
(|f(x)-2|) पहले (f(x)-2) को अऋणात्मक बनाता है। इसलिए (f(x)<2) वाला भाग रेखा (y=2) के सापेक्ष ऊपर मुड़ता है।
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फलन (y=f(|x-1|)) के ग्राफ में सममिति किस रेखा के सापेक्ष होगी?
The graph of (y=f(|x-1|)) will be symmetric about which line?
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A (x=1)
B (x=0)
C (y=1)
D (y=0)
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Step 1
Concept
In (|x-1|), inputs at equal distance from (x=1) become equal. Hence the graph is symmetric about (x=1).
Step 2
Why this answer is correct
The correct answer is A. (x=1). In (|x-1|), inputs at equal distance from (x=1) become equal. Hence the graph is symmetric about (x=1).
Step 3
Exam Tip
(|x-1|) में (x=1) से समान दूरी वाले इनपुट समान होते हैं। इसलिए ग्राफ (x=1) के सापेक्ष सममित होता है।
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