फलन (f(x)=|x-3|+2) के आलेख का शीर्ष कौन सा है?
What is the vertex of the graph of (f(x)=|x-3|+2)?
#standard-functions
#modulus-graph
#vertex
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A ((3,2))
B ((-3,2))
C ((3,-2))
D ((2,3))
Explanation opens after your attempt
Correct Answer
A. ((3,2))
Step 1
Concept
(|x-3|) becomes zero at (x=3), and (2) is added outside. So the vertex is ((3,2)).
Step 2
Why this answer is correct
The correct answer is A. ((3,2)). (|x-3|) becomes zero at (x=3), and (2) is added outside. So the vertex is ((3,2)).
Step 3
Exam Tip
(|x-3|) शून्य (x=3) पर होता है और बाहर (2) जुड़ता है। इसलिए शीर्ष ((3,2)) है।
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फलन (f(x)=-(x+2)2 +5) का अधिकतम मान क्या है?
What is the maximum value of (f(x)=-(x+2)2 +5)?
#standard-functions
#quadratic-graph
#maximum
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A (5)
B (2)
C (-2)
D कोई अधिकतम नहीं / No maximum
Explanation opens after your attempt
Step 1
Concept
Since ((x+2)2 \ge 0), (-(x+2)2 +5\le 5). The maximum value (5) occurs when (x=-2).
Step 2
Why this answer is correct
The correct answer is A. (5). Since ((x+2)2 \ge 0), (-(x+2)2 +5\le 5). The maximum value (5) occurs when (x=-2).
Step 3
Exam Tip
((x+2)2 \ge 0), इसलिए (-(x+2)2 +5\le 5)। अधिकतम मान (5) तब मिलता है जब (x=-2)।
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फलन (f(x)=\sqrt{x-4}+1) का डोमेन क्या है?
What is the domain of (f(x)=\sqrt{x-4}+1)?
#standard-functions
#square-root-graph
#domain
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A \(x\ge 4\)
B \(x\le 4\)
C (x>1)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 4\)
Step 1
Concept
For a real square root, \(x-4\ge 0\) is required. Hence the domain is \(x\ge 4\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 4\). For a real square root, \(x-4\ge 0\) is required. Hence the domain is \(x\ge 4\).
Step 3
Exam Tip
वास्तविक वर्गमूल के लिए \(x-4\ge 0\) चाहिए। इसलिए डोमेन \(x\ge 4\) है।
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फलन (f(x)=\sqrt{7-x}) का डोमेन क्या है?
What is the domain of (f(x)=\sqrt{7-x})?
#standard-functions
#square-root-graph
#domain-inequality
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A \(x\le 7\)
B \(x\ge 7\)
C (x<0)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(x\le 7\)
Step 1
Concept
For real values, \(7-x\ge 0\), so \(x\le 7\). The expression inside a square root must not be negative.
Step 2
Why this answer is correct
The correct answer is A. \(x\le 7\). For real values, \(7-x\ge 0\), so \(x\le 7\). The expression inside a square root must not be negative.
Step 3
Exam Tip
वास्तविक मान के लिए \(7-x\ge 0\), इसलिए \(x\le 7\)। वर्गमूल में अंदर की राशि ऋणात्मक नहीं होनी चाहिए।
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फलन (f(x)=\frac{1}{x+3}-2) का ऊर्ध्व आसिम्प्टोट कौन सा है?
What is the vertical asymptote of (f(x)=\frac{1}{x+3}-2)?
#standard-functions
#reciprocal-graph
#vertical-asymptote
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A (x=-3)
B (x=3)
C (y=-2)
D (y=0)
Explanation opens after your attempt
Step 1
Concept
The denominator (x+3) cannot be zero. Therefore the vertical asymptote is (x=-3).
Step 2
Why this answer is correct
The correct answer is A. (x=-3). The denominator (x+3) cannot be zero. Therefore the vertical asymptote is (x=-3).
Step 3
Exam Tip
हर (x+3) शून्य होने पर फलन परिभाषित नहीं रहता। इसलिए ऊर्ध्व आसिम्प्टोट (x=-3) है।
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फलन (f(x)=\frac{2}{x-1}+4) का क्षैतिज आसिम्प्टोट कौन सा है?
What is the horizontal asymptote of (f(x)=\frac{2}{x-1}+4)?
#standard-functions
#reciprocal-graph
#horizontal-asymptote
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A (y=4)
B (x=1)
C (y=0)
D (x=4)
Explanation opens after your attempt
Step 1
Concept
For large (|x|), \(\frac{2}{x-1}\) approaches (0). So the horizontal asymptote is (y=4).
Step 2
Why this answer is correct
The correct answer is A. (y=4). For large (|x|), \(\frac{2}{x-1}\) approaches (0). So the horizontal asymptote is (y=4).
Step 3
Exam Tip
बड़े (|x|) पर \(\frac{2}{x-1}\) (0) के पास जाता है। इसलिए क्षैतिज आसिम्प्टोट (y=4) है।
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फलन (f(x)=|2x-6|) का न्यूनतम मान किस (x) पर मिलता है?
At which (x)-value does (f(x)=|2x-6|) attain its minimum value?
#standard-functions
#modulus-graph
#minimum-point
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A (x=3)
B (x=6)
C (x=-3)
D (x=0)
Explanation opens after your attempt
Step 1
Concept
A modulus is minimum (0) when the inside expression is (0). From (2x-6=0), we get (x=3).
Step 2
Why this answer is correct
The correct answer is A. (x=3). A modulus is minimum (0) when the inside expression is (0). From (2x-6=0), we get (x=3).
Step 3
Exam Tip
मापांक का न्यूनतम मान (0) तब होता है जब अंदर की राशि (0) हो। (2x-6=0) से (x=3) मिलता है।
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फलन (f(x)=|x+1|-4) के (x)-अक्ष प्रतिच्छेद कौन से हैं?
What are the (x)-intercepts of (f(x)=|x+1|-4)?
#standard-functions
#modulus-graph
#x-intercepts
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A (x=3) और (x=-5) / (x=3) and (x=-5)
B (x=4) और (x=-4) / (x=4) and (x=-4)
C (x=1) और (x=-1) / (x=1) and (x=-1)
D (x=5) और (x=-3) / (x=5) and (x=-3)
Explanation opens after your attempt
Correct Answer
A. (x=3) और (x=-5) / (x=3) and (x=-5)
Step 1
Concept
On the (x)-axis, (|x+1|-4=0), so (|x+1|=4). This gives (x=3) and (x=-5).
Step 2
Why this answer is correct
The correct answer is A. (x=3) और (x=-5) / (x=3) and (x=-5). On the (x)-axis, (|x+1|-4=0), so (|x+1|=4). This gives (x=3) and (x=-5).
Step 3
Exam Tip
(x)-अक्ष पर (|x+1|-4=0), इसलिए (|x+1|=4)। इससे (x=3) और (x=-5) मिलते हैं।
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फलन (f(x)=(x-3)2 -4) के (x)-अक्ष प्रतिच्छेद कौन से हैं?
What are the (x)-intercepts of (f(x)=(x-3)2 -4)?
#standard-functions
#quadratic-graph
#x-intercepts
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A (x=1) और (x=5) / (x=1) and (x=5)
B (x=-1) और (x=5) / (x=-1) and (x=5)
C (x=3) और (x=4) / (x=3) and (x=4)
D (x=2) और (x=6) / (x=2) and (x=6)
Explanation opens after your attempt
Correct Answer
A. (x=1) और (x=5) / (x=1) and (x=5)
Step 1
Concept
Putting (y=0) gives ((x-3)2 =4). Thus \(x-3=\pm 2\), so (x=1,5).
Step 2
Why this answer is correct
The correct answer is A. (x=1) और (x=5) / (x=1) and (x=5). Putting (y=0) gives ((x-3)2 =4). Thus \(x-3=\pm 2\), so (x=1,5).
Step 3
Exam Tip
(y=0) रखने पर ((x-3)2 =4) मिलता है। इसलिए \(x-3=\pm 2\) और (x=1,5) हैं।
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फलन (f(x)=(x+1)2 +3) का परिसर क्या है?
What is the range of (f(x)=(x+1)2 +3)?
#standard-functions
#quadratic-graph
#range
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A \([3,\infty\))
B (\(3,\infty\))
C (\(-\infty,3]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([3,\infty\))
Step 1
Concept
Since ((x+1)2 \ge 0), (f(x)\ge 3). Hence the range is \([3,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([3,\infty\)). Since ((x+1)2 \ge 0), (f(x)\ge 3). Hence the range is \([3,\infty\)).
Step 3
Exam Tip
((x+1)2 \ge 0), इसलिए (f(x)\ge 3)। अतः परिसर \([3,\infty\)) है।
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फलन (f(x)=-(x-4)2 -1) का परिसर क्या है?
What is the range of (f(x)=-(x-4)2 -1)?
#standard-functions
#quadratic-graph
#negative-range
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A (\(-\infty,-1]\)
B \([-1,\infty\))
C (\(-\infty,\infty\))
D (\(-\infty,-4]\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-1]\)
Step 1
Concept
Since ((x-4)2 \ge 0), (-(x-4)2 -1\le -1). Therefore the range is (\(-\infty,-1]\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-1]\). Since ((x-4)2 \ge 0), (-(x-4)2 -1\le -1). Therefore the range is (\(-\infty,-1]\).
Step 3
Exam Tip
((x-4)2 \ge 0), इसलिए (-(x-4)2 -1\le -1)। अतः परिसर (\(-\infty,-1]\) है।
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फलन (f(x)=3|x-2|) का आलेख (y=|x-2|) की तुलना में कैसा है?
How is the graph of (f(x)=3|x-2|) compared with (y=|x-2|)?
#standard-functions
#modulus-graph
#vertical-stretch
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A ऊर्ध्व रूप से खिंचा / Vertically stretched
B ऊर्ध्व रूप से दबा / Vertically compressed
C (3) इकाई दाईं ओर / Shifted (3) units right
D (3) इकाई नीचे / Shifted (3) units down
Explanation opens after your attempt
Correct Answer
A. ऊर्ध्व रूप से खिंचा / Vertically stretched
Step 1
Concept
The coefficient (3) multiplies every (y)-value by (3). So the graph is vertically stretched.
Step 2
Why this answer is correct
The correct answer is A. ऊर्ध्व रूप से खिंचा / Vertically stretched. The coefficient (3) multiplies every (y)-value by (3). So the graph is vertically stretched.
Step 3
Exam Tip
गुणांक (3) हर (y)-मान को (3) गुना करता है। इसलिए आलेख ऊर्ध्व रूप से खिंचता है।
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फलन (f(x)=\frac{1}{2}x-2 ) का आलेख \(y=x^2\) की तुलना में कैसा होगा?
How is the graph of (f(x)=\frac{1}{2}x-2 ) compared with \(y=x^2\)?
#standard-functions
#quadratic-graph
#vertical-compression
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A ऊर्ध्व रूप से दबा हुआ / Vertically compressed
B ऊर्ध्व रूप से खिंचा हुआ / Vertically stretched
C बाईं ओर खिसका हुआ / Shifted left
D नीचे खिसका हुआ / Shifted down
Explanation opens after your attempt
Correct Answer
A. ऊर्ध्व रूप से दबा हुआ / Vertically compressed
Step 1
Concept
The coefficient \(\frac{1}{2}\) halves every (y)-value. So the parabola looks wider.
Step 2
Why this answer is correct
The correct answer is A. ऊर्ध्व रूप से दबा हुआ / Vertically compressed. The coefficient \(\frac{1}{2}\) halves every (y)-value. So the parabola looks wider.
Step 3
Exam Tip
गुणांक \(\frac{1}{2}\) हर (y)-मान को आधा करता है। इसलिए परवलय चौड़ा दिखता है।
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फलन (f(x)=-2x-2 ) का आलेख \(y=x^2\) से कैसे संबंधित है?
How is the graph of (f(x)=-2x-2 ) related to \(y=x^2\)?
#standard-functions
#quadratic-graph
#reflection-stretch
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A (x)-अक्ष में प्रतिबिंब और ऊर्ध्व खिंचाव / Reflection in (x)-axis and vertical stretch
B केवल दाईं ओर खिसकाव / Only shift right
C केवल ऊपर खिसकाव / Only shift up
D (y)-अक्ष में प्रतिबिंब / Reflection in (y)-axis
Explanation opens after your attempt
Correct Answer
A. (x)-अक्ष में प्रतिबिंब और ऊर्ध्व खिंचाव / Reflection in (x)-axis and vertical stretch
Step 1
Concept
The negative sign reflects in the (x)-axis, and (2) gives a vertical stretch. So the parabola opens downward and becomes narrower.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष में प्रतिबिंब और ऊर्ध्व खिंचाव / Reflection in (x)-axis and vertical stretch. The negative sign reflects in the (x)-axis, and (2) gives a vertical stretch. So the parabola opens downward and becomes narrower.
Step 3
Exam Tip
ऋण चिह्न (x)-अक्ष में प्रतिबिंब देता है और (2) ऊर्ध्व खिंचाव देता है। इसलिए परवलय नीचे खुलता है और संकरा होता है।
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फलन (f(x)=\sqrt{x+2}-3) का प्रारंभिक बिंदु कौन सा है?
What is the starting point of (f(x)=\sqrt{x+2}-3)?
#standard-functions
#square-root-graph
#start-point
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A ((-2,-3))
B ((2,-3))
C ((-2,3))
D ((0,-3))
Explanation opens after your attempt
Correct Answer
A. ((-2,-3))
Step 1
Concept
\(\sqrt{x+2}\) starts when (x+2=0), that is (x=-2). The outside (-3) makes the starting point ((-2,-3)).
Step 2
Why this answer is correct
The correct answer is A. ((-2,-3)). \(\sqrt{x+2}\) starts when (x+2=0), that is (x=-2). The outside (-3) makes the starting point ((-2,-3)).
Step 3
Exam Tip
\(\sqrt{x+2}\) तब शुरू होता है जब (x+2=0), अर्थात (x=-2)। बाहर (-3) होने से प्रारंभिक बिंदु ((-2,-3)) है।
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फलन (f(x)=2\sqrt{x-1}+3) का न्यूनतम मान क्या है?
What is the minimum value of (f(x)=2\sqrt{x-1}+3)?
#standard-functions
#square-root-graph
#minimum
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A (3)
B (2)
C (1)
D (0)
Explanation opens after your attempt
Step 1
Concept
Since \(\sqrt{x-1}\ge 0\), \(2\sqrt{x-1}+3\ge 3\). The minimum value (3) occurs at (x=1).
Step 2
Why this answer is correct
The correct answer is A. (3). Since \(\sqrt{x-1}\ge 0\), \(2\sqrt{x-1}+3\ge 3\). The minimum value (3) occurs at (x=1).
Step 3
Exam Tip
\(\sqrt{x-1}\ge 0\), इसलिए \(2\sqrt{x-1}+3\ge 3\)। न्यूनतम मान (3) (x=1) पर मिलता है।
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फलन (f(x)=\sqrt{5-x}+2) का परिसर क्या है?
What is the range of (f(x)=\sqrt{5-x}+2)?
#standard-functions
#square-root-graph
#range
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A \([2,\infty\))
B (\(-\infty,2]\)
C ([0,2])
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([2,\infty\))
Step 1
Concept
Since \(\sqrt{5-x}\ge 0\), (f(x)\ge 2). The square root can grow as (x) decreases, so the range is \([2,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([2,\infty\)). Since \(\sqrt{5-x}\ge 0\), (f(x)\ge 2). The square root can grow as (x) decreases, so the range is \([2,\infty\)).
Step 3
Exam Tip
\(\sqrt{5-x}\ge 0\), इसलिए (f(x)\ge 2)। (x) घटाने पर वर्गमूल बड़ा हो सकता है, इसलिए परिसर \([2,\infty\)) है।
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फलन (f(x)=\frac{1}{(x-2)2 }) का डोमेन क्या है?
What is the domain of (f(x)=\frac{1}{(x-2)2 })?
#standard-functions
#reciprocal-square-graph
#domain
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A \(x\in\mathbb{R}, x\neq 2\)
B \(x\in\mathbb{R}, x\neq 0\)
C (x>2)
D (x<2)
Explanation opens after your attempt
Correct Answer
A. \(x\in\mathbb{R}, x\neq 2\)
Step 1
Concept
The denominator ((x-2)2 ) cannot be zero. Therefore (x=2) is removed from the domain.
Step 2
Why this answer is correct
The correct answer is A. \(x\in\mathbb{R}, x\neq 2\). The denominator ((x-2)2 ) cannot be zero. Therefore (x=2) is removed from the domain.
Step 3
Exam Tip
हर ((x-2)2 ) शून्य नहीं हो सकता। इसलिए (x=2) को डोमेन से हटाते हैं।
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फलन (f(x)=\frac{1}{(x+1)2 }+3) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{(x+1)2 }+3)?
#standard-functions
#reciprocal-square-graph
#range
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A (\(3,\infty\))
B \([3,\infty\))
C (\(-\infty,3\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(3,\infty\))
Step 1
Concept
(\frac{1}{(x+1)2 }>0), so the function is greater than (3) but never equal to (3). Hence the range is (\(3,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (\(3,\infty\)). (\frac{1}{(x+1)2 }>0), so the function is greater than (3) but never equal to (3). Hence the range is (\(3,\infty\)).
Step 3
Exam Tip
(\frac{1}{(x+1)2 }>0), इसलिए फलन (3) से बड़ा है लेकिन (3) के बराबर नहीं। अतः परिसर (\(3,\infty\)) है।
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फलन (f(x)=\frac{-1}{(x-3)2 }) का परिसर क्या है?
What is the range of (f(x)=\frac{-1}{(x-3)2 })?
#standard-functions
#reciprocal-square-graph
#negative-range
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A (\(-\infty,0\))
B (\(0,\infty\))
C (\(-\infty,0]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,0\))
Step 1
Concept
(\frac{1}{(x-3)2 }) is positive, so its negative is always less than (0). It never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,0\)). (\frac{1}{(x-3)2 }) is positive, so its negative is always less than (0). It never becomes (0).
Step 3
Exam Tip
(\frac{1}{(x-3)2 }) धनात्मक होता है, इसलिए उसका ऋणात्मक मान हमेशा (0) से कम है। यह कभी (0) नहीं बनता।
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फलन (f(x)=\operatorname{sgn}(x-5)) में (x=5) पर मान क्या है?
What is the value of (f(x)=\operatorname{sgn}(x-5)) at (x=5)?
#standard-functions
#signum-graph
#value
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A (0)
B (1)
C (-1)
D (5)
Explanation opens after your attempt
Step 1
Concept
At (x=5), (x-5=0), so (\operatorname{sgn}(0)=0). Check the zero point separately in signum.
Step 2
Why this answer is correct
The correct answer is A. (0). At (x=5), (x-5=0), so (\operatorname{sgn}(0)=0). Check the zero point separately in signum.
Step 3
Exam Tip
(x=5) पर (x-5=0), इसलिए (\operatorname{sgn}(0)=0)। साइनम में शून्य बिंदु अलग से जांचें।
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फलन (f(x)=\operatorname{sgn}(2x+6)) में (x<-3) होने पर मान क्या होगा?
For (f(x)=\operatorname{sgn}(2x+6)), what is the value when (x<-3)?
#standard-functions
#signum-graph
#interval
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A (-1)
B (0)
C (1)
D (2)
Explanation opens after your attempt
Step 1
Concept
If (x<-3), then (2x+6<0). So the value of the signum function is (-1).
Step 2
Why this answer is correct
The correct answer is A. (-1). If (x<-3), then (2x+6<0). So the value of the signum function is (-1).
Step 3
Exam Tip
यदि (x<-3), तो (2x+6<0)। इसलिए साइनम फलन का मान (-1) होगा।
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फलन (f(x)=\operatorname{sgn}(4-x)) में (x<4) होने पर मान क्या है?
For (f(x)=\operatorname{sgn}(4-x)), what is the value when (x<4)?
#standard-functions
#signum-graph
#inequality
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A (1)
B (0)
C (-1)
D (4)
Explanation opens after your attempt
Step 1
Concept
If (x<4), then (4-x>0). Therefore (\operatorname{sgn}(4-x)=1).
Step 2
Why this answer is correct
The correct answer is A. (1). If (x<4), then (4-x>0). Therefore (\operatorname{sgn}(4-x)=1).
Step 3
Exam Tip
यदि (x<4), तो (4-x>0)। इसलिए (\operatorname{sgn}(4-x)=1) है।
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फलन (f(x)=\lfloor x+2\rfloor) में (x=1.6) पर मान क्या है?
What is the value of (f(x)=\lfloor x+2\rfloor) at (x=1.6)?
#standard-functions
#greatest-integer-graph
#value
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A (3)
B (2)
C (4)
D (1.6)
Explanation opens after your attempt
Step 1
Concept
Here (x+2=3.6) and \(\lfloor 3.6\rfloor=3\). First find the inside value and then apply the greatest integer function.
Step 2
Why this answer is correct
The correct answer is A. (3). Here (x+2=3.6) and \(\lfloor 3.6\rfloor=3\). First find the inside value and then apply the greatest integer function.
Step 3
Exam Tip
(x+2=3.6) और \(\lfloor 3.6\rfloor=3\)। पहले अंदर का मान निकालें फिर ग्रेटेस्ट इंटीजर लगाएं।
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फलन (f(x)=\lfloor 2x\rfloor) में (x=1.4) पर मान क्या होगा?
What is the value of (f(x)=\lfloor 2x\rfloor) at (x=1.4)?
#standard-functions
#greatest-integer-graph
#numerical
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A (2)
B (3)
C (2.8)
D (1)
Explanation opens after your attempt
Step 1
Concept
Here (2x=2.8), and \(\lfloor 2.8\rfloor=2\). For a decimal, take the greatest integer less than or equal to it.
Step 2
Why this answer is correct
The correct answer is A. (2). Here (2x=2.8), and \(\lfloor 2.8\rfloor=2\). For a decimal, take the greatest integer less than or equal to it.
Step 3
Exam Tip
(2x=2.8) और \(\lfloor 2.8\rfloor=2\)। दशमलव के लिए सबसे बड़ी छोटी या बराबर पूर्णांक संख्या लें।
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फलन (f(x)=\lfloor x\rfloor) के लिए \(x\in[3,4\)) पर आलेख कौन सा भाग है?
For (f(x)=\lfloor x\rfloor), which graph part appears on \(x\in[3,4\))?
#standard-functions
#greatest-integer-graph
#step-interval
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A क्षैतिज रेखा (y=3) / Horizontal line (y=3)
B क्षैतिज रेखा (y=4) / Horizontal line (y=4)
C रेखा (y=x) / Line (y=x)
D परवलय \(y=x^2\) / Parabola \(y=x^2\)
Explanation opens after your attempt
Correct Answer
A. क्षैतिज रेखा (y=3) / Horizontal line (y=3)
Step 1
Concept
For every \(x\in[3,4\)), \(\lfloor x\rfloor=3\). So the graph on that interval is (y=3).
Step 2
Why this answer is correct
The correct answer is A. क्षैतिज रेखा (y=3) / Horizontal line (y=3). For every \(x\in[3,4\)), \(\lfloor x\rfloor=3\). So the graph on that interval is (y=3).
Step 3
Exam Tip
हर \(x\in[3,4\)) के लिए \(\lfloor x\rfloor=3\)। इसलिए आलेख उस अंतराल पर (y=3) है।
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फलन (f(x)={x}) में (x=-1.2) पर मान क्या है?
What is the value of the fractional part function (f(x)={x}) at (x=-1.2)?
#standard-functions
#fractional-part-graph
#negative-value
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A (0.8)
B (-0.2)
C (0.2)
D (-1)
Explanation opens after your attempt
Step 1
Concept
\({x}=x-\lfloor x\rfloor\), and \(\lfloor -1.2\rfloor=-2\). Hence ({-1.2}=-1.2-(-2)=0.8).
Step 2
Why this answer is correct
The correct answer is A. (0.8). \({x}=x-\lfloor x\rfloor\), and \(\lfloor -1.2\rfloor=-2\). Hence ({-1.2}=-1.2-(-2)=0.8).
Step 3
Exam Tip
\({x}=x-\lfloor x\rfloor\) और \(\lfloor -1.2\rfloor=-2\)। इसलिए ({-1.2}=-1.2-(-2)=0.8)।
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फलन (f(x)={x+1}) का (x=2.3) पर मान क्या है?
What is the value of (f(x)={x+1}) at (x=2.3)?
#standard-functions
#fractional-part-graph
#value
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A (0.3)
B (3.3)
C (0.7)
D (1.3)
Explanation opens after your attempt
Step 1
Concept
Here (x+1=3.3), and ({3.3}=0.3). The fractional part is obtained by removing the integer part.
Step 2
Why this answer is correct
The correct answer is A. (0.3). Here (x+1=3.3), and ({3.3}=0.3). The fractional part is obtained by removing the integer part.
Step 3
Exam Tip
(x+1=3.3) और ({3.3}=0.3)। भिन्नात्मक भाग पूर्णांक भाग हटाने से मिलता है।
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रेखा (y=3x-6) का (x)-अक्ष प्रतिच्छेद क्या है?
What is the (x)-intercept of the line (y=3x-6)?
#standard-functions
#linear-graph
#x-intercept
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A (x=2)
B (x=3)
C (x=-2)
D (x=6)
Explanation opens after your attempt
Step 1
Concept
On the (x)-axis, (y=0), so (3x-6=0). This gives (x=2).
Step 2
Why this answer is correct
The correct answer is A. (x=2). On the (x)-axis, (y=0), so (3x-6=0). This gives (x=2).
Step 3
Exam Tip
(x)-अक्ष पर (y=0), इसलिए (3x-6=0)। इससे (x=2) मिलता है।
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रेखा (2x+y=8) का (y)-अक्ष प्रतिच्छेद क्या है?
What is the (y)-intercept of the line (2x+y=8)?
#standard-functions
#linear-graph
#y-intercept
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A (8)
B (4)
C (2)
D (-8)
Explanation opens after your attempt
Step 1
Concept
On the (y)-axis, (x=0), so (y=8). Therefore the (y)-intercept is (8).
Step 2
Why this answer is correct
The correct answer is A. (8). On the (y)-axis, (x=0), so (y=8). Therefore the (y)-intercept is (8).
Step 3
Exam Tip
(y)-अक्ष पर (x=0), इसलिए (y=8)। अतः (y)-अक्ष प्रतिच्छेद (8) है।
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रेखा \(y=-\frac{1}{2}x+3\) की ढाल क्या है?
What is the slope of the line \(y=-\frac{1}{2}x+3\)?
#standard-functions
#linear-graph
#slope
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A \(-\frac{1}{2}\)
B (3)
C \(\frac{1}{2}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{2}\)
Step 1
Concept
In (y=mx+c), (m) is the slope. Here \(m=-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). In (y=mx+c), (m) is the slope. Here \(m=-\frac{1}{2}\).
Step 3
Exam Tip
(y=mx+c) में (m) ढाल होता है। यहां \(m=-\frac{1}{2}\) है।
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रेखा (4x-2y=6) की ढाल क्या है?
What is the slope of the line (4x-2y=6)?
#standard-functions
#linear-graph
#slope-conversion
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A (2)
B (-2)
C (3)
D \(\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
From (4x-2y=6), we get (y=2x-3). So the slope is (2).
Step 2
Why this answer is correct
The correct answer is A. (2). From (4x-2y=6), we get (y=2x-3). So the slope is (2).
Step 3
Exam Tip
(4x-2y=6) से (y=2x-3) मिलता है। इसलिए ढाल (2) है।
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फलन (f(x)=|x|-|x-2|) का (x=3) पर मान क्या है?
What is the value of (f(x)=|x|-|x-2|) at (x=3)?
#standard-functions
#modulus-graph
#value
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A (2)
B (4)
C (0)
D (-2)
Explanation opens after your attempt
Step 1
Concept
(f(3)=|3|-|1|=3-1=2). In modulus questions, substitute the inside values first.
Step 2
Why this answer is correct
The correct answer is A. (2). (f(3)=|3|-|1|=3-1=2). In modulus questions, substitute the inside values first.
Step 3
Exam Tip
(f(3)=|3|-|1|=3-1=2)। मापांक में पहले अंदर का मान लगाएं।
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फलन (f(x)=|x-1|+|x+1|) में (x=0) पर मान क्या है?
What is the value of (f(x)=|x-1|+|x+1|) at (x=0)?
#standard-functions
#modulus-graph
#evaluation
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A (2)
B (0)
C (1)
D (-2)
Explanation opens after your attempt
Step 1
Concept
(f(0)=|-1|+|1|=1+1=2). Modulus always gives a non-negative value.
Step 2
Why this answer is correct
The correct answer is A. (2). (f(0)=|-1|+|1|=1+1=2). Modulus always gives a non-negative value.
Step 3
Exam Tip
(f(0)=|-1|+|1|=1+1=2)। मापांक हमेशा गैरऋणात्मक मान देता है।
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फलन (f(x)=x-2 -6x+9) का शीर्ष कौन सा है?
What is the vertex of (f(x)=x-2 -6x+9)?
#standard-functions
#quadratic-graph
#perfect-square
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A ((3,0))
B ((-3,0))
C ((0,9))
D ((6,9))
Explanation opens after your attempt
Correct Answer
A. ((3,0))
Step 1
Concept
(x-2 -6x+9=(x-3)2 ). Therefore the vertex of the parabola is ((3,0)).
Step 2
Why this answer is correct
The correct answer is A. ((3,0)). (x-2 -6x+9=(x-3)2 ). Therefore the vertex of the parabola is ((3,0)).
Step 3
Exam Tip
(x-2 -6x+9=(x-3)2 )। इसलिए परवलय का शीर्ष ((3,0)) है।
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फलन (f(x)=x-2 +4x+7) का न्यूनतम मान क्या है?
What is the minimum value of (f(x)=x-2 +4x+7)?
#standard-functions
#quadratic-graph
#completing-square
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A (3)
B (7)
C (-3)
D (4)
Explanation opens after your attempt
Step 1
Concept
(x-2 +4x+7=(x+2)2 +3). Therefore the minimum value is (3).
Step 2
Why this answer is correct
The correct answer is A. (3). (x-2 +4x+7=(x+2)2 +3). Therefore the minimum value is (3).
Step 3
Exam Tip
(x-2 +4x+7=(x+2)2 +3)। इसलिए न्यूनतम मान (3) है।
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फलन (f(x)=x-2 -4x) के (x)-अक्ष प्रतिच्छेद कौन से हैं?
What are the (x)-intercepts of (f(x)=x-2 -4x)?
#standard-functions
#quadratic-graph
#zeros
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A (x=0) और (x=4) / (x=0) and (x=4)
B (x=-4) और (x=4) / (x=-4) and (x=4)
C (x=2) और (x=-2) / (x=2) and (x=-2)
D (x=0) और (x=-4) / (x=0) and (x=-4)
Explanation opens after your attempt
Correct Answer
A. (x=0) और (x=4) / (x=0) and (x=4)
Step 1
Concept
(x-2 -4x=x(x-4)). At (y=0), we get (x=0) or (x=4).
Step 2
Why this answer is correct
The correct answer is A. (x=0) और (x=4) / (x=0) and (x=4). (x-2 -4x=x(x-4)). At (y=0), we get (x=0) or (x=4).
Step 3
Exam Tip
(x-2 -4x=x(x-4))। (y=0) पर (x=0) या (x=4) मिलता है।
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फलन (f(x)=x-3 -8) का (x)-अक्ष प्रतिच्छेद क्या है?
What is the (x)-intercept of (f(x)=x-3 -8)?
#standard-functions
#cubic-graph
#x-intercept
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A (x=2)
B (x=-2)
C (x=8)
D (x=-8)
Explanation opens after your attempt
Step 1
Concept
On the (x)-axis, \(x^3-8=0\), so \(x^3=8\). This gives (x=2).
Step 2
Why this answer is correct
The correct answer is A. (x=2). On the (x)-axis, \(x^3-8=0\), so \(x^3=8\). This gives (x=2).
Step 3
Exam Tip
(x)-अक्ष पर \(x^3-8=0\), इसलिए \(x^3=8\)। इससे (x=2) मिलता है।
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फलन (f(x)=(x+3)3 -1) में मोड़ का केंद्रीय बिंदु कौन सा है?
What is the central inflection point of (f(x)=(x+3)3 -1)?
#standard-functions
#cubic-graph
#translation
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A ((-3,-1))
B ((3,-1))
C ((-3,1))
D ((0,-1))
Explanation opens after your attempt
Correct Answer
A. ((-3,-1))
Step 1
Concept
The central point of \(y=x^3\) is ((0,0)). The graph ((x+3)3 -1) shifts it to ((-3,-1)).
Step 2
Why this answer is correct
The correct answer is A. ((-3,-1)). The central point of \(y=x^3\) is ((0,0)). The graph ((x+3)3 -1) shifts it to ((-3,-1)).
Step 3
Exam Tip
\(y=x^3\) का केंद्रीय बिंदु ((0,0)) है। ((x+3)3 -1) इसे ((-3,-1)) पर खिसकाता है।
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फलन (f(x)=|x-2 -4|) के शून्यक कौन से हैं?
What are the zeros of (f(x)=|x-2 -4|)?
#standard-functions
#modulus-graph
#zeros
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A (x=2) और (x=-2) / (x=2) and (x=-2)
B (x=4) और (x=-4) / (x=4) and (x=-4)
C केवल (x=0) / Only (x=0)
D कोई शून्यक नहीं / No zero
Explanation opens after your attempt
Correct Answer
A. (x=2) और (x=-2) / (x=2) and (x=-2)
Step 1
Concept
A modulus is zero only when the inside expression is zero. From \(x^2-4=0\), we get \(x=\pm 2\).
Step 2
Why this answer is correct
The correct answer is A. (x=2) और (x=-2) / (x=2) and (x=-2). A modulus is zero only when the inside expression is zero. From \(x^2-4=0\), we get \(x=\pm 2\).
Step 3
Exam Tip
मापांक शून्य तभी होता है जब अंदर की राशि शून्य हो। \(x^2-4=0\) से \(x=\pm 2\) मिलता है।
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किस समीकरण का आलेख फलन का आलेख नहीं है?
Which equation does not represent the graph of a function?
#standard-functions
#vertical-line-test
#sideways-parabola
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A \(x=y^2\)
B \(y=x^2\)
C (y=|x|)
D \(y=\sqrt{x}\)
Explanation opens after your attempt
Correct Answer
A. \(x=y^2\)
Step 1
Concept
For \(x=y^2\), many positive (x)-values give two (y)-values. So it fails the vertical line test.
Step 2
Why this answer is correct
The correct answer is A. \(x=y^2\). For \(x=y^2\), many positive (x)-values give two (y)-values. So it fails the vertical line test.
Step 3
Exam Tip
\(x=y^2\) में कई धनात्मक (x)-मानों के लिए दो (y)-मान मिलते हैं। इसलिए यह ऊर्ध्व रेखा परीक्षण में असफल है।
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कौन सा रूप \(y=x^2\) के आलेख को (2) इकाई बाईं और (3) इकाई ऊपर खिसकाता है?
Which form shifts the graph of \(y=x^2\) (2) units left and (3) units up?
#standard-functions
#graph-transformation
#quadratic
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A (y=(x+2)2 +3)
B (y=(x-2)2 +3)
C (y=(x+3)2 +2)
D (y=(x-2)2 -3)
Explanation opens after your attempt
Correct Answer
A. (y=(x+2)2 +3)
Step 1
Concept
For (2) units left, use (x+2), and for (3) units up, use outside (+3). Thus the correct form is (y=(x+2)2 +3).
Step 2
Why this answer is correct
The correct answer is A. (y=(x+2)2 +3). For (2) units left, use (x+2), and for (3) units up, use outside (+3). Thus the correct form is (y=(x+2)2 +3).
Step 3
Exam Tip
बाईं ओर (2) इकाई के लिए (x+2) और ऊपर (3) इकाई के लिए बाहर (+3) होता है। इसलिए सही रूप (y=(x+2)2 +3) है।
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कौन सा फलन (y=|x|) को (4) इकाई दाईं ओर और (1) इकाई नीचे खिसकाता है?
Which function shifts (y=|x|) (4) units right and (1) unit down?
#standard-functions
#graph-transformation
#modulus
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A (y=|x-4|-1)
B (y=|x+4|-1)
C (y=|x-1|-4)
D (y=|x+4|+1)
Explanation opens after your attempt
Correct Answer
A. (y=|x-4|-1)
Step 1
Concept
For (4) units right, use (x-4), and for (1) unit down, use outside (-1). So (y=|x-4|-1) is correct.
Step 2
Why this answer is correct
The correct answer is A. (y=|x-4|-1). For (4) units right, use (x-4), and for (1) unit down, use outside (-1). So (y=|x-4|-1) is correct.
Step 3
Exam Tip
दाईं ओर (4) इकाई के लिए अंदर (x-4) और नीचे (1) इकाई के लिए बाहर (-1) होता है। इसलिए (y=|x-4|-1) सही है।
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फलन (f(x)=\frac{1}{x}) और (g(x)=\frac{1}{x}+3) के आलेखों में क्या संबंध है?
What is the relation between the graphs of (f(x)=\frac{1}{x}) and (g(x)=\frac{1}{x}+3)?
#standard-functions
#reciprocal-graph
#vertical-shift
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A (g) का आलेख (3) इकाई ऊपर है / Graph of (g) is (3) units up
B (g) का आलेख (3) इकाई नीचे है / Graph of (g) is (3) units down
C (g) का आलेख (3) इकाई दाईं ओर है / Graph of (g) is (3) units right
D (g) का आलेख (3) इकाई बाईं ओर है / Graph of (g) is (3) units left
Explanation opens after your attempt
Correct Answer
A. (g) का आलेख (3) इकाई ऊपर है / Graph of (g) is (3) units up
Step 1
Concept
Since (g(x)=f(x)+3), every (y)-value increases by (3). Hence the graph shifts (3) units up.
Step 2
Why this answer is correct
The correct answer is A. (g) का आलेख (3) इकाई ऊपर है / Graph of (g) is (3) units up. Since (g(x)=f(x)+3), every (y)-value increases by (3). Hence the graph shifts (3) units up.
Step 3
Exam Tip
(g(x)=f(x)+3), इसलिए हर (y)-मान (3) बढ़ता है। अतः आलेख (3) इकाई ऊपर खिसकता है।
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फलन (f(x)=\frac{1}{x-2}) का आलेख \(y=\frac{1}{x}\) से कैसे संबंधित है?
How is the graph of (f(x)=\frac{1}{x-2}) related to \(y=\frac{1}{x}\)?
#standard-functions
#reciprocal-graph
#horizontal-shift
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A (2) इकाई दाईं ओर / (2) units right
B (2) इकाई बाईं ओर / (2) units left
C (2) इकाई ऊपर / (2) units up
D (2) इकाई नीचे / (2) units down
Explanation opens after your attempt
Correct Answer
A. (2) इकाई दाईं ओर / (2) units right
Step 1
Concept
The inside term (x-2) shifts the graph (2) units right. Therefore the vertical asymptote becomes (x=2).
Step 2
Why this answer is correct
The correct answer is A. (2) इकाई दाईं ओर / (2) units right. The inside term (x-2) shifts the graph (2) units right. Therefore the vertical asymptote becomes (x=2).
Step 3
Exam Tip
अंदर (x-2) होने से आलेख (2) इकाई दाईं ओर खिसकता है। इसलिए ऊर्ध्व आसिम्प्टोट (x=2) हो जाता है।
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फलन (f(x)=|x-2|+|x+2|) का (x=1) पर मान क्या है?
What is the value of (f(x)=|x-2|+|x+2|) at (x=1)?
#standard-functions
#modulus-graph
#evaluation
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A (4)
B (2)
C (6)
D (0)
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Step 1
Concept
(f(1)=|-1|+|3|=1+3=4). In modulus questions, evaluate each part separately.
Step 2
Why this answer is correct
The correct answer is A. (4). (f(1)=|-1|+|3|=1+3=4). In modulus questions, evaluate each part separately.
Step 3
Exam Tip
(f(1)=|-1|+|3|=1+3=4)। मापांक वाले प्रश्नों में हर भाग अलग से निकालें।
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फलन (f(x)=|x-1|-|x+3|) में (x=2) पर मान क्या है?
What is the value of (f(x)=|x-1|-|x+3|) at (x=2)?
#standard-functions
#modulus-graph
#evaluation
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A (-4)
B (4)
C (2)
D (0)
Explanation opens after your attempt
Step 1
Concept
(f(2)=|1|-|5|=1-5=-4). Evaluate each modulus part separately.
Step 2
Why this answer is correct
The correct answer is A. (-4). (f(2)=|1|-|5|=1-5=-4). Evaluate each modulus part separately.
Step 3
Exam Tip
(f(2)=|1|-|5|=1-5=-4)। मापांक वाले प्रत्येक भाग का मान अलग निकालें।
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फलन (f(x)=\sqrt{2x+6}) का डोमेन क्या है?
What is the domain of (f(x)=\sqrt{2x+6})?
#standard-functions
#square-root-graph
#domain
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A \(x\ge -3\)
B \(x\le -3\)
C \(x\ge 3\)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge -3\)
Step 1
Concept
For a real square root, \(2x+6\ge 0\) is required. This gives \(x\ge -3\).
Step 2
Why this answer is correct
The correct answer is A. \(x\ge -3\). For a real square root, \(2x+6\ge 0\) is required. This gives \(x\ge -3\).
Step 3
Exam Tip
वास्तविक वर्गमूल के लिए \(2x+6\ge 0\) चाहिए। इससे \(x\ge -3\) मिलता है।
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फलन (f(x)=\left\lfloor\frac{x-1}{2}\right\rfloor) में (x=5.8) पर मान क्या होगा?
For (f(x)=\left\lfloor\frac{x-1}{2}\right\rfloor), what is the value at (x=5.8)?
#standard-functions
#greatest-integer-graph
#value
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A (2)
B (3)
C (2.4)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(\frac{5.8-1}{2}=2.4\) and \(\lfloor2.4\rfloor=2\). First find the inside value and then apply the greatest integer function.
Step 2
Why this answer is correct
The correct answer is A. (2). \(\frac{5.8-1}{2}=2.4\) and \(\lfloor2.4\rfloor=2\). First find the inside value and then apply the greatest integer function.
Step 3
Exam Tip
\(\frac{5.8-1}{2}=2.4\) और \(\lfloor2.4\rfloor=2\)। पहले अंदर का मान निकालकर ग्रेटेस्ट इंटीजर लगाएं।
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फलन (f(x)=-(x-1)2 +4) के (x)-अक्ष प्रतिच्छेद कौन से हैं?
What are the (x)-intercepts of (f(x)=-(x-1)2 +4)?
#standard-functions
#quadratic-graph
#x-intercepts
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A (x=-1) और (x=3) / (x=-1) and (x=3)
B (x=1) और (x=4) / (x=1) and (x=4)
C (x=-3) और (x=1) / (x=-3) and (x=1)
D (x=0) और (x=4) / (x=0) and (x=4)
Explanation opens after your attempt
Correct Answer
A. (x=-1) और (x=3) / (x=-1) and (x=3)
Step 1
Concept
Putting (y=0) gives ((x-1)2 =4). Hence \(x-1=\pm2\), so (x=-1,3).
Step 2
Why this answer is correct
The correct answer is A. (x=-1) और (x=3) / (x=-1) and (x=3). Putting (y=0) gives ((x-1)2 =4). Hence \(x-1=\pm2\), so (x=-1,3).
Step 3
Exam Tip
(y=0) रखने पर ((x-1)2 =4) मिलता है। इसलिए \(x-1=\pm2\) और (x=-1,3) हैं।
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