फलन (f(x)=\sqrt{x-2}+\sqrt{5-x}) का अधिकतम संभव प्रांत क्या है?
What is the maximum possible domain of the function (f(x)=\sqrt{x-2}+\sqrt{5-x})?
#relations-functions
#domain
#range
#real-valued-functions
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A ( [2,5] )
B ( (2,5) )
C ( \(-\infty,2]\cup[5,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( [2,5] )
Step 1
Concept
For both square roots, \(x-2\ge 0\) and \(5-x\ge 0\) must hold. In exams, take the intersection of all conditions.
Step 2
Why this answer is correct
The correct answer is A. ( [2,5] ). For both square roots, \(x-2\ge 0\) and \(5-x\ge 0\) must hold. In exams, take the intersection of all conditions.
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(x-2\ge 0\) और \(5-x\ge 0\) होना चाहिए। परीक्षा में सभी शर्तों का छेदन लें।
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फलन (f(x)=\frac{1}{\sqrt{x-2 -9}}) का प्रांत क्या है?
What is the domain of the function (f(x)=\frac{1}{\sqrt{x-2 -9}})?
#relations-functions
#domain
#radical
#denominator
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A ( \(-\infty,-3\)\cup\(3,\infty\) )
B ( \(-\infty,-3]\cup[3,\infty\) )
C ( [-3,3] )
D \( \mathbb{R}\setminus{-3,3} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,-3\)\cup\(3,\infty\) )
Step 1
Concept
The square root is in the denominator, so \(x^2-9>0\) is required. A denominator can never be zero.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3\)\cup\(3,\infty\) ). The square root is in the denominator, so \(x^2-9>0\) is required. A denominator can never be zero.
Step 3
Exam Tip
हर में वर्गमूल है इसलिए \(x^2-9>0\) होना चाहिए। हर में शून्य कभी स्वीकार्य नहीं होता।
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फलन (f(x)=\sqrt{4-x-2 }) का परिसर क्या है?
What is the range of the function (f(x)=\sqrt{4-x-2 })?
#relations-functions
#range
#radical
#semicircle
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A ( [0,2] )
B ( [-2,2] )
C ( [0,4] )
D ( \(-\infty,2] \)
Explanation opens after your attempt
Correct Answer
A. ( [0,2] )
Step 1
Concept
This is the upper semicircle and the maximum of \(4-x^2\) is (4). A square root value is never negative.
Step 2
Why this answer is correct
The correct answer is A. ( [0,2] ). This is the upper semicircle and the maximum of \(4-x^2\) is (4). A square root value is never negative.
Step 3
Exam Tip
यह ऊपरी अर्धवृत्त है और \(4-x^2\) का अधिकतम (4) है। वर्गमूल का मान ऋणात्मक नहीं होता।
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यदि (f(x)=\frac{x+1}{x-2}) है, तो (f) का परिसर क्या है?
If (f(x)=\frac{x+1}{x-2}), what is the range of (f)?
#relations-functions
#range
#rational-function
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A \( \mathbb{R}\setminus{1} \)
B \( \mathbb{R}\setminus{2} \)
C \( \mathbb{R} \)
D \( \mathbb{R}\setminus{-1} \)
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{1} \)
Step 1
Concept
From \(y=\frac{x+1}{x-2}\), \(x=\frac{2y+1}{y-1}\), so \(y\ne 1\). For a linear fractional function, isolate the impossible value.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{1} \). From \(y=\frac{x+1}{x-2}\), \(x=\frac{2y+1}{y-1}\), so \(y\ne 1\). For a linear fractional function, isolate the impossible value.
Step 3
Exam Tip
\(y=\frac{x+1}{x-2}\) से \(x=\frac{2y+1}{y-1}\), इसलिए \(y\ne 1\)। रैखिक भिन्न फलन में असंभव मान अलग करें।
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फलन (f(x)=|x-3|+|x+1|) का न्यूनतम मान क्या है?
What is the minimum value of (f(x)=|x-3|+|x+1|)?
#relations-functions
#range
#modulus
#minimum
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A (4)
B (2)
C (0)
D (6)
Explanation opens after your attempt
Step 1
Concept
The sum of distances from (-1) and (3) is minimized at (4). On the interval ([-1,3]), the function stays constant.
Step 2
Why this answer is correct
The correct answer is A. (4). The sum of distances from (-1) and (3) is minimized at (4). On the interval ([-1,3]), the function stays constant.
Step 3
Exam Tip
दो बिंदुओं (-1) और (3) से दूरी का योग न्यूनतम (4) होता है। बीच के अंतराल ([-1,3]) पर फलन स्थिर रहता है।
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फलन (f(x)=\frac{1}{1+|x|}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{1+|x|})?
#relations-functions
#range
#modulus
#rational
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A ( (0,1] )
B ( [0,1] )
C \( [1,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( (0,1] )
Step 1
Concept
The denominator \(1+|x|\ge 1\), so the maximum value is (1). As (x) grows, the value approaches (0) but never becomes (0).
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). The denominator \(1+|x|\ge 1\), so the maximum value is (1). As (x) grows, the value approaches (0) but never becomes (0).
Step 3
Exam Tip
हर \(1+|x|\ge 1\) है, इसलिए अधिकतम मान (1) है। (x) बढ़ने पर मान (0) के पास जाता है पर (0) नहीं बनता।
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फलन (f(x)=\sqrt{x-2 +4x+7}) का परिसर क्या है?
What is the range of (f(x)=\sqrt{x-2 +4x+7})?
#relations-functions
#range
#quadratic
#radical
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A \( [\sqrt{3},\infty\) )
B \( [3,\infty\) )
C \( [0,\infty\) )
D ( \(-\infty,\infty\) )
Explanation opens after your attempt
Correct Answer
A. \( [\sqrt{3},\infty\) )
Step 1
Concept
Since (x-2 +4x+7=(x+2)2 +3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \( [\sqrt{3},\infty\) ). Since (x-2 +4x+7=(x+2)2 +3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).
Step 3
Exam Tip
(x-2 +4x+7=(x+2)2 +3), इसलिए अंदर का न्यूनतम (3) है। वर्गमूल लेने पर न्यूनतम \(\sqrt{3}\) मिलेगा।
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फलन (f(x)=\frac{\sqrt{x+1}}{x-2 -4}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{\sqrt{x+1}}{x-2 -4})?
#relations-functions
#domain
#radical
#rational
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A \( [-1,\infty\)\setminus{2} )
B \( [-1,\infty\)\setminus{-2,2} )
C ( \(-1,\infty\) )
D \( \mathbb{R}\setminus{-2,2} \)
Explanation opens after your attempt
Correct Answer
A. \( [-1,\infty\)\setminus{2} )
Step 1
Concept
The square root needs \(x\ge -1\), and the denominator needs \(x\ne \pm 2\). Since (-2) is already outside the domain, only (2) is removed.
Step 2
Why this answer is correct
The correct answer is A. \( [-1,\infty\)\setminus{2} ). The square root needs \(x\ge -1\), and the denominator needs \(x\ne \pm 2\). Since (-2) is already outside the domain, only (2) is removed.
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge -1\) और हर के लिए \(x\ne \pm 2\) चाहिए। (-2) पहले से प्रांत में नहीं है, इसलिए केवल (2) हटेगा।
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फलन (f(x)=\sqrt{x-1}+\sqrt{x+2}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{x-1}+\sqrt{x+2})?
#relations-functions
#domain
#radical
#intersection
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A \( [1,\infty\) )
B \( [-2,\infty\) )
C ( [-2,1] )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [1,\infty\) )
Step 1
Concept
Both square roots need \(x\ge 1\) and \(x\ge -2\). The stricter combined condition is \(x\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. \( [1,\infty\) ). Both square roots need \(x\ge 1\) and \(x\ge -2\). The stricter combined condition is \(x\ge 1\).
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(x\ge 1\) और \(x\ge -2\) चाहिए। संयुक्त शर्त अधिक कठोर \(x\ge 1\) है।
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फलन (f(x)=\frac{2x-3}{x+4}) के लिए कौन सा मान परिसर में नहीं आएगा?
Which value will not occur in the range of (f(x)=\frac{2x-3}{x+4})?
#relations-functions
#range
#linear-fractional
#missing-value
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A (2)
B (-4)
C (-3)
D (0)
Explanation opens after your attempt
Step 1
Concept
From \(y=\frac{2x-3}{x+4}\), \(x=\frac{-3-4y}{y-2}\), so (y=2) is impossible. In such fractions, the ratio of leading coefficients often gives the missing value.
Step 2
Why this answer is correct
The correct answer is A. (2). From \(y=\frac{2x-3}{x+4}\), \(x=\frac{-3-4y}{y-2}\), so (y=2) is impossible. In such fractions, the ratio of leading coefficients often gives the missing value.
Step 3
Exam Tip
\(y=\frac{2x-3}{x+4}\) से \(x=\frac{-3-4y}{y-2}\), इसलिए (y=2) असंभव है। अनुपात में प्रमुख गुणांकों का अनुपात अक्सर छूटा मान देता है।
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फलन (f(x)=x-2 -6x+11) का परिसर क्या है, जब \(x\in[1,5]\)?
What is the range of (f(x)=x-2 -6x+11) when \(x\in[1,5]\)?
#relations-functions
#range
#quadratic
#closed-interval
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A ( [2,6] )
B ( [1,5] )
C ( [0,6] )
D ( [2,11] )
Explanation opens after your attempt
Correct Answer
A. ( [2,6] )
Step 1
Concept
Since (f(x)=(x-3)2 +2), the minimum is (2). At the endpoints (x=1) and (x=5), the maximum is (6).
Step 2
Why this answer is correct
The correct answer is A. ( [2,6] ). Since (f(x)=(x-3)2 +2), the minimum is (2). At the endpoints (x=1) and (x=5), the maximum is (6).
Step 3
Exam Tip
(f(x)=(x-3)2 +2), इसलिए न्यूनतम (2) है। सिरा (x=1) और (x=5) पर अधिकतम (6) मिलता है।
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फलन (f(x)=|x-2|-|x+2|) का परिसर क्या है?
What is the range of (f(x)=|x-2|-|x+2|)?
#relations-functions
#range
#modulus
#distance
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A ( [-4,4] )
B ( (-4,4) )
C ( [0,4] )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( [-4,4] )
Step 1
Concept
The difference of distances from (2) and (-2) stays between (-4) and (4). Removing signs interval-wise is the safe method.
Step 2
Why this answer is correct
The correct answer is A. ( [-4,4] ). The difference of distances from (2) and (-2) stays between (-4) and (4). Removing signs interval-wise is the safe method.
Step 3
Exam Tip
बिंदुओं (2) और (-2) से दूरी के अंतर का मान (-4) से (4) तक रहता है। अंतरालों पर चिह्न हटाकर जांचना सुरक्षित तरीका है।
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फलन (f(x)=\frac{1}{x-2 +4x+5}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{x-2 +4x+5})?
#relations-functions
#range
#rational
#quadratic
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A ( (0,1] )
B \( [1,\infty\) )
C ( \(-\infty,1] \)
D ( \(0,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( (0,1] )
Step 1
Concept
The denominator ((x+2)2 +1\ge 1), so the maximum value is (1). As the denominator grows, the value approaches (0).
Step 2
Why this answer is correct
The correct answer is A. ( (0,1] ). The denominator ((x+2)2 +1\ge 1), so the maximum value is (1). As the denominator grows, the value approaches (0).
Step 3
Exam Tip
हर ((x+2)2 +1\ge 1) है, इसलिए अधिकतम मान (1) है। हर बड़ा होने पर मान (0) के पास जाता है।
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फलन (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{x}+\frac{1}{\sqrt{4-x}})?
#relations-functions
#domain
#radical
#denominator
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A ( [0,4) )
B ( [0,4] )
C ( (0,4) )
D ( \(-\infty,4\) )
Explanation opens after your attempt
Correct Answer
A. ( [0,4) )
Step 1
Concept
For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).
Step 2
Why this answer is correct
The correct answer is A. ( [0,4) ). For \(\sqrt{x}\), \(x\ge 0\), and for the denominator square root, (4-x>0) is needed. Together, \(0\le x<4\).
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर वाले वर्गमूल के लिए (4-x>0) चाहिए। दोनों मिलाकर \(0\le x<4\) मिलता है।
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फलन (f(x)=\frac{x-2 -4}{x-2}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{x-2 -4}{x-2})?
#relations-functions
#domain
#rational
#removable-discontinuity
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A \( \mathbb{R}\setminus{2} \)
B \( \mathbb{R} \)
C \( \mathbb{R}\setminus{-2,2} \)
D ( \(-\infty,2\) )
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{2} \)
Step 1
Concept
Even after simplification, the original denominator requires \(x-2\ne 0\). Do not add the cancelled point back into the domain.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{2} \). Even after simplification, the original denominator requires \(x-2\ne 0\). Do not add the cancelled point back into the domain.
Step 3
Exam Tip
सरलीकरण के बाद भी मूल हर में \(x-2\ne 0\) रहना चाहिए। हटे हुए बिंदु को प्रांत में वापस न जोड़ें।
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फलन (f(x)=\frac{x-2 -4}{x-2}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2 -4}{x-2})?
#relations-functions
#range
#rational
#hole
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A \( \mathbb{R}\setminus{4} \)
B \( \mathbb{R}\setminus{2} \)
C \( \mathbb{R} \)
D \( [4,\infty\) )
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{4} \)
Step 1
Concept
For \(x\ne 2\), the function equals (x+2). Thus the value (4), which would occur at (x=2), is not in the range.
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{4} \). For \(x\ne 2\), the function equals (x+2). Thus the value (4), which would occur at (x=2), is not in the range.
Step 3
Exam Tip
\(x\ne 2\) पर फलन (x+2) के बराबर है। इसलिए (x=2) से मिलने वाला मान (4) परिसर में नहीं आता।
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फलन (f(x)=\sqrt{|x|-3}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{|x|-3})?
#relations-functions
#domain
#modulus
#radical
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A ( \(-\infty,-3]\cup[3,\infty\) )
B ( [-3,3] )
C \( [3,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,-3]\cup[3,\infty\) )
Step 1
Concept
The square root needs \(|x|-3\ge 0\). Hence \(|x|\ge 3\), so \(x\le -3\) or \(x\ge 3\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-3]\cup[3,\infty\) ). The square root needs \(|x|-3\ge 0\). Hence \(|x|\ge 3\), so \(x\le -3\) or \(x\ge 3\).
Step 3
Exam Tip
वर्गमूल के लिए \(|x|-3\ge 0\) चाहिए। इसलिए \(|x|\ge 3\), यानी \(x\le -3\) या \(x\ge 3\)।
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यदि (f(x)=\frac{3}{2-\sqrt{x}}), तो प्रांत क्या है?
If (f(x)=\frac{3}{2-\sqrt{x}}), what is the domain?
#relations-functions
#domain
#radical
#rational
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A \( [0,\infty\)\setminus{4} )
B ( [0,4) )
C ( \(0,\infty\) )
D \( \mathbb{R}\setminus{4} \)
Explanation opens after your attempt
Correct Answer
A. \( [0,\infty\)\setminus{4} )
Step 1
Concept
For \(\sqrt{x}\), \(x\ge 0\), and for the denominator, \(2-\sqrt{x}\ne 0\) is needed. Hence (x=4) is removed.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\)\setminus{4} ). For \(\sqrt{x}\), \(x\ge 0\), and for the denominator, \(2-\sqrt{x}\ne 0\) is needed. Hence (x=4) is removed.
Step 3
Exam Tip
\(\sqrt{x}\) के लिए \(x\ge 0\) और हर के लिए \(2-\sqrt{x}\ne 0\) चाहिए। इसलिए (x=4) हटाया जाएगा।
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फलन (f(x)=\frac{1}{|x|-2}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{1}{|x|-2})?
#relations-functions
#domain
#modulus
#rational
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A \( \mathbb{R}\setminus{-2,2} \)
B \( \mathbb{R}\setminus{2} \)
C ( \(-\infty,-2\)\cup\(2,\infty\) )
D ( [-2,2] )
Explanation opens after your attempt
Correct Answer
A. \( \mathbb{R}\setminus{-2,2} \)
Step 1
Concept
The denominator must not be zero, so \(|x|-2\ne 0\). This gives \(x\ne -2\) and \(x\ne 2\).
Step 2
Why this answer is correct
The correct answer is A. \( \mathbb{R}\setminus{-2,2} \). The denominator must not be zero, so \(|x|-2\ne 0\). This gives \(x\ne -2\) and \(x\ne 2\).
Step 3
Exam Tip
हर शून्य नहीं होना चाहिए, इसलिए \(|x|-2\ne 0\)। इससे \(x\ne -2\) और \(x\ne 2\) मिलता है।
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फलन (f(x)=\frac{|x|}{1+|x|}) का परिसर क्या है?
What is the range of (f(x)=\frac{|x|}{1+|x|})?
#relations-functions
#range
#modulus
#rational
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A ( [0,1) )
B ( (0,1) )
C ( [0,1] )
D ( (-1,1) )
Explanation opens after your attempt
Correct Answer
A. ( [0,1) )
Step 1
Concept
Since \(|x|\ge 0\), the minimum is (0). The value approaches (1) but never equals it.
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). Since \(|x|\ge 0\), the minimum is (0). The value approaches (1) but never equals it.
Step 3
Exam Tip
\(|x|\ge 0\) होने से न्यूनतम (0) है। मान (1) के पास जाता है पर बराबर नहीं होता।
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यदि (f(x)=x+\frac{1}{x}) और \(x\ne 0\), तो (f) का परिसर क्या है?
If (f(x)=x+\frac{1}{x}) and \(x\ne 0\), what is the range of (f)?
#relations-functions
#range
#inequality
#rational
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A ( \(-\infty,-2]\cup[2,\infty\) )
B \( [2,\infty\) )
C \( \mathbb{R}\setminus{0} \)
D ( (-2,2) )
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,-2]\cup[2,\infty\) )
Step 1
Concept
For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\) ). For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.
Step 3
Exam Tip
धनात्मक (x) पर मान कम से कम (2) और ऋणात्मक (x) पर अधिक से अधिक (-2) होता है। दोनों स्थितियों को अलग जांचें।
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फलन (f(x)=\sqrt{x-2 -4x+4}) का सरल परिसर क्या है?
What is the simple range of (f(x)=\sqrt{x-2 -4x+4})?
#relations-functions
#range
#radical
#modulus
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A \( [0,\infty\) )
B ( \(0,\infty\) )
C \( [2,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [0,\infty\) )
Step 1
Concept
(\sqrt{x-2 -4x+4}=\sqrt{(x-2)2 }=|x-2|). Its minimum is (0), and it takes all non-negative values.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). (\sqrt{x-2 -4x+4}=\sqrt{(x-2)2 }=|x-2|). Its minimum is (0), and it takes all non-negative values.
Step 3
Exam Tip
(\sqrt{x-2 -4x+4}=\sqrt{(x-2)2 }=|x-2|)। इसका न्यूनतम (0) है और यह सभी अनऋणात्मक मान लेता है।
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फलन (f(x)=\frac{1}{\sqrt{16-x-2 }}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{\sqrt{16-x-2 }})?
#relations-functions
#range
#radical
#reciprocal
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A \( \left[\frac{1}{4},\infty\right\) )
B ( \left\(0,\frac{1}{4}\right] \)
C ( \(0,\infty\) )
D \( [4,\infty\) )
Explanation opens after your attempt
Correct Answer
A. \( \left[\frac{1}{4},\infty\right\) )
Step 1
Concept
The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Step 2
Why this answer is correct
The correct answer is A. \( \left[\frac{1}{4},\infty\right\) ). The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Step 3
Exam Tip
\(\sqrt{16-x^2}\) का अधिकतम (4) है, इसलिए भिन्न का न्यूनतम \(\frac{1}{4}\) है। किनारों के पास हर (0) की ओर जाता है।
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यदि (f(x)=\lfloor x\rfloor) और \(x\in[1,4\)), तो (f) का परिसर क्या है?
If (f(x)=\lfloor x\rfloor) and \(x\in[1,4\)), what is the range of (f)?
#relations-functions
#range
#greatest-integer-function
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A ( {1,2,3} )
B ( {1,2,3,4} )
C ( [1,4) )
D ( {0,1,2,3} )
Explanation opens after your attempt
Correct Answer
A. ( {1,2,3} )
Step 1
Concept
The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.
Step 2
Why this answer is correct
The correct answer is A. ( {1,2,3} ). The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.
Step 3
Exam Tip
महत्तम पूर्णांक फलन (x) से बड़ा नहीं सबसे बड़ा पूर्णांक देता है। (4) शामिल नहीं है, इसलिए मान (4) नहीं आएगा।
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यदि (f(x)={x}) भिन्नांश फलन है और \(x\in\mathbb{R}\), तो परिसर क्या है?
If (f(x)={x}) is the fractional part function and \(x\in\mathbb{R}\), what is the range?
#relations-functions
#range
#fractional-part-function
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A ( [0,1) )
B ( (0,1) )
C ( [0,1] )
D \( \mathbb{Z} \)
Explanation opens after your attempt
Correct Answer
A. ( [0,1) )
Step 1
Concept
The fractional part is always at least (0) and less than (1). At integer (x), the value is (0).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). The fractional part is always at least (0) and less than (1). At integer (x), the value is (0).
Step 3
Exam Tip
भिन्नांश भाग हमेशा (0) से बड़ा या बराबर और (1) से छोटा होता है। पूर्णांक (x) पर मान (0) मिलता है।
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फलन (f(x)=\sqrt{\frac{x-1}{x+2}}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{\frac{x-1}{x+2}})?
#relations-functions
#domain
#radical
#rational-inequality
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A ( \(-\infty,-2\)\cup[1,\infty) )
B ( [-2,1] )
C ( (-2,1) )
D \( \mathbb{R}\setminus{-2} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,-2\)\cup[1,\infty) )
Step 1
Concept
The expression inside the square root must satisfy \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (x<-2) or \(x\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-2\)\cup[1,\infty) ). The expression inside the square root must satisfy \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (x<-2) or \(x\ge 1\).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x-1}{x+2}\ge 0\) और \(x\ne -2\) चाहिए। संकेत सारणी से (x<-2) या \(x\ge 1\) मिलता है।
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फलन (f(x)=\sqrt{\frac{x+3}{5-x}}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{\frac{x+3}{5-x}})?
#relations-functions
#domain
#radical
#rational-inequality
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A ( [-3,5) )
B ( (-\infty,-3]\cup\(5,\infty\) )
C ( [-3,5] )
D ( (-3,5) )
Explanation opens after your attempt
Correct Answer
A. ( [-3,5) )
Step 1
Concept
The square root needs \(\frac{x+3}{5-x}\ge 0\) and \(x\ne 5\). Sign checking gives ([-3,5)).
Step 2
Why this answer is correct
The correct answer is A. ( [-3,5) ). The square root needs \(\frac{x+3}{5-x}\ge 0\) and \(x\ne 5\). Sign checking gives ([-3,5)).
Step 3
Exam Tip
वर्गमूल के अंदर \(\frac{x+3}{5-x}\ge 0\) और \(x\ne 5\) चाहिए। संकेत जांच से ([-3,5)) मिलता है।
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यदि (f(x)=\sqrt{x-a}) का प्रांत \([7,\infty\)) है, तो (a) का मान क्या है?
If the domain of (f(x)=\sqrt{x-a}) is \([7,\infty\)), what is the value of (a)?
#relations-functions
#domain
#parameter
#radical
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A (7)
B (-7)
C (0)
D (14)
Explanation opens after your attempt
Step 1
Concept
For \(\sqrt{x-a}\), \(x-a\ge 0\), so \(x\ge a\). Comparing with the given domain gives (a=7).
Step 2
Why this answer is correct
The correct answer is A. (7). For \(\sqrt{x-a}\), \(x-a\ge 0\), so \(x\ge a\). Comparing with the given domain gives (a=7).
Step 3
Exam Tip
\(\sqrt{x-a}\) के लिए \(x-a\ge 0\), यानी \(x\ge a\)। दिए गए प्रांत से (a=7) है।
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यदि (f(x)=\frac{1}{x-a}) का प्रांत \(\mathbb{R}\setminus{5}\) है, तो (a) क्या है?
If the domain of (f(x)=\frac{1}{x-a}) is \(\mathbb{R}\setminus{5}\), what is (a)?
#relations-functions
#domain
#parameter
#rational
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A (5)
B (-5)
C (0)
D (1)
Explanation opens after your attempt
Step 1
Concept
The denominator becomes zero at (x=a), so that value is removed. The removed value is (5), hence (a=5).
Step 2
Why this answer is correct
The correct answer is A. (5). The denominator becomes zero at (x=a), so that value is removed. The removed value is (5), hence (a=5).
Step 3
Exam Tip
हर शून्य होने पर (x=a) हटता है। दिए गए हटे हुए मान से (a=5) है।
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यदि (f(x)=ax-2 +4) का परिसर \([4,\infty\)) है, तो (a) के लिए कौन सी शर्त सही है?
If the range of (f(x)=ax-2 +4) is \([4,\infty\)), which condition on (a) is correct?
#relations-functions
#range
#parameter
#quadratic
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A (a>0)
B (a<0)
C (a=0)
D \(a\ne 0\)
Explanation opens after your attempt
Step 1
Concept
An upward-opening parabola gives \([4,\infty\)) only when (a>0). If (a=0), the range is only ({4}).
Step 2
Why this answer is correct
The correct answer is A. (a>0). An upward-opening parabola gives \([4,\infty\)) only when (a>0). If (a=0), the range is only ({4}).
Step 3
Exam Tip
ऊपर खुलने वाला परवलय तभी \([4,\infty\)) देता है जब (a>0)। (a=0) पर परिसर केवल ({4}) होगा।
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फलन (f(x)=\frac{x-2 +2x+5}{x-2 +2x+2}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2 +2x+5}{x-2 +2x+2})?
#relations-functions
#range
#rational
#substitution
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A ( (1,4] )
B ( [1,4] )
C \( [4,\infty\) )
D ( \(-\infty,1\) )
Explanation opens after your attempt
Correct Answer
A. ( (1,4] )
Step 1
Concept
Put (t=(x+1)2 \ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). Hence the values are greater than (1) and up to (4).
Step 2
Why this answer is correct
The correct answer is A. ( (1,4] ). Put (t=(x+1)2 \ge 0), then \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). Hence the values are greater than (1) and up to (4).
Step 3
Exam Tip
मान (t=(x+1)2 \ge 0) रखें, तो \(f=\frac{t+4}{t+1}=1+\frac{3}{t+1}\)। इसलिए मान (1) से बड़े और (4) तक हैं।
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फलन (f(x)=\sqrt{2x-x-2 }+3) का परिसर क्या है?
What is the range of (f(x)=\sqrt{2x-x-2 }+3)?
#relations-functions
#range
#radical
#quadratic
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A ( [3,4] )
B ( [0,1] )
C ( [2,4] )
D \( [3,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( [3,4] )
Step 1
Concept
Since (2x-x-2 =1-(x-1)2 ), its value lies from (0) to (1). After square root and adding (3), the range is ([3,4]).
Step 2
Why this answer is correct
The correct answer is A. ( [3,4] ). Since (2x-x-2 =1-(x-1)2 ), its value lies from (0) to (1). After square root and adding (3), the range is ([3,4]).
Step 3
Exam Tip
(2x-x-2 =1-(x-1)2 ), जिसका मान (0) से (1) तक है। वर्गमूल के बाद (3) जोड़ने से परिसर ([3,4]) है।
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फलन (f(x)=\frac{1}{x-2 -6x+10}) का अधिकतम मान क्या है?
What is the maximum value of (f(x)=\frac{1}{x-2 -6x+10})?
#relations-functions
#range
#maximum
#rational
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A (1)
B \(\frac{1}{10}\)
C \(\frac{1}{6}\)
D (10)
Explanation opens after your attempt
Step 1
Concept
The denominator (x-2 -6x+10=(x-3)2 +1) has minimum (1). Therefore the maximum fraction is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). The denominator (x-2 -6x+10=(x-3)2 +1) has minimum (1). Therefore the maximum fraction is (1).
Step 3
Exam Tip
हर (x-2 -6x+10=(x-3)2 +1) का न्यूनतम (1) है। इसलिए भिन्न का अधिकतम (1) होगा।
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यदि (f(x)=\sqrt{x+4}+\sqrt{4-x}), तो (f) का प्रांत क्या है?
If (f(x)=\sqrt{x+4}+\sqrt{4-x}), what is the domain of (f)?
#relations-functions
#domain
#radical
#intersection
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A ( [-4,4] )
B ( (-4,4) )
C ( \(-\infty,-4]\cup[4,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( [-4,4] )
Step 1
Concept
The conditions are \(x+4\ge 0\) and \(4-x\ge 0\). Their intersection is ([-4,4]).
Step 2
Why this answer is correct
The correct answer is A. ( [-4,4] ). The conditions are \(x+4\ge 0\) and \(4-x\ge 0\). Their intersection is ([-4,4]).
Step 3
Exam Tip
शर्तें \(x+4\ge 0\) और \(4-x\ge 0\) हैं। इनका छेदन ([-4,4]) है।
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यदि (f(x)=\sqrt{x+4}+\sqrt{4-x}), तो (f) का परिसर क्या है?
If (f(x)=\sqrt{x+4}+\sqrt{4-x}), what is the range of (f)?
#relations-functions
#range
#radical
#symmetry
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A \( [2\sqrt{2},4] \)
B ( [0,4] )
C \( [4,\infty\) )
D ( [2,4] )
Explanation opens after your attempt
Correct Answer
A. \( [2\sqrt{2},4] \)
Step 1
Concept
At the endpoints the value is \(2\sqrt{2}\), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \( [2\sqrt{2},4] \). At the endpoints the value is \(2\sqrt{2}\), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.
Step 3
Exam Tip
सिरों पर मान \(2\sqrt{2}\) और (x=0) पर अधिकतम (4) है। सममिति वाले वर्गमूल फलनों में मध्य बिंदु भी जांचें।
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फलन (f(x)=\log_{10}\(4-x^2\)) का प्रांत क्या है?
What is the domain of (f(x)=\log_{10}\(4-x^2\))?
#relations-functions
#domain
#logarithm
#inequality
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A ( (-2,2) )
B ( [-2,2] )
C ( \(-\infty,-2\)\cup\(2,\infty\) )
D \( \mathbb{R}\setminus{-2,2} \)
Explanation opens after your attempt
Correct Answer
A. ( (-2,2) )
Step 1
Concept
The logarithm input must satisfy \(4-x^2>0\). Thus \(x^2<4\), so (-2<x<2).
Step 2
Why this answer is correct
The correct answer is A. ( (-2,2) ). The logarithm input must satisfy \(4-x^2>0\). Thus \(x^2<4\), so (-2<x<2).
Step 3
Exam Tip
लघुगणक के अंदर \(4-x^2>0\) होना चाहिए। इसलिए \(x^2<4\), यानी (-2<x<2)।
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फलन (f(x)=\log_{3}(x-1)+\log_{3}(5-x)) का प्रांत क्या है?
What is the domain of (f(x)=\log_{3}(x-1)+\log_{3}(5-x))?
#relations-functions
#domain
#logarithm
#intersection
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A ( (1,5) )
B ( [1,5] )
C ( \(-\infty,1\)\cup\(5,\infty\) )
D \( \mathbb{R}\setminus{1,5} \)
Explanation opens after your attempt
Correct Answer
A. ( (1,5) )
Step 1
Concept
Both logarithm inputs must be positive. Hence (x>1) and (x<5), so (1<x<5).
Step 2
Why this answer is correct
The correct answer is A. ( (1,5) ). Both logarithm inputs must be positive. Hence (x>1) and (x<5), so (1<x<5).
Step 3
Exam Tip
दोनों लघुगणकों के अंदर धनात्मक होना चाहिए। इसलिए (x>1) और (x<5), अर्थात (1<x<5)।
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फलन (f(x)=\frac{1}{\log_{2}x}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{1}{\log_{2}x})?
#relations-functions
#domain
#logarithm
#denominator
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A ( \(0,\infty\)\setminus{1} )
B ( \(0,\infty\) )
C \( \mathbb{R}\setminus{1} \)
D ( \(1,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( \(0,\infty\)\setminus{1} )
Step 1
Concept
The logarithm needs (x>0), and the denominator needs \(\log_{2}x\ne 0\). Since \(\log_{2}x=0\) at (x=1), remove (1).
Step 2
Why this answer is correct
The correct answer is A. ( \(0,\infty\)\setminus{1} ). The logarithm needs (x>0), and the denominator needs \(\log_{2}x\ne 0\). Since \(\log_{2}x=0\) at (x=1), remove (1).
Step 3
Exam Tip
लघुगणक के लिए (x>0) और हर के लिए \(\log_{2}x\ne 0\) चाहिए। \(\log_{2}x=0\) पर (x=1) हटेगा।
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फलन (f(x)=|2x-1|+3) का परिसर क्या है?
What is the range of (f(x)=|2x-1|+3)?
#relations-functions
#range
#modulus
#minimum
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A \( [3,\infty\) )
B ( \(3,\infty\) )
C \( [1,\infty\) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. \( [3,\infty\) )
Step 1
Concept
Since \(|2x-1|\ge 0\), the least value is (3). It occurs at \(x=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( [3,\infty\) ). Since \(|2x-1|\ge 0\), the least value is (3). It occurs at \(x=\frac{1}{2}\).
Step 3
Exam Tip
\(|2x-1|\ge 0\), इसलिए सबसे छोटा मान (3) है। यह \(x=\frac{1}{2}\) पर मिलता है।
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फलन (f(x)=5-\sqrt{x-2 +1}) का परिसर क्या है?
What is the range of (f(x)=5-\sqrt{x-2 +1})?
#relations-functions
#range
#radical
#maximum
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A ( \(-\infty,4] \)
B \( [4,\infty\) )
C ( \(-\infty,5] \)
D ( [0,4] )
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,4] \)
Step 1
Concept
Since \(\sqrt{x^2+1}\ge 1\), \(5-\sqrt{x^2+1}\le 4\). As (x) grows, the value decreases without bound.
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,4] \). Since \(\sqrt{x^2+1}\ge 1\), \(5-\sqrt{x^2+1}\le 4\). As (x) grows, the value decreases without bound.
Step 3
Exam Tip
\(\sqrt{x^2+1}\ge 1\), इसलिए \(5-\sqrt{x^2+1}\le 4\)। (x) बड़ा होने पर मान असीम रूप से घटता है।
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यदि (f(x)=\frac{x}{|x|}), तो (f) का प्रांत और परिसर कौन सा है?
If (f(x)=\frac{x}{|x|}), which are the domain and range of (f)?
#relations-functions
#domain
#range
#signum
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A प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( {-1,1} ) / domain \( \mathbb{R}\setminus{0} \), range ( {-1,1} )
B प्रांत \( \mathbb{R} \), परिसर ( {-1,1} ) / domain \( \mathbb{R} \), range ( {-1,1} )
C प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( [0,1] ) / domain \( \mathbb{R}\setminus{0} \), range ( [0,1] )
D प्रांत \( \mathbb{R} \), परिसर \( \mathbb{R} \) / domain \( \mathbb{R} \), range \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( {-1,1} ) / domain \( \mathbb{R}\setminus{0} \), range ( {-1,1} )
Step 1
Concept
At (x=0), the denominator becomes (0). For positive (x), the value is (1), and for negative (x), it is (-1).
Step 2
Why this answer is correct
The correct answer is A. प्रांत \( \mathbb{R}\setminus{0} \), परिसर ( {-1,1} ) / domain \( \mathbb{R}\setminus{0} \), range ( {-1,1} ). At (x=0), the denominator becomes (0). For positive (x), the value is (1), and for negative (x), it is (-1).
Step 3
Exam Tip
(x=0) पर हर (0) हो जाता है। धनात्मक (x) पर मान (1) और ऋणात्मक (x) पर (-1) है।
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फलन (f(x)=\sqrt{x-2 -1}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{x-2 -1})?
#relations-functions
#domain
#radical
#inequality
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A ( \(-\infty,-1]\cup[1,\infty\) )
B ( [-1,1] )
C ( (-1,1) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,-1]\cup[1,\infty\) )
Step 1
Concept
The square root needs \(x^2-1\ge 0\). Thus \(|x|\ge 1\), so \(x\le -1\) or \(x\ge 1\).
Step 2
Why this answer is correct
The correct answer is A. ( \(-\infty,-1]\cup[1,\infty\) ). The square root needs \(x^2-1\ge 0\). Thus \(|x|\ge 1\), so \(x\le -1\) or \(x\ge 1\).
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-1\ge 0\) चाहिए। इसलिए \(|x|\ge 1\), यानी \(x\le -1\) या \(x\ge 1\)।
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फलन (f(x)=\sqrt{x-2 -1}) का परिसर क्या है?
What is the range of (f(x)=\sqrt{x-2 -1})?
#relations-functions
#range
#radical
#inequality
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A \( [0,\infty\) )
B \( [1,\infty\) )
C ( \(-\infty,\infty\) )
D ( \(-\infty,0] \)
Explanation opens after your attempt
Correct Answer
A. \( [0,\infty\) )
Step 1
Concept
On the domain, the minimum of \(x^2-1\) is (0). As (x) grows, the square root takes all larger non-negative values.
Step 2
Why this answer is correct
The correct answer is A. \( [0,\infty\) ). On the domain, the minimum of \(x^2-1\) is (0). As (x) grows, the square root takes all larger non-negative values.
Step 3
Exam Tip
प्रांत में \(x^2-1\) का न्यूनतम (0) है। (x) बड़ा होने पर वर्गमूल सभी बड़े अनऋणात्मक मान लेता है।
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फलन (f(x)=\frac{4x}{x-2 +4}) का परिसर क्या है?
What is the range of (f(x)=\frac{4x}{x-2 +4})?
#relations-functions
#range
#rational
#inequality
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A ( [-1,1] )
B ( (-1,1) )
C ( [0,1] )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( [-1,1] )
Step 1
Concept
\(|4x|\le x^2+4\) because ((x-2)2 \ge 0) and ((x+2)2 \ge 0). Hence values lie from (-1) to (1), and both endpoints occur.
Step 2
Why this answer is correct
The correct answer is A. ( [-1,1] ). \(|4x|\le x^2+4\) because ((x-2)2 \ge 0) and ((x+2)2 \ge 0). Hence values lie from (-1) to (1), and both endpoints occur.
Step 3
Exam Tip
\(|4x|\le x^2+4\) क्योंकि ((x-2)2 \ge 0) और ((x+2)2 \ge 0)। इसलिए मान (-1) से (1) तक हैं और दोनों मिलते हैं।
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फलन (f(x)=\frac{\sqrt{(x-2)(6-x)}}{x-4}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{\sqrt{(x-2)(6-x)}}{x-4})?
#relations-functions
#domain
#radical
#denominator
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A ( [2,4)\cup(4,6] )
B ( [2,6] )
C \((2,4)\cup(4,6)\)
D \( \mathbb{R}\setminus{4} \)
Explanation opens after your attempt
Correct Answer
A. ( [2,4)\cup(4,6] )
Step 1
Concept
For the square root, ((x-2)(6-x)\ge 0), so \(x\in[2,6]\). Because of the denominator, (x=4) is removed.
Step 2
Why this answer is correct
The correct answer is A. ( [2,4)\cup(4,6] ). For the square root, ((x-2)(6-x)\ge 0), so \(x\in[2,6]\). Because of the denominator, (x=4) is removed.
Step 3
Exam Tip
वर्गमूल के लिए ((x-2)(6-x)\ge 0), इसलिए \(x\in[2,6]\) है। हर के कारण (x=4) हटेगा।
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फलन (f(x)=\frac{x-2 }{x-2 -4}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2 }{x-2 -4})?
#relations-functions
#range
#rational
#quadratic
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A ( (-\infty,0]\cup\(1,\infty\) )
B \( \mathbb{R}\setminus{1} \)
C \( [0,\infty\) )
D ( \(-\infty,1\) )
Explanation opens after your attempt
Correct Answer
A. ( (-\infty,0]\cup\(1,\infty\) )
Step 1
Concept
Put \(t=x^2\ge 0\) and \(t\ne 4\). Then \(f=\frac{t}{t-4}\), giving (\(-\infty,0]\) for \(t\in[0,4\)) and (\(1,\infty\)) for (t>4).
Step 2
Why this answer is correct
The correct answer is A. ( (-\infty,0]\cup\(1,\infty\) ). Put \(t=x^2\ge 0\) and \(t\ne 4\). Then \(f=\frac{t}{t-4}\), giving (\(-\infty,0]\) for \(t\in[0,4\)) and (\(1,\infty\)) for (t>4).
Step 3
Exam Tip
मान \(t=x^2\ge 0\) रखें और \(t\ne 4\)। तब \(f=\frac{t}{t-4}\), जिससे \(t\in[0,4\)) पर (\(-\infty,0]\) और (t>4) पर (\(1,\infty\)) मिलता है।
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फलन (f(x)=\log_{e}\left\(\frac{x-2}{7-x}\right\)) का प्रांत क्या है?
What is the domain of (f(x)=\log_{e}\left\(\frac{x-2}{7-x}\right\))?
#relations-functions
#domain
#logarithm
#rational-inequality
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A ( (2,7) )
B ( \(-\infty,2\)\cup\(7,\infty\) )
C \( \mathbb{R}\setminus{2,7} \)
D ( [2,7] )
Explanation opens after your attempt
Correct Answer
A. ( (2,7) )
Step 1
Concept
The logarithm input must satisfy \(\frac{x-2}{7-x}>0\). A sign check gives only (2<x<7).
Step 2
Why this answer is correct
The correct answer is A. ( (2,7) ). The logarithm input must satisfy \(\frac{x-2}{7-x}>0\). A sign check gives only (2<x<7).
Step 3
Exam Tip
लघुगणक के अंदर \(\frac{x-2}{7-x}>0\) होना चाहिए। संकेत जांच से केवल (2<x<7) मिलता है।
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फलन (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2}) का परिसर क्या है?
What is the range of (f(x)=\frac{\sqrt{x+1}}{\sqrt{x+1}+2})?
#relations-functions
#range
#radical
#rational
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A ( [0,1) )
B ( (0,1) )
C ( [0,1] )
D \( [1,\infty\) )
Explanation opens after your attempt
Correct Answer
A. ( [0,1) )
Step 1
Concept
Put \(t=\sqrt{x+1}\ge 0\), then \(f=\frac{t}{t+2}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1) but never equals (1).
Step 2
Why this answer is correct
The correct answer is A. ( [0,1) ). Put \(t=\sqrt{x+1}\ge 0\), then \(f=\frac{t}{t+2}\). At (t=0), the value is (0), and as \(t\to\infty\), it approaches (1) but never equals (1).
Step 3
Exam Tip
\(t=\sqrt{x+1}\ge 0\) रखने पर \(f=\frac{t}{t+2}\)। (t=0) पर मान (0) और \(t\to\infty\) पर मान (1) के पास जाता है पर (1) नहीं होता।
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फलन (f(x)=3+\frac{1}{x-2 +2x+2}) का परिसर क्या है?
What is the range of (f(x)=3+\frac{1}{x-2 +2x+2})?
#relations-functions
#range
#rational
#quadratic
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A ( (3,4] )
B ( [3,4] )
C \( [4,\infty\) )
D ( (0,4] )
Explanation opens after your attempt
Correct Answer
A. ( (3,4] )
Step 1
Concept
The denominator (x-2 +2x+2=(x+1)2 +1\ge 1). So (\frac{1}{x-2 +2x+2}\in(0,1]), hence the range is ((3,4]).
Step 2
Why this answer is correct
The correct answer is A. ( (3,4] ). The denominator (x-2 +2x+2=(x+1)2 +1\ge 1). So (\frac{1}{x-2 +2x+2}\in(0,1]), hence the range is ((3,4]).
Step 3
Exam Tip
हर (x-2 +2x+2=(x+1)2 +1\ge 1) है। इसलिए (\frac{1}{x-2 +2x+2}\in(0,1]), अतः परिसर ((3,4]) है।
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फलन (f(x)=\sqrt{1-|x-2|}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{1-|x-2|})?
#relations-functions
#domain
#modulus
#radical
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A ( [1,3] )
B ( \(-\infty,1]\cup[3,\infty\) )
C ( (1,3) )
D \( \mathbb{R} \)
Explanation opens after your attempt
Correct Answer
A. ( [1,3] )
Step 1
Concept
The square root needs \(1-|x-2|\ge 0\). Thus \(|x-2|\le 1\), so \(1\le x\le 3\).
Step 2
Why this answer is correct
The correct answer is A. ( [1,3] ). The square root needs \(1-|x-2|\ge 0\). Thus \(|x-2|\le 1\), so \(1\le x\le 3\).
Step 3
Exam Tip
वर्गमूल के लिए \(1-|x-2|\ge 0\) चाहिए। इसलिए \(|x-2|\le 1\), यानी \(1\le x\le 3\)।
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