फलन (f(x)=\sqrt{3x-12}) का प्रांत क्या है?
What is the domain of the function (f(x)=\sqrt{3x-12})?
#relations-functions
#domain
#square-root
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A \(x\ge 4\)
B (x>4)
C \(x\le 4\)
D \(x\in\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(x\ge 4\)
Step 1
Concept
The expression inside the square root must satisfy \(3x-12\ge 0\), so \(x\ge 4\). In exams, always keep the radicand non-negative.
Step 2
Why this answer is correct
The correct answer is A. \(x\ge 4\). The expression inside the square root must satisfy \(3x-12\ge 0\), so \(x\ge 4\). In exams, always keep the radicand non-negative.
Step 3
Exam Tip
वर्गमूल के अंदर \(3x-12\ge 0\) होना चाहिए, इसलिए \(x\ge 4\)। परीक्षा में वर्गमूल के लिए अंदर का भाग हमेशा अशून्य या धनात्मक रखें।
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फलन (f(x)=\sqrt{25-x-2 }) का प्रांत चुनिए।
Choose the domain of (f(x)=\sqrt{25-x-2 }).
#relations-functions
#domain
#inequality
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A ([-5,5])
B (\(-\infty,-5]\cup[5,\infty\))
C ((-5,5))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. ([-5,5])
Step 1
Concept
Here \(25-x^2\ge 0\), so \(x^2\le 25\) and \(x\in[-5,5]\). In such questions, solve the inequality carefully.
Step 2
Why this answer is correct
The correct answer is A. ([-5,5]). Here \(25-x^2\ge 0\), so \(x^2\le 25\) and \(x\in[-5,5]\). In such questions, solve the inequality carefully.
Step 3
Exam Tip
यहां \(25-x^2\ge 0\), इसलिए \(x^2\le 25\) और \(x\in[-5,5]\)। ऐसे प्रश्नों में असमानता को सावधानी से हल करें।
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फलन (f(x)=\frac{1}{\sqrt{x-2}}) का प्रांत क्या होगा?
What will be the domain of (f(x)=\frac{1}{\sqrt{x-2}})?
#relations-functions
#domain
#denominator
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A (x>2)
B \(x\ge 2\)
C (x<2)
D \(x\ne 2\)
Explanation opens after your attempt
Step 1
Concept
The square root is in the denominator, so (x-2>0) is required. Hence the domain is (\(2,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. (x>2). The square root is in the denominator, so (x-2>0) is required. Hence the domain is (\(2,\infty\)).
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए (x-2>0) चाहिए। इसलिए प्रांत (\(2,\infty\)) है।
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फलन (f(x)=\frac{x+1}{x-2 -9}) का प्रांत ज्ञात कीजिए।
Find the domain of (f(x)=\frac{x+1}{x-2 -9}).
#relations-functions
#domain
#rational-function
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A \(\mathbb{R}-{-3,3}\)
B \(\mathbb{R}-{3}\)
C \(\mathbb{R}-{-3}\)
D ({-3,3})
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-3,3}\)
Step 1
Concept
The denominator must satisfy \(x^2-9\ne 0\), so \(x\ne -3,3\). In exams, remove values that make the denominator zero.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-3,3}\). The denominator must satisfy \(x^2-9\ne 0\), so \(x\ne -3,3\). In exams, remove values that make the denominator zero.
Step 3
Exam Tip
हर \(x^2-9\ne 0\) होना चाहिए, इसलिए \(x\ne -3,3\)। परीक्षा में हर को शून्य बनाने वाले मान हटाएं।
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फलन (f(x)=\frac{\sqrt{x+1}}{x-4}) का प्रांत कौन सा है?
Which is the domain of (f(x)=\frac{\sqrt{x+1}}{x-4})?
#relations-functions
#domain
#combined-condition
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A \([-1,\infty\)-{4})
B (\(-1,\infty\)-{4})
C ([-1,4))
D \(\mathbb{R}-{4}\)
Explanation opens after your attempt
Correct Answer
A. \([-1,\infty\)-{4})
Step 1
Concept
For the square root \(x+1\ge 0\) and for the denominator \(x\ne 4\) are required. Thus the intersection is \([-1,\infty\)-{4}).
Step 2
Why this answer is correct
The correct answer is A. \([-1,\infty\)-{4}). For the square root \(x+1\ge 0\) and for the denominator \(x\ne 4\) are required. Thus the intersection is \([-1,\infty\)-{4}).
Step 3
Exam Tip
वर्गमूल के लिए \(x+1\ge 0\) और हर के लिए \(x\ne 4\) चाहिए। इसलिए दोनों शर्तों का प्रतिच्छेद \([-1,\infty\)-{4}) है।
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फलन (f(x)=(x-2)2 +5) का परिसर क्या है?
What is the range of (f(x)=(x-2)2 +5)?
#relations-functions
#range
#quadratic
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A \([5,\infty\))
B (\(5,\infty\))
C (\(-\infty,5]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([5,\infty\))
Step 1
Concept
Since ((x-2)2 \ge 0), the minimum value is (5). Hence the range is \([5,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([5,\infty\)). Since ((x-2)2 \ge 0), the minimum value is (5). Hence the range is \([5,\infty\)).
Step 3
Exam Tip
क्योंकि ((x-2)2 \ge 0), न्यूनतम मान (5) है। अतः परिसर \([5,\infty\)) है।
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फलन (f(x)=7-(x+3)2 ) का परिसर चुनिए।
Choose the range of (f(x)=7-(x+3)2 ).
#relations-functions
#range
#maximum
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A (\(-\infty,7]\)
B \([7,\infty\))
C (\(-\infty,7\))
D \(\mathbb{R}-{7}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,7]\)
Step 1
Concept
Since ((x+3)2 \ge 0), (7-(x+3)2 \le 7). The maximum value is (7).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,7]\). Since ((x+3)2 \ge 0), (7-(x+3)2 \le 7). The maximum value is (7).
Step 3
Exam Tip
क्योंकि ((x+3)2 \ge 0), (7-(x+3)2 \le 7)। अधिकतम मान (7) है।
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यदि (f(x)=\frac{2x+1}{x-3}), तो (f) का प्रांत क्या है?
If (f(x)=\frac{2x+1}{x-3}), what is the domain of (f)?
#relations-functions
#domain
#rational
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A \(\mathbb{R}-{3}\)
B \(\mathbb{R}-{-3}\)
C (\(3,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{3}\)
Step 1
Concept
The denominator \(x-3\ne 0\), so \(x\ne 3\). For rational functions, check the denominator condition first.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{3}\). The denominator \(x-3\ne 0\), so \(x\ne 3\). For rational functions, check the denominator condition first.
Step 3
Exam Tip
हर \(x-3\ne 0\), इसलिए \(x\ne 3\)। रैशनल फलन में हर की शर्त सबसे पहले देखें।
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फलन (f(x)=\frac{2x+1}{x-3}) का परिसर क्या है?
What is the range of (f(x)=\frac{2x+1}{x-3})?
#relations-functions
#range
#rational
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A \(\mathbb{R}-{2}\)
B \(\mathbb{R}-{3}\)
C (\(-\infty,2]\)
D \([2,\infty\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{2}\)
Step 1
Concept
If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{2}\). If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).
Step 3
Exam Tip
यदि \(y=\frac{2x+1}{x-3}\), तो \(x=\frac{3y+1}{y-2}\), अतः \(y\ne 2\)। इसलिए परिसर \(\mathbb{R}-{2}\) है।
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फलन (f(x)=\frac{1}{x-2 +4}) का परिसर ज्ञात कीजिए।
Find the range of (f(x)=\frac{1}{x-2 +4}).
#relations-functions
#range
#fraction
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A (\(0,\frac{1}{4}]\)
B \([0,\frac{1}{4}]\)
C (\(-\infty,\frac{1}{4}]\)
D \([\frac{1}{4},\infty\))
Explanation opens after your attempt
Correct Answer
A. (\(0,\frac{1}{4}]\)
Step 1
Concept
Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.
Step 2
Why this answer is correct
The correct answer is A. (\(0,\frac{1}{4}]\). Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.
Step 3
Exam Tip
क्योंकि \(x^2+4\ge 4\), इसलिए (0<f(x)\le \frac{1}{4})। शून्य प्राप्त नहीं होता।
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फलन (f(x)=\sqrt{9-(x-1)2 }) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{9-(x-1)2 })?
#relations-functions
#domain
#square-root-shift
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A ([-2,4])
B ([-4,2])
C (\(-\infty,-2]\cup[4,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. ([-2,4])
Step 1
Concept
The condition (9-(x-1)2 \ge 0) gives ((x-1)2 \le 9). Hence \(x\in[-2,4]\).
Step 2
Why this answer is correct
The correct answer is A. ([-2,4]). The condition (9-(x-1)2 \ge 0) gives ((x-1)2 \le 9). Hence \(x\in[-2,4]\).
Step 3
Exam Tip
शर्त (9-(x-1)2 \ge 0) से ((x-1)2 \le 9) मिलता है। इसलिए \(x\in[-2,4]\)।
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फलन (f(x)=\sqrt{9-(x-1)2 }) का परिसर चुनिए।
Choose the range of (f(x)=\sqrt{9-(x-1)2 }).
#relations-functions
#range
#square-root
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A ([0,3])
B ([-3,3])
C ((0,3])
D ([0,9])
Explanation opens after your attempt
Correct Answer
A. ([0,3])
Step 1
Concept
A square root is never negative and the maximum inside value is (9). Hence the range is ([0,3]).
Step 2
Why this answer is correct
The correct answer is A. ([0,3]). A square root is never negative and the maximum inside value is (9). Hence the range is ([0,3]).
Step 3
Exam Tip
वर्गमूल का मान ऋणात्मक नहीं होता और अंदर का अधिकतम मान (9) है। इसलिए परिसर ([0,3]) है।
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यदि (f(x)=|x-4|+2), तो इसका परिसर क्या है?
If (f(x)=|x-4|+2), what is its range?
#relations-functions
#range
#modulus
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A \([2,\infty\))
B (\(2,\infty\))
C (\(-\infty,2]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([2,\infty\))
Step 1
Concept
Since \(|x-4|\ge 0\), the minimum value is (2). Hence the range is \([2,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. \([2,\infty\)). Since \(|x-4|\ge 0\), the minimum value is (2). Hence the range is \([2,\infty\)).
Step 3
Exam Tip
क्योंकि \(|x-4|\ge 0\), न्यूनतम मान (2) है। अतः परिसर \([2,\infty\)) है।
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फलन (f(x)=\frac{1}{|x|+1}) का परिसर कौन सा है?
Which is the range of (f(x)=\frac{1}{|x|+1})?
#relations-functions
#range
#modulus-fraction
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A ((0,1])
B ([0,1])
C \([1,\infty\))
D (\(-\infty,1]\)
Explanation opens after your attempt
Correct Answer
A. ((0,1])
Step 1
Concept
Since \(|x|+1\ge 1\), (0<f(x)\le 1). The value (0) is only a limiting value and is not attained.
Step 2
Why this answer is correct
The correct answer is A. ((0,1]). Since \(|x|+1\ge 1\), (0<f(x)\le 1). The value (0) is only a limiting value and is not attained.
Step 3
Exam Tip
क्योंकि \(|x|+1\ge 1\), इसलिए (0<f(x)\le 1)। (0) केवल सीमा मान है, प्राप्त नहीं होता।
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फलन (f(x)=\frac{x-2 -1}{x-2 +1}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2 -1}{x-2 +1})?
#relations-functions
#range
#quadratic-rational
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A ([-1,1))
B ((-1,1])
C ((-1,1))
D ([-1,1])
Explanation opens after your attempt
Correct Answer
A. ([-1,1))
Step 1
Concept
Let \(t=x^2\ge 0\), then \(f=\frac{t-1}{t+1}\). At (t=0), (-1) occurs and (1) is not attained.
Step 2
Why this answer is correct
The correct answer is A. ([-1,1)). Let \(t=x^2\ge 0\), then \(f=\frac{t-1}{t+1}\). At (t=0), (-1) occurs and (1) is not attained.
Step 3
Exam Tip
मान लें \(t=x^2\ge 0\), तब \(f=\frac{t-1}{t+1}\)। (t=0) पर (-1) मिलता है और (1) प्राप्त नहीं होता।
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फलन (f(x)=\sqrt{x-2 -16}) का प्रांत ज्ञात कीजिए।
Find the domain of (f(x)=\sqrt{x-2 -16}).
#relations-functions
#domain
#radical-inequality
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A (\(-\infty,-4]\cup[4,\infty\))
B ([-4,4])
C ((-4,4))
D \(\mathbb{R}-{-4,4}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-4]\cup[4,\infty\))
Step 1
Concept
For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). 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 \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).
Step 3
Exam Tip
वर्गमूल के लिए \(x^2-16\ge 0\), इसलिए \(x^2\ge 16\)। अतः \(x\le -4\) या \(x\ge 4\)।
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फलन (f(x)=\sqrt{x-2 -16}) का परिसर चुनिए।
Choose the range of (f(x)=\sqrt{x-2 -16}).
#relations-functions
#range
#radical
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A \([0,\infty\))
B (\(0,\infty\))
C \([4,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([0,\infty\))
Step 1
Concept
On the domain, \(x^2-16\ge 0\) and the minimum square-root value is (0). For large (|x|), the value grows without bound.
Step 2
Why this answer is correct
The correct answer is A. \([0,\infty\)). On the domain, \(x^2-16\ge 0\) and the minimum square-root value is (0). For large (|x|), the value grows without bound.
Step 3
Exam Tip
प्रांत में \(x^2-16\ge 0\) और वर्गमूल का न्यूनतम मान (0) है। बड़े (|x|) पर मान असीमित बढ़ता है।
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फलन (f(x)=\log_2\(x^2-5x+6\)) का प्रांत क्या है?
What is the domain of (f(x)=\log_2\(x^2-5x+6\))?
#relations-functions
#domain
#logarithm
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A (\(-\infty,2\)\cup\(3,\infty\))
B ((2,3))
C ([2,3])
D \(\mathbb{R}-{2,3}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,2\)\cup\(3,\infty\))
Step 1
Concept
For a logarithm, \(x^2-5x+6>0\) is needed. Since ((x-2)(x-3)>0), (x<2) or (x>3).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,2\)\cup\(3,\infty\)). For a logarithm, \(x^2-5x+6>0\) is needed. Since ((x-2)(x-3)>0), (x<2) or (x>3).
Step 3
Exam Tip
लघुगणक के लिए \(x^2-5x+6>0\) चाहिए। ((x-2)(x-3)>0), इसलिए (x<2) या (x>3)।
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फलन (f(x)=\log_5(4-x)) का प्रांत कौन सा है?
Which is the domain of (f(x)=\log_5(4-x))?
#relations-functions
#domain
#log
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A (\(-\infty,4\))
B (\(4,\infty\))
C (\(-\infty,4]\)
D \(\mathbb{R}-{4}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,4\))
Step 1
Concept
The logarithm argument must satisfy (4-x>0). Therefore (x<4).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,4\)). The logarithm argument must satisfy (4-x>0). Therefore (x<4).
Step 3
Exam Tip
लघुगणक का आर्गुमेंट (4-x>0) होना चाहिए। इसलिए (x<4)।
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फलन (f(x)=\frac{1}{\log_3 x}) का प्रांत क्या है?
What is the domain of (f(x)=\frac{1}{\log_3 x})?
#relations-functions
#domain
#log-denominator
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A (\(0,\infty\)-{1})
B (\(0,\infty\))
C \(\mathbb{R}-{1}\)
D (\(1,\infty\))
Explanation opens after your attempt
Correct Answer
A. (\(0,\infty\)-{1})
Step 1
Concept
For the logarithm (x>0), and for the denominator \(\log_3 x\ne 0\), so \(x\ne 1\). Keep both conditions together.
Step 2
Why this answer is correct
The correct answer is A. (\(0,\infty\)-{1}). For the logarithm (x>0), and for the denominator \(\log_3 x\ne 0\), so \(x\ne 1\). Keep both conditions together.
Step 3
Exam Tip
लघुगणक के लिए (x>0) और हर के लिए \(\log_3 x\ne 0\), यानी \(x\ne 1\)। दोनों शर्तें साथ रखें।
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यदि (f(x)=\sqrt{\frac{x-1}{x+2}}), तो प्रांत कौन सा है?
If (f(x)=\sqrt{\frac{x-1}{x+2}}), which is the domain?
#relations-functions
#domain
#sign-chart
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A (\(-\infty,-2\)\cup[1,\infty))
B ((-2,1])
C ([-2,1])
D \(\mathbb{R}-{-2}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-2\)\cup[1,\infty))
Step 1
Concept
The condition is \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (\(-\infty,-2\)\cup[1,\infty)).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-2\)\cup[1,\infty)). The condition is \(\frac{x-1}{x+2}\ge 0\) and \(x\ne -2\). A sign chart gives (\(-\infty,-2\)\cup[1,\infty)).
Step 3
Exam Tip
शर्त \(\frac{x-1}{x+2}\ge 0\) और \(x\ne -2\) है। साइन चार्ट से (\(-\infty,-2\)\cup[1,\infty)) मिलता है।
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फलन (f(x)=\sqrt{\frac{5-x}{x+1}}) का प्रांत ज्ञात कीजिए।
Find the domain of (f(x)=\sqrt{\frac{5-x}{x+1}}).
#relations-functions
#domain
#rational-radical
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A ((-1,5])
B ([-1,5])
C (\(-\infty,-1\)\cup[5,\infty))
D (\(-\infty,5]\)
Explanation opens after your attempt
Correct Answer
A. ((-1,5])
Step 1
Concept
The condition is \(\frac{5-x}{x+1}\ge 0\) and \(x\ne -1\). Sign testing gives ((-1,5]).
Step 2
Why this answer is correct
The correct answer is A. ((-1,5]). The condition is \(\frac{5-x}{x+1}\ge 0\) and \(x\ne -1\). Sign testing gives ((-1,5]).
Step 3
Exam Tip
शर्त \(\frac{5-x}{x+1}\ge 0\) और \(x\ne -1\) है। साइन परीक्षण से ((-1,5]) मिलता 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
#complete-square
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A ((0,1])
B \([1,\infty\))
C (\(0,\infty\))
D ([0,1])
Explanation opens after your attempt
Correct Answer
A. ((0,1])
Step 1
Concept
The denominator (x-2 -4x+5=(x-2)2 +1\ge 1). Hence (0<f(x)\le 1).
Step 2
Why this answer is correct
The correct answer is A. ((0,1]). The denominator (x-2 -4x+5=(x-2)2 +1\ge 1). Hence (0<f(x)\le 1).
Step 3
Exam Tip
हर (x-2 -4x+5=(x-2)2 +1\ge 1)। इसलिए (0<f(x)\le 1)।
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फलन (f(x)=\frac{x}{x-2 +1}) का परिसर चुनिए।
Choose the range of (f(x)=\frac{x}{x-2 +1}).
#relations-functions
#range
#discriminant-method
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A \([-\frac{1}{2},\frac{1}{2}]\)
B ((-1,1))
C \([-\frac{1}{4},\frac{1}{4}]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([-\frac{1}{2},\frac{1}{2}]\)
Step 1
Concept
If \(y=\frac{x}{x^2+1}\), then the discriminant of \(yx^2-x+y=0\) must satisfy \(1-4y^2\ge 0\). Thus \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \([-\frac{1}{2},\frac{1}{2}]\). If \(y=\frac{x}{x^2+1}\), then the discriminant of \(yx^2-x+y=0\) must satisfy \(1-4y^2\ge 0\). Thus \(-\frac{1}{2}\le y\le \frac{1}{2}\).
Step 3
Exam Tip
यदि \(y=\frac{x}{x^2+1}\), तो \(yx^2-x+y=0\) में विविक्तकर \(1-4y^2\ge 0\) चाहिए। अतः \(-\frac{1}{2}\le y\le \frac{1}{2}\)।
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यदि (f(x)=x+\frac{1}{x}), (x>0), तो (f) का परिसर क्या है?
If (f(x)=x+\frac{1}{x}), (x>0), what is the range of (f)?
#relations-functions
#range
#am-gm
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A \([2,\infty\))
B (\(2,\infty\))
C (\(-\infty,-2]\cup[2,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. \([2,\infty\))
Step 1
Concept
For (x>0), \(x+\frac{1}{x}\ge 2\). Equality occurs at (x=1).
Step 2
Why this answer is correct
The correct answer is A. \([2,\infty\)). For (x>0), \(x+\frac{1}{x}\ge 2\). Equality occurs at (x=1).
Step 3
Exam Tip
(x>0) के लिए \(x+\frac{1}{x}\ge 2\) होता है। बराबरी (x=1) पर आती है।
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यदि (f(x)=x+\frac{4}{x}), (x<0), तो (f) का परिसर चुनिए।
If (f(x)=x+\frac{4}{x}), (x<0), choose the range of (f).
#relations-functions
#range
#negative-domain
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A (\(-\infty,-4]\)
B \([4,\infty\))
C (\(-\infty,-2]\)
D \(\mathbb{R}-{0}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-4]\)
Step 1
Concept
For negative (x), put (x=-t), (t>0), then (f=-\(t+\frac{4}{t}\)\le -4). Hence the range is (\(-\infty,-4]\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,-4]\). For negative (x), put (x=-t), (t>0), then (f=-\(t+\frac{4}{t}\)\le -4). Hence the range is (\(-\infty,-4]\).
Step 3
Exam Tip
ऋणात्मक (x) के लिए (x=-t), (t>0), तब (f=-\(t+\frac{4}{t}\)\le -4)। इसलिए परिसर (\(-\infty,-4]\) है।
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फलन (f(x)=\frac{x-2}{x+5}) का परिसर ज्ञात कीजिए।
Find the range of (f(x)=\frac{x-2}{x+5}).
#relations-functions
#range
#linear-fractional
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A \(\mathbb{R}-{1}\)
B \(\mathbb{R}-{-5}\)
C (\(-\infty,1]\)
D \([1,\infty\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{1}\)
Step 1
Concept
If \(y=\frac{x-2}{x+5}\), then \(x=\frac{-5y-2}{y-1}\), so \(y\ne 1\). Hence the range is \(\mathbb{R}-{1}\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{1}\). If \(y=\frac{x-2}{x+5}\), then \(x=\frac{-5y-2}{y-1}\), so \(y\ne 1\). Hence the range is \(\mathbb{R}-{1}\).
Step 3
Exam Tip
यदि \(y=\frac{x-2}{x+5}\), तो \(x=\frac{-5y-2}{y-1}\), इसलिए \(y\ne 1\)। अतः परिसर \(\mathbb{R}-{1}\) है।
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फलन (f(x)=\sqrt{x+2}+\sqrt{6-x}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{x+2}+\sqrt{6-x})?
#relations-functions
#domain
#two-radicals
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A ([-2,6])
B (\(-\infty,-2]\cup[6,\infty\))
C ((-2,6))
D ([2,6])
Explanation opens after your attempt
Correct Answer
A. ([-2,6])
Step 1
Concept
For both square roots, \(x+2\ge 0\) and \(6-x\ge 0\) are required. Hence \(x\in[-2,6]\).
Step 2
Why this answer is correct
The correct answer is A. ([-2,6]). For both square roots, \(x+2\ge 0\) and \(6-x\ge 0\) are required. Hence \(x\in[-2,6]\).
Step 3
Exam Tip
दोनों वर्गमूलों के लिए \(x+2\ge 0\) और \(6-x\ge 0\) चाहिए। इसलिए \(x\in[-2,6]\)।
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फलन (f(x)=\sqrt{x+2}+\sqrt{6-x}) का अधिकतम मान क्या है?
What is the maximum value of (f(x)=\sqrt{x+2}+\sqrt{6-x})?
#relations-functions
#range
#maximum-radical
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A (4)
B \(2\sqrt{2}\)
C \(2\sqrt{3}\)
D (8)
Explanation opens after your attempt
Step 1
Concept
On ([-2,6]), the sum is symmetric and maximum occurs at (x=2). Then the value is \(\sqrt{4}+\sqrt{4}=4\).
Step 2
Why this answer is correct
The correct answer is A. (4). On ([-2,6]), the sum is symmetric and maximum occurs at (x=2). Then the value is \(\sqrt{4}+\sqrt{4}=4\).
Step 3
Exam Tip
प्रांत ([-2,6]) पर योग सममित है और अधिकतम (x=2) पर मिलता है। तब मान \(\sqrt{4}+\sqrt{4}=4\) है।
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फलन (f(x)=\frac{1}{\sqrt{4-x-2 }}) का प्रांत चुनिए।
Choose the domain of (f(x)=\frac{1}{\sqrt{4-x-2 }}).
#relations-functions
#domain
#strict-inequality
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A ((-2,2))
B ([-2,2])
C (\(-\infty,-2\)\cup\(2,\infty\))
D \(\mathbb{R}-{-2,2}\)
Explanation opens after your attempt
Correct Answer
A. ((-2,2))
Step 1
Concept
The square root is in the denominator, so \(4-x^2>0\) is required. This gives (-2<x<2).
Step 2
Why this answer is correct
The correct answer is A. ((-2,2)). The square root is in the denominator, so \(4-x^2>0\) is required. This gives (-2<x<2).
Step 3
Exam Tip
हर में वर्गमूल है, इसलिए \(4-x^2>0\) चाहिए। इससे (-2<x<2) मिलता है।
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फलन (f(x)=\frac{1}{\sqrt{4-x-2 }}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{\sqrt{4-x-2 }})?
#relations-functions
#range
#reciprocal-radical
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A \([\frac{1}{2},\infty\))
B (\(0,\frac{1}{2}]\)
C (\(\frac{1}{2},\infty\))
D \([0,\infty\))
Explanation opens after your attempt
Correct Answer
A. \([\frac{1}{2},\infty\))
Step 1
Concept
The maximum value of \(\sqrt{4-x^2}\) is (2), so the minimum value of (f) is \(\frac{1}{2}\). Near the endpoints, the value grows without bound.
Step 2
Why this answer is correct
The correct answer is A. \([\frac{1}{2},\infty\)). The maximum value of \(\sqrt{4-x^2}\) is (2), so the minimum value of (f) is \(\frac{1}{2}\). Near the endpoints, the value grows without bound.
Step 3
Exam Tip
\(\sqrt{4-x^2}\) का अधिकतम मान (2) है, इसलिए (f) का न्यूनतम मान \(\frac{1}{2}\) है। किनारों के पास मान असीमित बढ़ता है।
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फलन (f(x)=\sqrt{x-1}+\frac{1}{x-5}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{x-1}+\frac{1}{x-5})?
#relations-functions
#domain
#mixed-function
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A \([1,\infty\)-{5})
B (\(1,\infty\)-{5})
C ([1,5))
D \(\mathbb{R}-{5}\)
Explanation opens after your attempt
Correct Answer
A. \([1,\infty\)-{5})
Step 1
Concept
For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).
Step 2
Why this answer is correct
The correct answer is A. \([1,\infty\)-{5}). For the square root \(x\ge 1\), and for the denominator \(x\ne 5\) are needed. Hence the domain is \([1,\infty\)-{5}).
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge 1\) और हर के लिए \(x\ne 5\) चाहिए। इसलिए प्रांत \([1,\infty\)-{5}) है।
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फलन (f(x)=\frac{\sqrt{x-3}}{\sqrt{7-x}}) का प्रांत चुनिए।
Choose the domain of (f(x)=\frac{\sqrt{x-3}}{\sqrt{7-x}}).
#relations-functions
#domain
#two-square-roots
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A ([3,7))
B ((3,7])
C ([3,7])
D (\(-\infty,3]\cup[7,\infty\))
Explanation opens after your attempt
Correct Answer
A. ([3,7))
Step 1
Concept
For the numerator square root \(x-3\ge 0\), and for the denominator square root (7-x>0). Thus \(x\in[3,7\)).
Step 2
Why this answer is correct
The correct answer is A. ([3,7)). For the numerator square root \(x-3\ge 0\), and for the denominator square root (7-x>0). Thus \(x\in[3,7\)).
Step 3
Exam Tip
ऊपर के वर्गमूल के लिए \(x-3\ge 0\) और हर के वर्गमूल के लिए (7-x>0) चाहिए। अतः \(x\in[3,7\))।
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यदि (f(x)=x-2 -6x+11), तो (f) का न्यूनतम मान क्या है?
If (f(x)=x-2 -6x+11), what is the minimum value of (f)?
#relations-functions
#range
#minimum
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A (2)
B (3)
C (11)
D (0)
Explanation opens after your attempt
Step 1
Concept
Since (x-2 -6x+11=(x-3)2 +2), the minimum value is (2). Completing the square is the fastest method here.
Step 2
Why this answer is correct
The correct answer is A. (2). Since (x-2 -6x+11=(x-3)2 +2), the minimum value is (2). Completing the square is the fastest method here.
Step 3
Exam Tip
(x-2 -6x+11=(x-3)2 +2), इसलिए न्यूनतम मान (2) है। पूर्ण वर्ग बनाना ऐसे प्रश्नों में सबसे तेज विधि है।
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फलन (f(x)=-2x-2 +8x+1) का परिसर क्या है?
What is the range of (f(x)=-2x-2 +8x+1)?
#relations-functions
#range
#parabola
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A (\(-\infty,9]\)
B \([9,\infty\))
C (\(-\infty,1]\)
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,9]\)
Step 1
Concept
(f(x)=-2(x-2)2 +9), so the maximum value is (9). A downward-opening parabola has range (\(-\infty,9]\).
Step 2
Why this answer is correct
The correct answer is A. (\(-\infty,9]\). (f(x)=-2(x-2)2 +9), so the maximum value is (9). A downward-opening parabola has range (\(-\infty,9]\).
Step 3
Exam Tip
(f(x)=-2(x-2)2 +9), इसलिए अधिकतम मान (9) है। नीचे की ओर खुलने वाले परवलय का परिसर (\(-\infty,9]\) होता है।
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यदि (f(x)=\frac{3}{x-2 -2x+2}+1), तो (f) का परिसर क्या है?
If (f(x)=\frac{3}{x-2 -2x+2}+1), what is the range of (f)?
#relations-functions
#range
#shifted-rational
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A ((1,4])
B ([1,4])
C \([4,\infty\))
D (\(-\infty,4]\)
Explanation opens after your attempt
Correct Answer
A. ((1,4])
Step 1
Concept
\(The denominator (x^2-2x+2=(x-1)^2+1\ge 1), so (\frac{3}{\)denominator\(}\in(0,3]). Hence the range is ((1,4]).\)
Step 2
Why this answer is correct
\(The correct answer is A. ((1,4]). The denominator (x^2-2x+2=(x-1)^2+1\ge 1), so (\frac{3}{\)denominator\(}\in(0,3]). Hence the range is ((1,4]).\)
Step 3
Exam Tip
\(हर (x^2-2x+2=(x-1)^2+1\ge 1), इसलिए (\frac{3}{\)हर}\in(0,3])। इसलिए परिसर ((1,4]) है।
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फलन (f(x)=\frac{x-2 +2}{x-2 +5}) का परिसर चुनिए।
Choose the range of (f(x)=\frac{x-2 +2}{x-2 +5}).
#relations-functions
#range
#rational-square
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A \([\frac{2}{5},1\))
B (\(\frac{2}{5},1]\)
C \([\frac{2}{5},1]\)
D ((0,1))
Explanation opens after your attempt
Correct Answer
A. \([\frac{2}{5},1\))
Step 1
Concept
Let \(t=x^2\ge 0\), then \(f=\frac{t+2}{t+5}\). It is \(\frac{2}{5}\) at (t=0) and approaches (1) but never equals (1).
Step 2
Why this answer is correct
The correct answer is A. \([\frac{2}{5},1\)). Let \(t=x^2\ge 0\), then \(f=\frac{t+2}{t+5}\). It is \(\frac{2}{5}\) at (t=0) and approaches (1) but never equals (1).
Step 3
Exam Tip
मान लें \(t=x^2\ge 0\), तब \(f=\frac{t+2}{t+5}\)। यह (t=0) पर \(\frac{2}{5}\) है और (1) तक पहुंचता है पर (1) नहीं होता।
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यदि (f(x)=\frac{x-2 +1}{x-2 -1}), तो प्रांत क्या है?
If (f(x)=\frac{x-2 +1}{x-2 -1}), what is the domain?
#relations-functions
#domain
#rational-square-denominator
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A \(\mathbb{R}-{-1,1}\)
B \(\mathbb{R}-{1}\)
C \(\mathbb{R}-{-1}\)
D ((-1,1))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-1,1}\)
Step 1
Concept
The denominator must satisfy \(x^2-1\ne 0\). Therefore \(x\ne -1\) and \(x\ne 1\).
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-1,1}\). The denominator must satisfy \(x^2-1\ne 0\). Therefore \(x\ne -1\) and \(x\ne 1\).
Step 3
Exam Tip
हर \(x^2-1\ne 0\) होना चाहिए। इसलिए \(x\ne -1\) और \(x\ne 1\)।
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फलन (f(x)=\frac{x-2 +1}{x-2 -1}) का परिसर क्या है?
What is the range of (f(x)=\frac{x-2 +1}{x-2 -1})?
#relations-functions
#range
#advanced-rational
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A ((-\infty,-1]\cup\(1,\infty\))
B ((-1,1))
C \(\mathbb{R}-{1}\)
D \([1,\infty\))
Explanation opens after your attempt
Correct Answer
A. ((-\infty,-1]\cup\(1,\infty\))
Step 1
Concept
If \(t=x^2\ge 0\) and \(t\ne 1\), then \(y=\frac{t+1}{t-1}\). Using \(t=\frac{y+1}{y-1}\ge 0\) gives the range ((-\infty,-1]\cup\(1,\infty\)).
Step 2
Why this answer is correct
The correct answer is A. ((-\infty,-1]\cup\(1,\infty\)). If \(t=x^2\ge 0\) and \(t\ne 1\), then \(y=\frac{t+1}{t-1}\). Using \(t=\frac{y+1}{y-1}\ge 0\) gives the range ((-\infty,-1]\cup\(1,\infty\)).
Step 3
Exam Tip
यदि \(t=x^2\ge 0\) और \(t\ne 1\), तो \(y=\frac{t+1}{t-1}\)। इससे \(t=\frac{y+1}{y-1}\ge 0\) देकर परिसर ((-\infty,-1]\cup\(1,\infty\)) मिलता है।
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फलन (f(x)=\sqrt{2x-x-2 }) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{2x-x-2 })?
#relations-functions
#domain
#quadratic-radical
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A ([0,2])
B (\(-\infty,0]\cup[2,\infty\))
C ((0,2))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. ([0,2])
Step 1
Concept
The condition \(2x-x^2\ge 0\) gives (x(2-x)\ge 0). Hence \(x\in[0,2]\).
Step 2
Why this answer is correct
The correct answer is A. ([0,2]). The condition \(2x-x^2\ge 0\) gives (x(2-x)\ge 0). Hence \(x\in[0,2]\).
Step 3
Exam Tip
शर्त \(2x-x^2\ge 0\) से (x(2-x)\ge 0) मिलता है। इसलिए \(x\in[0,2]\)।
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फलन (f(x)=\sqrt{2x-x-2 }) का परिसर चुनिए।
Choose the range of (f(x)=\sqrt{2x-x-2 }).
#relations-functions
#range
#semicircle
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A ([0,1])
B ([-1,1])
C ([0,2])
D ((0,1])
Explanation opens after your attempt
Correct Answer
A. ([0,1])
Step 1
Concept
(2x-x-2 =1-(x-1)2 ), whose maximum value is (1). The square root gives the range ([0,1]).
Step 2
Why this answer is correct
The correct answer is A. ([0,1]). (2x-x-2 =1-(x-1)2 ), whose maximum value is (1). The square root gives the range ([0,1]).
Step 3
Exam Tip
(2x-x-2 =1-(x-1)2 ), जिसका अधिकतम मान (1) है। वर्गमूल से परिसर ([0,1]) मिलता है।
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यदि (f(x)=\frac{1}{x-1}+\frac{1}{x+1}), तो प्रांत क्या है?
If (f(x)=\frac{1}{x-1}+\frac{1}{x+1}), what is the domain?
#relations-functions
#domain
#sum-rational
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A \(\mathbb{R}-{-1,1}\)
B \(\mathbb{R}-{0}\)
C ((-1,1))
D (\(-\infty,-1\)\cup\(1,\infty\))
Explanation opens after your attempt
Correct Answer
A. \(\mathbb{R}-{-1,1}\)
Step 1
Concept
Both denominators must be non-zero, so \(x\ne 1\) and \(x\ne -1\). In sums of rational functions, check every denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\mathbb{R}-{-1,1}\). Both denominators must be non-zero, so \(x\ne 1\) and \(x\ne -1\). In sums of rational functions, check every denominator.
Step 3
Exam Tip
दोनों हर शून्य नहीं होने चाहिए, इसलिए \(x\ne 1\) और \(x\ne -1\)। योग वाले रैशनल फलन में सभी हर जांचें।
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फलन (f(x)=\frac{1}{\sqrt{x-2 +6x+10}}) का परिसर कौन सा है?
Which is the range of (f(x)=\frac{1}{\sqrt{x-2 +6x+10}})?
#relations-functions
#range
#reciprocal-square-root
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A ((0,1])
B \([1,\infty\))
C ([0,1])
D (\(1,\infty\))
Explanation opens after your attempt
Correct Answer
A. ((0,1])
Step 1
Concept
Since (x-2 +6x+10=(x+3)2 +1\ge 1), the denominator has minimum value (1). Therefore (0<f(x)\le 1).
Step 2
Why this answer is correct
The correct answer is A. ((0,1]). Since (x-2 +6x+10=(x+3)2 +1\ge 1), the denominator has minimum value (1). Therefore (0<f(x)\le 1).
Step 3
Exam Tip
क्योंकि (x-2 +6x+10=(x+3)2 +1\ge 1), हर का न्यूनतम मान (1) है। इसलिए (0<f(x)\le 1)।
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फलन (f(x)=\frac{x}{\sqrt{x-2 +1}}) का परिसर क्या है?
What is the range of (f(x)=\frac{x}{\sqrt{x-2 +1}})?
#relations-functions
#range
#bounded-function
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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
The denominator is always greater than (|x|), so (-1<f(x)<1). The values (-1) and (1) are only limiting values.
Step 2
Why this answer is correct
The correct answer is A. ((-1,1)). The denominator is always greater than (|x|), so (-1<f(x)<1). The values (-1) and (1) are only limiting values.
Step 3
Exam Tip
हर हमेशा (|x|) से बड़ा होता है, इसलिए (-1<f(x)<1)। (-1) और (1) केवल सीमा मान हैं।
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यदि (f(x)=\sqrt{x}+\sqrt{1-x}), तो प्रांत क्या है?
If (f(x)=\sqrt{x}+\sqrt{1-x}), what is the domain?
#relations-functions
#domain
#intersection
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A ([0,1])
B ((0,1))
C (\(-\infty,0]\cup[1,\infty\))
D \(\mathbb{R}\)
Explanation opens after your attempt
Correct Answer
A. ([0,1])
Step 1
Concept
The conditions are \(x\ge 0\) and \(1-x\ge 0\). Together they give \(x\in[0,1]\).
Step 2
Why this answer is correct
The correct answer is A. ([0,1]). The conditions are \(x\ge 0\) and \(1-x\ge 0\). Together they give \(x\in[0,1]\).
Step 3
Exam Tip
शर्तें \(x\ge 0\) और \(1-x\ge 0\) हैं। दोनों मिलाकर \(x\in[0,1]\) मिलता है।
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फलन (f(x)=\sqrt{x}+\sqrt{1-x}) का परिसर चुनिए।
Choose the range of (f(x)=\sqrt{x}+\sqrt{1-x}).
#relations-functions
#range
#sum-radicals
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A \([1,\sqrt{2}]\)
B \([0,\sqrt{2}]\)
C (\(1,\sqrt{2}\))
D ([0,1])
Explanation opens after your attempt
Correct Answer
A. \([1,\sqrt{2}]\)
Step 1
Concept
On ([0,1]), the minimum at endpoints is (1) and the maximum at \(x=\frac{1}{2}\) is \(\sqrt{2}\). Hence the range is \([1,\sqrt{2}]\).
Step 2
Why this answer is correct
The correct answer is A. \([1,\sqrt{2}]\). On ([0,1]), the minimum at endpoints is (1) and the maximum at \(x=\frac{1}{2}\) is \(\sqrt{2}\). Hence the range is \([1,\sqrt{2}]\).
Step 3
Exam Tip
प्रांत ([0,1]) में न्यूनतम मान किनारों पर (1) और अधिकतम \(x=\frac{1}{2}\) पर \(\sqrt{2}\) है। इसलिए परिसर \([1,\sqrt{2}]\) है।
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फलन (f(x)=\frac{1}{x-2 +2x+2}) का अधिकतम मान क्या है?
What is the maximum value of (f(x)=\frac{1}{x-2 +2x+2})?
#relations-functions
#range
#maximum-reciprocal
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A (1)
B \(\frac{1}{2}\)
C (2)
D (0)
Explanation opens after your attempt
Step 1
Concept
The denominator (x-2 +2x+2=(x+1)2 +1) has minimum value (1). Hence the maximum value of the function is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). The denominator (x-2 +2x+2=(x+1)2 +1) has minimum value (1). Hence the maximum value of the function is (1).
Step 3
Exam Tip
हर (x-2 +2x+2=(x+1)2 +1) का न्यूनतम मान (1) है। इसलिए फलन का अधिकतम मान (1) है।
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यदि वास्तविक फलन (f(x)=\frac{\sqrt{x-2}}{x-2 -9}) दिया है, तो उसका प्रांत क्या है?
If the real function (f(x)=\frac{\sqrt{x-2}}{x-2 -9}) is given, what is its domain?
#relations-functions
#domain
#combined-restrictions
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A \([2,\infty\)-{3})
B \([2,\infty\)-{-3,3})
C (\(2,\infty\)-{3})
D \(\mathbb{R}-{-3,3}\)
Explanation opens after your attempt
Correct Answer
A. \([2,\infty\)-{3})
Step 1
Concept
For the square root \(x\ge 2\), and for the denominator \(x\ne -3,3\) are required. But (-3) is already outside \(x\ge 2\), so only (3) is removed.
Step 2
Why this answer is correct
The correct answer is A. \([2,\infty\)-{3}). For the square root \(x\ge 2\), and for the denominator \(x\ne -3,3\) are required. But (-3) is already outside \(x\ge 2\), so only (3) is removed.
Step 3
Exam Tip
वर्गमूल के लिए \(x\ge 2\) और हर के लिए \(x\ne -3,3\) चाहिए। लेकिन (-3) पहले ही \(x\ge 2\) में नहीं है, इसलिए केवल (3) हटेगा।
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फलन (f(x)=\sqrt{\frac{x-2 -4}{9-x-2 }}) का प्रांत क्या है?
What is the domain of (f(x)=\sqrt{\frac{x-2 -4}{9-x-2 }})?
#relations-functions
#domain
#radical-rational
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A (\(-3,-2]\cup[2,3\))
B ([-2,2])
C (\(-\infty,-3\)\cup\(3,\infty\))
D ((-3,3))
Explanation opens after your attempt
Correct Answer
A. (\(-3,-2]\cup[2,3\))
Step 1
Concept
The condition is \(\frac{x^2-4}{9-x^2}\ge 0\) and \(x\ne -3,3\). A sign chart gives (\(-3,-2]\cup[2,3\)).
Step 2
Why this answer is correct
The correct answer is A. (\(-3,-2]\cup[2,3\)). The condition is \(\frac{x^2-4}{9-x^2}\ge 0\) and \(x\ne -3,3\). A sign chart gives (\(-3,-2]\cup[2,3\)).
Step 3
Exam Tip
शर्त \(\frac{x^2-4}{9-x^2}\ge 0\) और \(x\ne -3,3\) है। साइन चार्ट से (\(-3,-2]\cup[2,3\)) मिलता है।
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यदि (f(x)=\frac{2x-2 +3}{x-2 +1}), तो (f) का परिसर क्या है?
If (f(x)=\frac{2x-2 +3}{x-2 +1}), what is the range of (f)?
#relations-functions
#range
#rational-function
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A ((2,3])
B ([2,3])
C \([3,\infty\))
D (\(-\infty,2\)\cup[3,\infty))
Explanation opens after your attempt
Correct Answer
A. ((2,3])
Step 1
Concept
Let \(t=x^2\ge 0\), then \(f=\frac{2t+3}{t+1}\) is decreasing. At (t=0), (3) is obtained and (2) is not attained.
Step 2
Why this answer is correct
The correct answer is A. ((2,3]). Let \(t=x^2\ge 0\), then \(f=\frac{2t+3}{t+1}\) is decreasing. At (t=0), (3) is obtained and (2) is not attained.
Step 3
Exam Tip
मान लें \(t=x^2\ge 0\), तब \(f=\frac{2t+3}{t+1}\) घटता है। (t=0) पर (3) मिलता है और (2) प्राप्त नहीं होता।
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