Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 371-380 of 600 questions.
What is the domain of (f(x)=\sqrt{x-2-4x+3})?
A. ( \(-\infty,1]\cup[3,\infty\) )
(x-2-4x+3=(x-1)(x-3)) और इसे \(\ge0\) चाहिए। ऊपर खुलने वाले द्विघात में बाहरी अंतराल चुने जाते हैं। / Since (x-2-4x+3=(x-1)(x-3)), it must be \(\ge0\). For an upward quadratic, the outside intervals are selected.
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If (f(x)=\frac{1}{\sqrt{16-x-2}}), what is the domain?
A. ( (-4,4) )
हर में वर्गमूल है इसलिए \(16-x^2>0\) चाहिए। यहाँ सिरों \(x=\pm4\) को शामिल नहीं किया जाएगा। / The square root is in the denominator, so \(16-x^2>0\) is required. The endpoints \(x=\pm4\) are not included here.
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What is the range of (f(x)=\frac{3}{2+\cos x})?
A. ( [1,3] )
\(\cos x\in[-1,1]\) इसलिए \(2+\cos x\in[1,3]\), अतः \(f(x)\in[1,3]\)। reciprocal लेते समय क्रम उलट सकता है। / Since \(\cos x\in[-1,1]\), \(2+\cos x\in[1,3]\), so \(f(x)\in[1,3]\). The order may reverse when taking reciprocals.
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If (f(x)=5-2\sin x), what is the range of (f)?
A. ( [3,7] )
\(\sin x\in[-1,1]\) होने से \(-2\sin x\in[-2,2]\) और कुल परिसर ( [3,7] ) है। गुणा और जोड़ को क्रम से लागू करें। / Since \(\sin x\in[-1,1]\), \(-2\sin x\in[-2,2]\), giving total range ([3,7]). Apply multiplication and addition step by step.
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What is the domain of (f(x)=\tan x)?
A. \( \mathbb{R}\setminus{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}} \)
\(\tan x=\frac{\sin x}{\cos x}\) है इसलिए \(\cos x\ne0\) होना चाहिए। \(\tan x\) में विषम \(\frac{\pi}{2}\) वाले कोण हटते हैं। / Since \(\tan x=\frac{\sin x}{\cos x}\), we need \(\cos x\ne0\). For \(\tan x\), odd multiples around \(\frac{\pi}{2}\) are excluded.
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If (f(x)=\frac{1}{\log x}), what is the domain of (f)?
A. ( \(0,\infty\)\setminus{1} )
\(\log x\) के लिए (x>0) और हर के लिए \(\log x\ne0\) चाहिए। इसलिए (x=1) अलग से हटेगा। / For \(\log x\), (x>0), and for the denominator \(\log x\ne0\). Therefore (x=1) must be excluded separately.
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What is the domain of (f(x)=\sqrt{\log_3(x-2)})?
A. \( [3,\infty\) )
वर्गमूल के लिए (\log_3(x-2)\ge0) चाहिए जिससे \(x-2\ge1\) मिलता है। साथ में (x-2>0) पहले से पूरा है। / The square root requires (\log_3(x-2)\ge0), giving \(x-2\ge1\). The condition (x-2>0) is already satisfied.
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If (f(x)=\frac{\sqrt{x-1}}{x-5}), what is the domain?
A. \( [1,\infty\)\setminus{5} )
\(\sqrt{x-1}\) के लिए \(x\ge1\) और हर के लिए \(x\ne5\) चाहिए। सिरा (x=1) मान्य है क्योंकि हर शून्य नहीं है। / For \(\sqrt{x-1}\), \(x\ge1\), and for the denominator \(x\ne5\). The endpoint (x=1) is valid because the denominator is not zero.
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What is the range of (f(x)=\frac{1}{x-2-6x+10})?
A. ( (0,1] )
हर ((x-3)2+1\ge1) है इसलिए अधिकतम \(\frac{1}{1}=1\) है और (0) प्राप्त नहीं होता। पूर्ण वर्ग से सीमा साफ मिलती है। / The denominator ((x-3)2+1\ge1), so the maximum is \(\frac{1}{1}=1\), and (0) is not attained. Completing the square gives the bounds clearly.
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If (f(x)=\sqrt{x}+\sqrt{6-x}), what is the domain?
A. ( [0,6] )
\(\sqrt{x}\) से \(x\ge0\) और \(\sqrt{6-x}\) से \(x\le6\) है। दोनों शर्तों का साझा भाग ( [0,6] ) है। / From \(\sqrt{x}\), \(x\ge0\), and from \(\sqrt{6-x}\), \(x\le6\). The common part is ([0,6]).
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