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 521-530 of 600 questions.
What is the domain of \(y=\sqrt{16-x^2}\)?
A. ([-4,4])
वर्गमूल के लिए \(16-x^2\ge 0\) चाहिए। इससे \(-4\le x\le 4\) मिलता है। / For the square root, \(16-x^2\ge 0\) is required. This gives \(-4\le x\le 4\).
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What is the range of the graph \(y=\sqrt{16-x^2}\)?
A. ([0,4])
यह वृत्त \(x^2+y^2=16\) का ऊपरी अर्धभाग है। इसलिए (y) का मान (0) से (4) तक है। / It is the upper semicircle of \(x^2+y^2=16\). Hence (y) ranges from (0) to (4).
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What is the vertical asymptote of the graph \(y=\frac{1}{x+5}\)?
A. (x=-5)
हर (x+5) शून्य होने पर फलन अपरिभाषित हो जाता है। इसलिए (x=-5) ऊर्ध्वाधर आसम्टोट है। / The denominator (x+5) makes the function undefined when it is zero. Therefore (x=-5) is the vertical asymptote.
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What is the horizontal asymptote of \(y=\frac{3}{x-2}+4\)?
A. (y=4)
\(\frac{3}{x-2}\) बड़े (|x|) पर (0) के पास जाता है। इसलिए ग्राफ (y=4) के पास जाता है। / The term \(\frac{3}{x-2}\) approaches (0) for large (|x|). Hence the graph approaches (y=4).
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Which (x) value is not in the domain of \(y=\frac{x-1}{x+2}\)?
A. (x=-2)
हर (x+2) शून्य नहीं हो सकता। (x+2=0) से (x=-2) निषिद्ध है। / The denominator (x+2) cannot be zero. From (x+2=0), (x=-2) is excluded.
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What is the horizontal asymptote of \(y=\frac{2x+1}{x-3}\)?
A. (y=2)
अंश और हर की डिग्री समान है इसलिए प्रमुख गुणांकों का अनुपात लें। \(\frac{2}{1}=2\) से आसम्टोट (y=2) है। / The numerator and denominator have equal degree, so take the ratio of leading coefficients. Since \(\frac{2}{1}=2\), the asymptote is (y=2).
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What is the range of \(y=\frac{1}{x^2+4}\)?
A. (\(0,\frac{1}{4}]\)
\(x^2+4\ge 4\) इसलिए अधिकतम मान \(\frac{1}{4}\) है। ग्राफ (0) के पास जाता है पर (0) नहीं लेता। / Since \(x^2+4\ge 4\), the maximum value is \(\frac{1}{4}\). The graph approaches (0) but never takes (0).
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What is the symmetry of the graph of (f(x)=\frac{x-2-1}{x-2+1})?
A. (y)-अक्ष के सापेक्ष सममितिSymmetry about the (y)-axis
(f(-x)=\frac{x-2-1}{x-2+1}=f(x)) है। इसलिए यह सम फलन है। / (f(-x)=\frac{x-2-1}{x-2+1}=f(x)). Therefore it is an even function.
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At which point does the graph \(y=x^3+2\) cut the (y)-axis?
A. ((0,2))
(y)-अक्ष पर (x=0) रखें। तब \(y=0^3+2=2\) मिलता है। / On the (y)-axis, set (x=0). Then \(y=0^3+2=2\).
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In (y=(x+2)3-1), where does the point ((0,0)) of \(y=x^3\) move?
A. ((-2,-1))
(x+2) ग्राफ को (2) इकाई बाईं ओर और (-1) नीचे ले जाता है। इसलिए ((0,0)) से ((-2,-1)) मिलता है। / The term (x+2) shifts the graph (2) units left and (-1) shifts it down. Hence ((0,0)) becomes ((-2,-1)).
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