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 211-220 of 600 questions.
What is the axis of symmetry of the graph of (f(x)=2x-2-8x+1)?
A. (x=2)
\(x=-\frac{b}{2a}\) में (a=2) और (b=-8) रखें। इससे (x=2) मिलता है। / Put (a=2) and (b=-8) in \(x=-\frac{b}{2a}\). This gives (x=2).
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What is the minimum value of (f(x)=x-2+2x+5)?
A. (4)
(x-2+2x+5=(x+1)2+4)। इसलिए न्यूनतम मान (4) है। / (x-2+2x+5=(x+1)2+4). Therefore the minimum value is (4).
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What is the maximum value of (f(x)=-x-2+6x-5)?
A. (4)
(-x-2+6x-5=-(x-3)2+4)। इसलिए अधिकतम मान (4) है। / (-x-2+6x-5=-(x-3)2+4). Therefore the maximum value is (4).
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What is the starting point of (f(x)=\sqrt{x+6}-2)?
A. ((-6,-2))
\(\sqrt{x+6}\) (x=-6) से शुरू होता है और बाहर (-2) जुड़ता है। इसलिए प्रारंभिक बिंदु ((-6,-2)) है। / \(\sqrt{x+6}\) starts at (x=-6), and (-2) is added outside. So the starting point is ((-6,-2)).
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What is the maximum value of (f(x)=3-\sqrt{x-1})?
A. (3)
\(\sqrt{x-1}\ge 0\), इसलिए \(3-\sqrt{x-1}\le 3\)। अधिकतम मान (3) (x=1) पर मिलता है। / Since \(\sqrt{x-1}\ge 0\), \(3-\sqrt{x-1}\le 3\). The maximum value (3) occurs at (x=1).
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What is the range of (f(x)=2\sqrt{x+1}-5)?
A. \([-5,\infty\))
\(\sqrt{x+1}\ge 0\), इसलिए \(2\sqrt{x+1}-5\ge -5\)। अतः परिसर \([-5,\infty\)) है। / Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).
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What is the vertical asymptote of (f(x)=\frac{1}{x-5}+2)?
A. (x=5)
हर (x-5) शून्य नहीं हो सकता। इसलिए ऊर्ध्व आसिम्प्टोट (x=5) है। / The denominator (x-5) cannot be zero. Therefore the vertical asymptote is (x=5).
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What is the horizontal asymptote of (f(x)=\frac{3}{x+2}-1)?
A. (y=-1)
बड़े (|x|) पर \(\frac{3}{x+2}\) (0) के पास जाता है। इसलिए क्षैतिज आसिम्प्टोट (y=-1) है। / For large (|x|), \(\frac{3}{x+2}\) approaches (0). So the horizontal asymptote is (y=-1).
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What is the range of (f(x)=\frac{1}{x+4}+6)?
A. (\(-\infty,6\)\cup\(6,\infty\))
\(\frac{1}{x+4}\) कभी (0) नहीं होता। इसलिए (f(x)) कभी (6) नहीं होता। / \(\frac{1}{x+4}\) is never (0). Therefore (f(x)) is never (6).
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Relative to its shifted asymptotes, in which branches does (f(x)=\frac{-2}{x-3}) lie?
A. द्वितीय और चतुर्थ जैसेLike second and fourth
\(\frac{-2}{x-3}\) में गुणांक ऋणात्मक है। इसलिए इसकी शाखाएं मूल \(-\frac{1}{x}\) जैसी विपरीत चतुर्थांशों में होती हैं। / The coefficient in \(\frac{-2}{x-3}\) is negative. So its branches are like \(-\frac{1}{x}\) in opposite quadrants.
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