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 471-480 of 600 questions.
What is the (x)-intercept of the graph of (y=|2x-6|)?
A. ( \left\(3,0\right\) )
( |2x-6|=0 ) तभी होगा जब (2x-6=0)। इसलिए (x=3) है। / ( |2x-6|=0 ) only when (2x-6=0). Hence (x=3).
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What is the slope of the line (3x+2y=12)?
A. \(-\frac{3}{2}\)
समीकरण को \(y=-\frac{3}{2}x+6\) लिखें। (x) का गुणांक ढाल होता है। / Write the equation as \(y=-\frac{3}{2}x+6\). The coefficient of (x) is the slope.
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Which statement is correct for the graph of (y=5)?
A. यह (x)-अक्ष के समानांतर रेखा हैIt is a line parallel to the (x)-axis
(y) का मान हर (x) के लिए (5) है। स्थिर फलन का ग्राफ क्षैतिज रेखा होता है। / The value of (y) is (5) for every (x). The graph of a constant function is a horizontal line.
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Which point lies on the graph of the identity function (f(x)=x)?
A. ( \left\(-4,-4\right\) )
पहचान फलन में (y=x) होता है। इसलिए दोनों निर्देशांक बराबर होने चाहिए। / In the identity function (y=x). Therefore both coordinates must be equal.
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For which (x) values do the graphs \(y=x^2\) and (y=2x+3) intersect?
A. (x=-1,3)
प्रतिच्छेद के लिए \(x^2=2x+3\) रखें। \(x^2-2x-3=0\) से (x=-1,3) मिलते हैं। / For intersection set \(x^2=2x+3\). From \(x^2-2x-3=0\), (x=-1,3).
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At which (x) values do the graphs \(y=x^2\) and (y=|x|) meet?
A. (x=-1,0,1)
\(x^2=|x|\) रखने पर (x=0) या (|x|=1) मिलता है। इसलिए (x=-1,0,1) हैं। / Putting \(x^2=|x|\) gives (x=0) or (|x|=1). Hence (x=-1,0,1).
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For (x>0), the graph of (f(x)=\frac{x}{|x|}) lies on which line?
A. (y=1)
(x>0) पर (|x|=x) होता है। इसलिए \(\frac{x}{|x|}=1\)। / For (x>0), (|x|=x). Hence \(\frac{x}{|x|}=1\).
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For (x>-1), what is (y) in the graph of (y=\min(x,-1))?
A. (y=-1)
जब (x>-1) हो तो (x) और (-1) में छोटा मान (-1) है। इसलिए ग्राफ (y=-1) है। / When (x>-1), the smaller value between (x) and (-1) is (-1). Hence the graph is (y=-1).
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What is the range of the graph (y=|x-2|+1)?
A. \( [1,\infty\) )
(|x-2|) का न्यूनतम मान (0) है। इसलिए (y) का न्यूनतम मान (1) है। / The minimum value of (|x-2|) is (0). Therefore the minimum value of (y) is (1).
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What is the range of the graph (y=-(x+4)2+9)?
A. ( \(-\infty,9] \)
परवलय नीचे खुलता है और शीर्ष पर (y=9) है। इसलिए रेंज ( \(-\infty,9] \) है। / The parabola opens downward and its vertex has (y=9). Hence the range is ( \(-\infty,9] \).
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