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 141-150 of 600 questions.
If (f(x)=x-2-2x) and (g(x)=x), what is ((f+3g)(x))?
A. \(x^2+x\)
((f+3g)(x)=x-2-2x+3x=x-2+x)। रैखिक संयोजन में गुणक सही लगाएं। / ((f+3g)(x)=x-2-2x+3x=x-2+x). Apply the multiplier correctly in a linear combination.
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If (f(x)=\frac{1}{x+4}) and (g(x)=x-2), what is the domain of ((f-g)(x))?
A. \(\mathbb{R}-{-4}\)
\(\frac{1}{x+4}\) में (x=-4) पर हर (0) है। अंतर का डोमेन भी \(\mathbb{R}-{-4}\) होगा। / In \(\frac{1}{x+4}\), the denominator is (0) at (x=-4). The domain of the difference is also \(\mathbb{R}-{-4}\).
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If (f(x)=x+1) and (g(x)=x-2+1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?
A. \(\mathbb{R}-{-1}\)
\(\frac{g}{f}\) में हर (f(x)=x+1) है, इसलिए (x=-1) हटेगा। भागफल में हर वाले फलन को ध्यान से देखें। / In \(\frac{g}{f}\), the denominator is (f(x)=x+1), so (x=-1) is excluded. Identify the denominator function carefully.
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If (f(x)=2x+3) and (g(x)=2x-3), what is ((fg)(x))?
A. \(4x^2-9\)
((2x+3)(2x-3)=(2x)2-32=4x-2-9)। यह अंतर वर्ग का सीधा प्रयोग है। / ((2x+3)(2x-3)=(2x)2-32=4x-2-9). This is a direct use of difference of squares.
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If (f(x)=x-2+1) and (g(x)=x-2+1), then (\left\(\frac{f}{g}\right\)(x)) equals which expression?
A. (1)
क्योंकि (f(x)=g(x)), इसलिए (\frac{f(x)}{g(x)}=1)। यहाँ \(x^2+1\) कभी (0) नहीं होता, इसलिए सभी वास्तविक (x) मान्य हैं। / Since (f(x)=g(x)), (\frac{f(x)}{g(x)}=1). Here \(x^2+1\) is never (0), so all real (x) are valid.
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If (f(x)=x-2+4x) and (g(x)=2x-2-x), what is ((f+g)(x))?
A. \(3x^2+3x\)
((f+g)(x)=x-2+4x+2x-2-x=3x-2+3x)। समान घात वाले पदों को ही जोड़ें। / ((f+g)(x)=x-2+4x+2x-2-x=3x-2+3x). Add only like terms with the same power.
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If (f(x)=5x+2) and (g(x)=2x-7), what is the value of ((f-g)(1))?
A. (12)
(f(1)=7) और (g(1)=-5), इसलिए ((f-g)(1)=7-(-5)=12)। ऋणात्मक मान घटाते समय सावधानी रखें। / (f(1)=7) and (g(1)=-5), so ((f-g)(1)=7-(-5)=12). Be careful while subtracting a negative value.
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If (f(x)=x+2) and (g(x)=x-2-2x), what is ((fg)(2))?
A. (0)
(f(2)=4) और (g(2)=0), इसलिए ((fg)(2)=4\cdot 0=0)। गुणन में किसी एक मान के (0) होने पर उत्पाद (0) होता है। / (f(2)=4) and (g(2)=0), so ((fg)(2)=4\cdot 0=0). In multiplication, if one value is (0), the product is (0).
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If (f(x)=\sqrt{x+2}) and (g(x)=\frac{1}{x-1}), what is the real domain of ((f+g)(x))?
A. \([-2,\infty\)-{1})
\(\sqrt{x+2}\) के लिए \(x\ge -2\) और \(\frac{1}{x-1}\) के लिए \(x\ne 1\) चाहिए। दोनों शर्तों को साथ लगाएं। / For \(\sqrt{x+2}\), \(x\ge -2\), and for \(\frac{1}{x-1}\), \(x\ne 1\). Apply both conditions together.
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If (f(x)=x-2-1) and (g(x)=x-2+1), what is ((g-f)(x))?
A. (2)
((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2)। घटाते समय दूसरे कोष्ठक के चिह्न बदलते हैं। / ((g-f)(x)=\(x^2+1\)-\(x^2-1\)=2). While subtracting, the signs inside the second bracket change.
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