Bihar BoardHindi Mediumकक्षा 12 साइन्सगणितसमाकलन1 Mark
MCQ
यदि f(x) = $\int_{0}^{x}$ t sin t dt, तब f'(x) है:
A
x cos x
B
x sin x
C
cos x + x sin x
D
sin x + x cos x
✓
Answer
$f(x)=\int_{0}^{x} t \sin t d t$ $\Rightarrow \int \mathrm{u} . \mathrm{v} \mathrm{d} \mathrm{x}$ = $\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$ f(x) = $[t]_{0}^{x} \int_{0}^{x} \sin t d t-\int_{0}^{x}\left\{\left(\frac{d}{d t} t\right) \cdot \int \sin t d t\right\} d t$ $=[t(-\cos t)]_{0}^{x}-\int_{0}^{x}(-\cos t) d t$ $=[-t(\cos t)+\sin t]_{0}^{x}$ = -x cos x + sin x - 0 $\Rightarrow$ f(x) = -x cos x + sin x $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$ = $-\left[\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}+\cos \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}\right]$ $\Rightarrow$ f'(x) = -[{x(-sin x)} + cos x] + cos x = x sin x - cos x + cos x = x sin x
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