Number of values of x satisfying ∫0xt2sin(x−t)dt=x2 in [0, 100]

Gives, ∫0xt2sin(x−t) dt=x2⇒ ∫0x(x−t)2sin t dt =x2 ⇒ x2∫0xsint dt−2x∫0xtsint+∫0xt2sint=x2 ⇒ x2 (1−cosx)−2x(−x cosx+sin x)+(−x2cosx+2xsin+cosx−1)=x2 ⇒ cosx=1 So, x=2nπ;n∈IHence, number of values of 0, 2π, 4π.,.......,30π

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Number of values of $x$ satisfying $\int_{0}^{x} t^{2} \sin (x - t) d t = x^{2} i n [0, 100]$

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answer is 16.

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Detailed Solution

Gives, $\int_{0}^{x} t^{2} \sin (x - t) d t = x^{2}$

$\Rightarrow \int_{0}^{x} {(x - t)}^{2} \sin t d t = x^{2} \Rightarrow x^{2} \int_{0}^{x} \sin t d t - 2 x \int_{0}^{x} t \sin t + \int_{0}^{x} t^{2} \sin t = x^{2} \Rightarrow x^{2} (1 - \cos x) - 2 x (- x \cos x + \sin x) + (- x^{2} \cos x + 2 x \sin + \cos x - 1) = x^{2} \Rightarrow \cos x = 1 S o, x = 2 n π; n \in I$