Let g be a differentiable function satisfying ∫0x(x−t+1)g(t)dt=x4+x2 for all x≥0. The value of ∫0112g'(x)+g(x)+10dx is equal to

x∫0xg(t)dt+∫0x(1−t)g(t)dt=x4+x2differentiate w.r.t xx g(x)+∫0xg(t)dt+(1−x)g(x)=4x3+2xAgain differentiate w.r.t xg'(x)+g(x)=12x2+2 Now, ∫0112g'+g+10dx=∫01dxx2+1=tan−1x]01=π4

Let g be a differentiable function satisfying ∫0x(x−t+1)g(t)dt=x4+x2 for all x≥0. The value of ∫0112g'(x)+g(x)+10dx is equal to

x∫0xg(t)dt+∫0x(1−t)g(t)dt=x4+x2differentiate w.r.t xx g(x)+∫0xg(t)dt+(1−x)g(x)=4x3+2xAgain differentiate w.r.t xg'(x)+g(x)=12x2+2 Now, ∫0112g'+g+10dx=∫01dxx2+1=tan−1x]01=π4

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Let g be a differentiable function satisfying $\int_{0}^{x} (x - t + 1) g (t) dt = x^{4} + x^{2}$ for all $x \geq 0$ . The value of $\int_{0}^{1} \frac{12}{g' (x) + g (x) + 10} dx$ is equal to

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$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

$x \int_{0}^{x} g (t) dt + \int_{0}^{x} (1 - t) g (t) dt = x^{4} + x^{2}$
differentiate w.r.t x
$x g (x) + \int_{0}^{x} g (t) dt + (1 - x) g (x) = 4 x^{3} + 2 x$
Again differentiate w.r.t x
$g' (x) + g (x) = 12 x^{2} + 2 Now, \int_{0}^{1} \frac{12}{g' + g + 10} dx = \int_{0}^{1} \frac{dx}{x^{2} + 1} = \tan^{- 1} {x]}_{0}^{1} = \frac{π}{4}$