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(B) Given $f(x) = \int_{0}^{x} 	an(t-x) dt - \int_{0}^{x} f(t) 	an t dt$.Using the Leibniz integral rule, we differentiate with respect to $x$:$f'(x

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(B) Given $f(x) = \int_{0}^{x} 	an(t-x) dt - \int_{0}^{x} f(t) 	an t dt$.Using the Leibniz integral rule, we differentiate with respect to $x$:$f'(x) = 	an(x-x) - 	an(0-x) - f(x) 	an x = 0 - (-	an x) - f(x) 	an x = 	an x (1 - f(x))$.This is a separable differential equation: $\frac{df}{1-f} = 	an x dx$.Integrating both sides: $-\ln|1-f| = \ln|\sec x| + C$.This simplifies to $\ln|1-f|^{-1} = \ln|\sec x| + C$, or $1-f = k \cos x$.At $x=0$, $f(0) = \int_{0}^{0} 	an(t) dt - \int_{0}^{0} f(t) 	an t dt = 0$.Substituting $x=0$ into $1-f(x) = k \cos x$, we get $1-0 = k(1)$, so $k=1$.Thus, $f(x) = 1 - \cos x$.Now, $f'(x) = \sin x$ and $f''(x) = \cos x$.We need to evaluate $f''(\pi/6) + f(\pi/6)$:$f''(\pi/6) = \cos(\pi/6) = \frac{\sqrt{3}}{2}$.$f(\pi/6) = 1 - \cos(\pi/6) = 1 - \frac{\sqrt{3}}{2}$.Therefore, $f''(\pi/6) + f(\pi/6) = \frac{\sqrt{3}}{2} + 1 - \frac{\sqrt{3}}{2} = 1$.

Column $I$	Column $II$
$(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$	$(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$
$(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$	$(q)$ $(0, \frac{\pi}{2})$
$(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies	$(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$
$(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$, where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

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