Skip to main content

### Updates ...

Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

For guest/sponsored article(s), please check this link.

### $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy}$ f(x)=?

The function f(x), that satisfies the condition $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy}$ is:

(A) $x + \frac{2}{3}(\pi - 2)\sin x$
(B) $x + (\pi + 2)\sin x$
(C) $x + \frac{\pi }{2}\sin x$
(D) $x + (\pi - 2)\sin x$

Solution

We have, $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} = x + \sin x\int\limits_0^{\pi /2} {\cos yf(y)dy} = x + \sin x.k$

Where, $k = \int\limits_0^{\pi /2} {\cos yf(y)dy} = \int\limits_0^{\pi /2} {\cos y(y + \sin y.k)dy}$

$\therefore k = \int\limits_0^{\pi /2} {y\cos ydy} + k\int\limits_0^{\pi /2} {\cos y\sin ydy}$

$\Rightarrow k = \left. {y\sin y} \right|_0^{\pi /2} - \int\limits_0^{\pi /2} {\sin ydy} + \frac{k}{2}\int\limits_0^{\pi /2} {sin2ydy} = \frac{\pi }{2} + \left. {\cos y} \right|_0^{\pi /2} - \frac{k}{2}.\frac{1}{2}\left. {\cos 2y} \right|_0^{\pi /2}$

$\Rightarrow k = \frac{\pi }{2} - 1 - \frac{k}{4}( - 1 - 1) = \frac{\pi }{2} - 1 + \frac{k}{2}$

$\Rightarrow \frac{k}{2} = \frac{\pi }{2} - 1 = \frac{{\pi - 2}}{2}$

$\therefore k = \pi - 2$

Now, $f(x) = x + \sin x.k = x + (\pi - 2)\sin x$

Answer: (D)

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$