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

### Solve for x & y$\sqrt x + y = 7$$x + \sqrt y = 11$

Putting $y = {(11 - x)^2}$ obtained from the 2nd equation into the 1st equation,

We have, $\sqrt x + {(11 - x)^2} = 7$

$\Rightarrow x = {[7 - {(11 - x)^2}]^2}$

$= 49 + {(11 - x)^4} - 14{(11 - x)^2}$

Let, $11 - x = t$

So, $11 - t = 49 + {t^4} - 14{t^2}$

$\Rightarrow {t^4} - 14{t^2} + t + 38 = 0$

t=2 satisfies the above equation. So, using factor theorem

${t^3}(t - 2) + 2{t^2}(t - 2) - 10t(t - 2) - 19(t - 2) = 0$

$\Rightarrow (t - 2)({t^3} + 2{t^2} - 10t - 19) = 0$ ........... (A)

Let, $f(t) = {t^3} + 2{t^2} - 10t - 19$

$t = 11 - x = \sqrt y \ge 0$

$y = 7 - \sqrt x , \Rightarrow y \le 7$

So, $t = \sqrt y \le \sqrt 7$

Thus, $0 \le t \le \sqrt 7$

Or, $0 \le t < 3$

Let us investigate the behaviour of f in the interval [0, 3].

f(0) = -19

f(1) = -26

f(2) = -23

f(3) = -4

The function decreases initially, then increases but is unable to cut the t-axis in the interval [0, 3).

t = 2 is the only solution as obtained earlier as per (A).

Now, $11 - x = 2$ or $x = 9$

$y = {t^2} = 4$

### 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, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$