Question

Observe the following lists

List-I	List-II
(A) Maximum value of $xy$ subject to $x+y=7$ is	1) $72$
(B) If $l^2+m^2=1$ , then the maximum value of $l+m$ is	2) $1$
(C) If $x+y=12$, then the minimum Value of $x^2+y^2$ is	3) $\sqrt{2}$
(D) Minimum value $x^2-8x+17$ is	4) $\frac{49}{4}$
	5) $0$

• A. A - 4, B -3, C -1, D -2.
• B. A - 4, B -3, C -2, D -1.
• C. A - 2, B -3, C -5, D -4.
• D. A - 2, B -3, C -1, D -4.

Answer 1

Answer: (A)

A - 4, B -3, C -1, D -2.

(A) use A.M. & G.M.
$\frac{x+y}{2}\ge (xy{)}^{\frac{1}{2}}$
$\left(\frac{7}{2}\right)\ge (xy{)}^{\frac{1}{2}}$
$\left(xy\right)\le (\frac{7}{2}{)}^2$
(B) $y=l+\sqrt{1-l^2}$

$\frac{dy}{dl}=1-\frac{l}{\sqrt{1-l^2}}$

$\frac{dy}{dl}=0$ when $l=\frac{1}{\sqrt{2}}$
So $m=\frac{1}{\sqrt{2}}$
$\Rightarrow l=\frac{1}{\sqrt{2}}$

$\Rightarrow l+m=\sqrt{2}$

(C) $s=x^2+(12-x{)}^2$
$\frac{ds}{dx}=2x-2(12-x)$
$\frac{ds}{dx}=0$ when $x=6$ $y=6$
$s=36+36=72$
(D) $f^{\prime }\left(x\right)=2x-8$
$f^{\prime }\left(x\right)=0$ at $x=y$
$f\left(y\right)=1$