Question

Match the functions given in List 1 with their corresponding greatest value given in List 2.

Answer 1

A)

$y=\sqrt{100-x^2}$ on [-6, 8]
$\frac{dy}{dx}=\frac{-2x}{2\sqrt{100-x^2}}$
$\therefore$ The greatest value is at $x=0$.
i.e., $y=10$

B)

$y=2\tan x-{\tan }^{2}x$ on $[0,\frac{\pi }{2}]$
$\frac{dy}{dx}=2{\sec }^{2}x-2\tan x{\sec }^{2}x$
$\frac{dy}{dx}=2{\sec }^{2}x(1-\tan x)$
$\therefore$ The greatest value is at $\tan x=1$.
i.e., $y=1$ (Since, $\frac{dy}{dx}$ changes sign from positive to negative)

C)

$y={\tan }^{-1}\frac{1-x}{1+x}$ on $[0,1]$

$\Rightarrow y={\tan }^{-1}\frac{\pi }{4}-{\tan }^{-1}x$

$\frac{dy}{dx}=\frac{-1}{1+x^2}<0$
$y$ is decreasing.

$\therefore$ The greatest value is at $x=0$.
i.e., $y=\frac{\pi }{4}$

D)

$y=\frac{a^2}{x}+\frac{b^2}{1-x}$ on (0, 1), a> 0, b > 0

$\frac{dy}{dx}=-\frac{a^2}{x^2}+\frac{b^2}{(1-x{)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{2a^2}{x^3}+\frac{2b^2}{(-x{)}^3}>0$ on $(0,1)$

$\therefore$ No greatest value on $(0,1)$.
$A\to 2,B\to 4,C\to 3,D\to 1$