Question

If the function $y = f ( x )$ is represented as,
$x =\phi( t )= t ^{3}-5 t ^{2}-20 t +7 $
$y =\psi( t )=4 t ^{3}-3 t ^{2}-18 t +3 \,\,(| t |<2), $ then:

• A. $y_{\max }=12$
• B. $y _{\max }=14$
• C. $y _{\min }=-67 / 4$
• D. $y_{\min }=-69 / 4$

Answer 1

Answer: (B), (D)

we have
$x=t^{5}-5 t^{3}-20 t+7$
and, $y=4 t^{3}-3 t^{2}-18 t+3$
$\therefore \frac{ dx }{ dt }=5 t ^{4}-15 t ^{2}-20$
and $\frac{ dy }{ dt }=12 t ^{2}-6 t -18$
$\Rightarrow \frac{d x}{d t}=5\left(t^{2}-4\right)\left(t^{2}+1\right)$ and
$\Rightarrow \frac{d y}{d t}=6(2 t-3)(t+1)$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{6(2 t-3)(t+1)}{5\left(t^{2}-4\right)\left(t^{2}+1\right)}$
Now, $ \frac{ dy }{ dx }=0 $
$\Rightarrow t =\frac{3}{2},-1$
clearly, these values satisfy $-2< t <2$.
Now, $\frac{ d ^{2} y }{ dx ^{2}}=\frac{ d }{ dx }\left(\frac{\frac{ dy }{ dt }}{\frac{ dx }{ dt }}\right)$
$=\frac{ d }{ dt }\left(\frac{\frac{ dy }{ dt }}{\frac{ dx }{ dt }}\right) \cdot \frac{ dt }{ dx }$
$=\frac{\frac{d x}{d t} \cdot \frac{d^{2} y}{d t^{2}}-\frac{d y}{d t} \cdot \frac{d^{2} x}{d t^{2}}}{\left(\frac{d x}{d t}\right)^{3}}$
$=\frac{\left(5 t ^{4}-15 t ^{2}-20\right)(24 t -6)-\left(12 t ^{2}-6 t -18\right)\left(20 t ^{3}-30 t \right)}{\left(5 t ^{4}-15 t ^{2}-20\right)^{3}}$
When $t=-1$, we have
$\frac{ d ^{2} y }{ dx ^{2}}=\frac{-30 \times-30}{(-30)^{3}}<0, x =31 $ and $y =14$
when $t =\frac{3}{2}$, we have
$\frac{ d ^{2} y }{ dx ^{2}}>0, x =\frac{-1033}{32}$ and $y =\frac{-69}{4}$
Hence, the function $y=f(x)$ attains a local maximum at $(31,14)$ and a local minimum at $(-1033 / 32,-69 / 4)$