Question

Consider the following statements
$S_1:$ If $f\left(x\right)=|x^2-5|x|+6|,$ then $f^{\prime }\left(\frac{-5}{2}\right)=0$
$S_2:$ The slope of tangent at $x=1$ on the curve $y={\sin }^{-1}\left(\cos \pi x\right)$ is $\pi$
$S_3:$ If $x=f\left(t\right),y=g\left(t\right),$ then $\frac{d^{2}x}{{dy}^2}=\frac{f^{\prime \prime }\left(t\right)}{g^{\prime \prime }\left(t\right)}$
Which of the following is/are correct about the truth value of above statements

• A. $S_1$ is true
• B. $S_2$ is false
• C. all are true
• D. $S_3$ is false

Answer 1

Answer: (A), (B), (D)

$S_3$ is false

$S_1:f\left(x\right)=|x^2+5x+6|,$ for $x<0$
$f\left(x\right)=-(x^2+5x+6),\text{if}-3\le x\le -2$
$f^{\prime }\left(x\right)=-2x-5$
$f^{\prime }\left(\frac{-5}{2}\right)=0$

$S_2:y={\sin }^{-1}\left(\cos \pi x\right)$
$=\frac{\pi }{2}-{\cos }^{-1}\left(\cos \pi x\right)=\{\begin{array}{cc}\frac{\pi }{2}-\pi x,& 0\le x\le 1\\ \pi x-\frac{3\pi }{2},& 1<x\le 2\end{array}$
Function is not differentiable at $x=1.$
Hence, no tangent can be drawn at this point.

$S_3:\frac{dx}{dt}=f^{\prime }\left(t\right),\frac{dy}{dt}=g^{\prime }\left(t\right)$
$\Rightarrow \frac{dx}{dy}=\frac{dx}{dt}\cdot \frac{dt}{dy}=\frac{f^{\prime }\left(t\right)}{g^{\prime }\left(t\right)}$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dt}\left(\frac{dx}{dy}\right)\cdot \frac{dt}{dy}$
$\Rightarrow \frac{d^{2}x}{d{y}^2}=\frac{f^{\prime }\left(t\right)g^{\prime \prime }\left(t\right)-f^{\prime \prime }\left(t\right)g^{\prime }\left(t\right)}{\left(f^{\prime }\right(t){)}^3}$