Question

Let
$f\left(x\right)=\left|\begin{array}{ccc}\frac{{\pi }^n}{x}& \sin \pi x& \cos \pi x\\ (-1{)}^{n}n!& -\sin \left(\frac{n\pi }{2}\right)& -\cos \left(\frac{n\pi }{2}\right)\\ -1& \frac{1}{\sqrt{2}}& \frac{\sqrt{3}}{2}\end{array}\right|$
then value of $\frac{d^n}{d{x}^n}\left[f\left(x\right)\right]atx=1$ is

• A. $0$
• B. $-1$
• C. $\frac{1}{\sqrt{2}}$
• D. independent of $x$

Answer 1

Answer: (A), (D)

The correct options are
A $0$
D independent of $x$

We know that
$\frac{d^n}{d{x}^n}\left[\frac{{\pi }^n}{x}\right]=\frac{(-1{)}^{n}n!{\pi }^n}{x^{n+1}}$;

$\frac{d^n}{d{x}^n}\left[\sin \pi x\right]={\pi }^n\sin (\pi x+\frac{n\pi }{2})$ and

$\frac{d^n}{d{x}^n}\left[\cos \pi x\right]={\pi }^n\cos (\pi x+\frac{n\pi }{2})$

Thus,

$f^n\left(x\right)=\left|\begin{array}{ccc}\frac{(-1{)}^{n}n!{\pi }^n}{x^{n+1}}& {\pi }^n\sin (\pi x+\frac{n\pi }{2})& {\pi }^n\cos (\pi x+\frac{n\pi }{2})\\ (-1{)}^{n}n!& -\sin \left(\frac{n\pi }{2}\right)& -\cos \left(\frac{n\pi }{2}\right)\\ -1& \frac{1}{\sqrt{2}}& \frac{\sqrt{3}}{2}\end{array}\right|$

$\Rightarrow f^n\left(1\right)=\left|\begin{array}{ccc}(-1{)}^{n}n!{\pi }^n& {\pi }^n\sin (\pi +\frac{n\pi }{2})& {\pi }^n\cos (\pi +\frac{n\pi }{2})\\ (-1{)}^{n}n!& -\sin \left(\frac{n\pi }{2}\right)& -\cos \left(\frac{n\pi }{2}\right)\\ -1& \frac{1}{\sqrt{2}}& \frac{\sqrt{3}}{2}\end{array}\right|$

$=0$ [$\because R_{1}and{R}_3$ are proportional]

This also show that $f^n\left(1\right)$ is independent of $a$