Question

If $f\left(x\right)=\left|\begin{array}{ccc}{x}^n& \sin x& \cos x\\ n!& \sin \frac{n\pi }{2}& \cos \frac{n\pi }{2}\\ a& a^2& a^3\end{array}\right|$, then the value of $\frac{d^n}{d{x}^n}\left(f\right(x\left)\right)$ at $x=0$ for $n=2m+1$ is

• A. dependent on $n$
• B. $0$
• C. $1$
• D. dependent on $a$

Answer 1

Answer: (B)

$0$

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

Now, $\frac{d^n}{d{x}^n}\left(f\right(x\left)\right)=$$\left|\begin{array}{ccc}n!& \sin (x+\frac{n\pi }{2})& \cos (x+\frac{n\pi }{2})\\ n!& \sin \frac{n\pi }{2}& \cos \frac{n\pi }{2}\\ a& a^2& a^3\end{array}\right|$ [ Ignoring the other two determinants which have det value $=0$]

or, ${\frac{d^n}{d{x}^n}\left(f\right(x\left)\right)|}_{x=0,n=2m+1}=$$\left|\begin{array}{ccc}(2m+1)!& \sin \left(\frac{(2m+1)\pi }{2}\right)& \cos \left(\frac{(2m+1)\pi }{2}\right)\\ (2m+1)!& \sin \frac{(2m+1)\pi }{2}& \cos \frac{(2m+1)\pi }{2}\\ a& a^2& a^3\end{array}\right|=0$. [ Since $R_1=R_2$]