Question

Assertion :If $f\left(x\right)=(\cos x+i\sin x)(\cos 2x+i\sin 2x)$$(\cos 3x+i\sin 3x)...(\cos nx+i\sin nx)$ and $f\left(1\right)=1$ then $f^{\prime \prime }\left(1\right)$ is equal to $-{\left(\frac{n(n+1)}{2}\right)}^2$. Reason: $f\left(x\right)=\cos \frac{n(n-1)}{2}x+i\sin \frac{n(n-1)}{2}x$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$f\left(x\right)=(\cos x+i\sin x)(\cos 2x+i\sin 2x)$

$\left(\cos 3x+i\sin 3x\right)...(\cos nx+i\sin nx)$

$=\cos (x+2x=3x+...+nx)+i\sin (x+2x+3x+...+nx)$

$=\cos \frac{n(n+1)}{2}x+i\sin \frac{n(n+1)}{2}x$

$\Rightarrow f^{\prime }\left(x\right)=\frac{n(n+1)}{2}[-\sin \frac{n(n+1)}{2}x+i\cos \frac{n(n+1)}{2}x]$

$\Rightarrow f^{\prime \prime }\left(x\right)=-{\left(\frac{n(n+1)}{2}\right)}^2\left(\cos \frac{n(n+1)}{2}x+i\sin \frac{n(n+1)}{2}x\right)=-{\left(\frac{n(n+1)}{2}\right)}^{2}f\left(x\right)$

$\therefore f^{\prime \prime }\left(x\right)=-{\left(\frac{n(n+1)}{2}\right)}^{2}f\left(1\right)=-{\left(\frac{n(n+1)}{2}\right)}^2$