Question

Let $P(x)$ be a polynomial of degree 6 defined on real numbers and graph of $y=P(x)$ is symmetrical w.r.t. $y$-axis. If $P(1)=2,P(2)=5,P(3)=10$ and $R(x)$ is remainder when $P(x)$ is divided by

• A. minimum value of $R(x)$ is 1
• B. minimum value of $R(x)$ is 0.
• C. ${\sin }^{-1}(\sin R(1))+{\cos }^{-1}(\cos R(1))+{\tan }^{-1}(\tan R(1))=2$
• D. ${\sin }^{-1}(\sin R(2))+{\cos }^{-1}(\cos R(2))+{\tan }^{-1}(\tan R(1))=2$

Answer 1

Answer: (A), (C)

(A) $\Theta$ Graph is symmetrical w.r.t. y-axis.
$\therefore P(x)$ is even function
$\therefore P(x)=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)+\left(x^2+1\right)$
dividing $P(x)$ by $(x-1)(x-2)(x-3)$, we get remainder as $R(x)=x^2+1$
$\therefore$ Minimum value of $R(x)=1$
(C)${\sin }^{-1}(\sin R(1))+{\cos }^{-1}(\cos R(1))+{\tan }^{-1}(\tan R(1))={\sin }^{-1}(\sin 2)+{\cos }^{-1}(\cos 2)+{\tan }^{-1}(\tan 2)$
$=\pi -2+2+2-\pi =2$
(D)${\sin }^{-1}(\sin R(2))+{\cos }^{-1}(\cos R(2))+{\tan }^{-1}(\tan R(1))={\sin }^{-1}(\sin 5)+{\cos }^{-1}(\cos 5)+{\tan }^{-1}(\tan 2)$
$=5-2\pi +2\pi -5+2-\pi =2-\pi$