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) $
$=5-2 \pi+2 \pi-5+2-\pi=2-\pi$