Let $p(x)$ be a real polynomial of least degree which has a local maximum at x= 1 and a local minimum at x = 3,. If $p(1) = 6$ and $p(3 ) = 2 $, then $p'(0)$ is
Answer & explanation
Correct answer: option 4
The correct answer is option (4) : 9
Since $x=1$ and $x=3$ are extreme points of p(x).
Therefore, x= 1 and x= 3 are roots of p'(x) = 0.
$∴p'(x0 = k(x_1) (x-3) = k(x^2-4x+3)$
$⇒ p(x) = k \left(\frac{x^3}{3}-2x^2+3x\right)+C$
Now, $p(1) = 6 $ and $p(3) = 2$
$⇒6 = k \left(\frac{1}{3} - 2+3\right) + C$ and $2= k(9-18+9)+C$
$⇒6+\frac{4k}{3}+C$ and $C= 2$
$⇒ C=2 $ and $k=3$
$p'(x) = 3(x^2 -4x+3)$
Hence, $p'(0)=9.$