Consider the function $f(x)=\frac{P(x)}{\sin (x-2)}, \quad x \neq 2$ \[=7, \quad x=2\] where $P(x)$ is a polynomial such that $P^{\prime \prime}(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to...... .

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Continuity and Differentiability

Consider the function

f (x) = \frac{P ( x )}{s i n ( x - 2 )}, x \neq = 2

= 7, x = 2

where

P (x)

is a polynomial such that

P^{''} (x)

is always a constant and

P (3) = 9

. If

f (x)

is continuous at

x = 2

, then

P (5)

is equal to...... .

Answer: 39

Since

P^{''} (x)

is constant,

P (x) = a x^{2} + b x + c

. Continuity at

x = 2

needs the numerator to vanish there, so

P (2) = 0

, and then

x \to 2 lim \frac{P ( x )}{sin ( x - 2 )} = x \to 2 lim \frac{P ^{'} ( x )}{cos ( x - 2 )} = P^{'} (2) = 7

Thus

4 a + 2 b + c = 0

and

4 a + b = 7

. Together with

P (3) = 9 a + 3 b + c = 9

, subtracting the first of these from the last gives

5 a + b = 9

, so

a = 2

b = - 1

and

c = - 6

. Hence

P (x) = 2 x^{2} - x - 6

and

P (5) = 50 - 5 - 6 = 39

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer