JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 18)
Consider the function
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 _____________._25th_July_Evening_Shift_en_18_1.png)
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
Explanation
$$f(x) = \left\{ {\matrix{
{{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr
{7,} & {x = 2} \cr
} } \right.$$
P''(x) = const. $$\Rightarrow$$ P(x) is a 2 degree polynomial
f(x) is cont. at x = 2
f(2+) = f(2$$-$$)
$$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$$
$$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$$
P(x) = (x $$-$$ 2)(ax + b)
P(3) = (3 $$-$$ 2)(3a + b) = 9 $$\Rightarrow$$ 3a + b = 9
a = 2, b = 3
P(5) = (5 $$-$$ 2)(2.5 + 3) = 3.13 = 39
P''(x) = const. $$\Rightarrow$$ P(x) is a 2 degree polynomial
f(x) is cont. at x = 2
f(2+) = f(2$$-$$)
$$\mathop {\lim }\limits_{x \to {2^ + }} {{P(x)} \over {\sin (x - 2)}} = 7$$
$$\mathop {\lim }\limits_{x \to {2^ + }} {{(x - 2)(ax + b)} \over {\sin (x - 2)}} = 7 \Rightarrow 2a + b = 7$$
P(x) = (x $$-$$ 2)(ax + b)
P(3) = (3 $$-$$ 2)(3a + b) = 9 $$\Rightarrow$$ 3a + b = 9
a = 2, b = 3
P(5) = (5 $$-$$ 2)(2.5 + 3) = 3.13 = 39
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