JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 4)

If a function f(x) defined by

$$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$$

be continuous for some $$a$$, b, c $$ \in $$ R and f'(0) + f'(2) = e, then the value of of $$a$$ is :
$${e \over {{e^2} - 3e - 13}}$$
$${1 \over {{e^2} - 3e + 13}}$$
$${e \over {{e^2} - 3e + 13}}$$
$${e \over {{e^2} + 3e + 13}}$$

Explanation

Given function,
$$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$$

For continuity at x = 1

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$$

$$ \Rightarrow $$ $$ae + b{e^{ - 1}} = c$$

$$ \Rightarrow $$ b = ce - $$a$$e2 .....(1)

For continuity at x = 3

$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)$$

$$ \Rightarrow $$ 9c = 9a + 6c

$$ \Rightarrow $$ c = 3a .......(2)

Also given, f'(0) + f'(2) = e

$$ \Rightarrow $$ (aex – bex )x=0 + (2cx)x=2 = e

$$ \Rightarrow $$ a – b + 4c = e ........(3)

From (1), (2) & (3)

a – 3ae + ae2 + 12a = e

$$ \Rightarrow $$ a(e2 + 13 – 3e) = e

$$ \Rightarrow $$ a = $${e \over {{e^2} - 3e + 13}}$$

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