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 :
Solution
Given function,
<br>$$f\left( x \right) = \left\{ {\matrix{
{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr
{c{x^2},} & {1 \le x \le 3} \cr
{a{x^2} + 2cx,} & {3 < x \le 4} \cr
} } \right.$$
<br><br>For continuity at x = 1
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$$
<br><br>$\Rightarrow$ $ae + b{e^{ - 1}} = c$
<br><br>$\Rightarrow$ b = ce - $a$e<sup>2</sup> .....(1)
<br><br>For continuity at x = 3
<br><br>$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)$$
<br><br>$\Rightarrow$ 9c = 9a + 6c
<br><br>$\Rightarrow$ c = 3a .......(2)
<br><br>Also given, f'(0) + f'(2) = e
<br><br>$\Rightarrow$ (ae<sup>x</sup>
– be<sup>x</sup>
)<sub>x=0</sub> + (2c<sup>x</sup>)<sub>x=2</sub> = e
<br><br>$\Rightarrow$ a – b + 4c = e ........(3)
<br><br>From (1), (2) & (3)
<br><br>a – 3ae + ae<sup>2</sup>
+ 12a = e
<br><br>$\Rightarrow$ a(e<sup>2</sup>
+ 13 – 3e) = e
<br><br>$\Rightarrow$ a = ${e \over {{e^2} - 3e + 13}}$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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