Verified Solution • 2025–26 Session moderate

QJEEM20Ma1630: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1630 Solution: If a function f(x) defined by f ( x ) = { a e x + b e − x , − 1 …

If a function f(x) defined by

f(x)={aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx,3<x≤4

be continuous for some aa , b, c ∈\in R and f'(0) + f'(2) = e, then the value of of aa is :

Multiple Choice Options

  • A ee2−3e−13{e \over {{e^2} - 3e - 13}}
  • B 1e2−3e+13{1 \over {{e^2} - 3e + 13}}
  • C ee2−3e+13{e \over {{e^2} - 3e + 13}}
  • D ee2+3e+13{e \over {{e^2} + 3e + 13}}

Step-by-Step Verified Solution

Given function,
f(x)={aex+be−x,−1≤x<1cx2,1≤x≤3ax2+2cx,3<x≤4

For continuity at x = 1

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

⇒\Rightarrow ae+be−1=cae + b{e^{ - 1}} = c

⇒\Rightarrow b = ce - aa e2 .....(1)

For continuity at x = 3

lim⁡x→3−f(x)=lim⁡x→3+f(x)\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 = ee2−3e+13{e \over {{e^2} - 3e + 13}}

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