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Let `h(x)` be differentiable for all `x` and let `f(x)=(k x+e^x)h(x)` , where `k` is some constant. If `h(0)=5,h^(prime)(0)=-2,a n df^(prime)(0)=18 ,` then the value of `k` is 5 (b) 4 (c) 3 (d) 2.2.

A

5

B

4

C

3

D

`2.2`

Text Solution

Verified by Experts

The correct Answer is:
C

f(x) = (kx + `e^(x))` h (x)
Diff.
f'(x) = (kx + `e^(x))h' (x) + h(x) (k + e^(x))`
put x = 0
18 = `-2 + 5 ` (k+1)
k = 3
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