# ISI2017-DCG-30

0 votes
62 views

If $f(x)=e^{5x}$ and $h(x)=f’’(x)+2f’(x)+f(x)+2$  then $h(0)$ equals

1. $38$
2. $8$
3. $4$
4. $0$
in Calculus
recategorized

## 2 Answers

1 vote

$f(x)=e^{5x}$

$f'(x)=5e^{5x}$

$f''(x)=25 e^{5x}$

$h(x)=25 e^{5x} + 10 e^{5x} + e^{5x} +2$

$h(0)= 25+10+1+2=38$

Option A)

1 vote

Answer : Option (A)

$f(x) = e^{5x}$

Differentiating wrt. x,

$f'(x) = 5 \cdot e^{5x}$

Differentiating wrt. x,

$f''(x) = 5 \cdot 5 \cdot e^{5x} = 25 \cdot e^{5x}$

Thus, anwer will be

$25 \cdot e^{0} + 10 \cdot e^0 + e^0 + 2 = 38$

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