Q13

Calculus Level 3

A twice differentiable function f ( x ) f(x) is defined for all real numbers and satisfies the following conditions :

{ f ( 0 ) = 2 f ′ ( 0 ) = − 5 f ′ ′ ( 0 ) = 3 \begin{cases} f(0)=2 \\ f'(0)=-5 \\ f''(0)=3 \end{cases}

The function g ( x ) g(x) is defined by g ( x ) = e a x + f ( x ) g(x)={ e }^{ ax }+f(x) for all x ∈ R x\in R , where a a is a constant.

If g ′ ( 0 ) + g ′ ′ ( 0 ) = 0 g'(0)+g''(0)=0 , then find the possible values of a a .

1 , − 2 1,-2 1 , 2 1,2 1 , − 1 1,-1 2 , − 2 2,-2

Sahil Bansal by Sahil Bansal , 21, India

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1 solution

Differentiating g ( x ) g(x) the first time gives, g ′ ( x ) = a e a x + f ′ ( x ) g'(x) = ae^{ax}+f'(x) At x = 0 x=0 ,we have , g ′ ( 0 ) = a × e 0 + f ′ ( 0 ) ⇒ g ′ ( 0 ) = a − 5 g'(0)=a\times e^{0}+f'(0) \Rightarrow g'(0)=a-5 Now differentiating g ′ ( x ) g'(x) we get, g ′ ′ ( x ) = a 2 e a x + f ′ ′ ( x ) g''(x)=a^{2}e^{ax}+f''(x) Again at x = 0 x=0 we have, g ′ ′ ( 0 ) = a 2 × e 0 + f ′ ′ ( 0 ) ⇒ g ′ ′ ( 0 ) = a 2 + 3 g''(0)=a^{2}\times e^{0}+f''(0) \Rightarrow g''(0) = a^{2}+3 Now its given that , g ′ ( 0 ) + g ′ ′ ( 0 ) = 0 ⇒ a − 5 + a 2 + 3 = 0 ⇒ a 2 + a − 2 = 0 g'(0)+g''(0) = 0 \Rightarrow a-5+a^{2}+3=0 \Rightarrow a^{2}+a-2=0 Solving the above quadratic we get the answer as, a = 1 , − 2 a=\boxed{\boxed{\boxed{1,-2}}}

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