# If f(x) = b\[{{\text{e}}^{{\text{ax}}}}\]+ a\[{{\text{e}}^{{\text{bx}}}}\], then ${{\text{f}}^{''}}$(0) =

$

{\text{A}}{\text{. 0}} \\

{\text{B}}{\text{. 2ab}} \\

{\text{C}}{\text{. ab(a + b)}} \\

{\text{D}}{\text{. ab}} \\

$

Answer

Verified

361.2k+ views

Hint - To solve this question we differentiate the equation f(x) twice. Then we substitute 0 in place of x to determine the answer.

Complete step-by-step answer:

Given f(x) = b\[{{\text{e}}^{{\text{ax}}}}\]+ a\[{{\text{e}}^{{\text{bx}}}}\]

⟹f’(x) = $\dfrac{{{\text{df}}}}{{{\text{dx}}}}$ = b\[{{\text{e}}^{{\text{ax}}}}\]$\dfrac{{\text{d}}}{{{\text{dx}}}}$(ax) +a\[{{\text{e}}^{{\text{bx}}}}\]$\dfrac{{\text{d}}}{{{\text{dx}}}}$(bx) = b\[{{\text{e}}^{{\text{ax}}}}\]a +a\[{{\text{e}}^{{\text{bx}}}}\]b = ab (\[{{\text{e}}^{{\text{ax}}}}\]+ \[{{\text{e}}^{{\text{bx}}}}\])

($\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{e}}^{\text{x}}}} \right) = {{\text{e}}^{\text{x}}}$)

⟹f “(x) = $\dfrac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{x}}^2}}}$ = ab ( a\[{{\text{e}}^{{\text{ax}}}}\] + \[{{\text{e}}^{{\text{bx}}}}\]b)

⟹f “(0) = ab(${{\text{e}}^0}$a+${{\text{e}}^0}$b) (${{\text{e}}^0}$=1)

⟹f”(0) = ab (a+b)

Note: In order to solve this type of question the key is to carefully differentiate the equation.

Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable.

The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.

Complete step-by-step answer:

Given f(x) = b\[{{\text{e}}^{{\text{ax}}}}\]+ a\[{{\text{e}}^{{\text{bx}}}}\]

⟹f’(x) = $\dfrac{{{\text{df}}}}{{{\text{dx}}}}$ = b\[{{\text{e}}^{{\text{ax}}}}\]$\dfrac{{\text{d}}}{{{\text{dx}}}}$(ax) +a\[{{\text{e}}^{{\text{bx}}}}\]$\dfrac{{\text{d}}}{{{\text{dx}}}}$(bx) = b\[{{\text{e}}^{{\text{ax}}}}\]a +a\[{{\text{e}}^{{\text{bx}}}}\]b = ab (\[{{\text{e}}^{{\text{ax}}}}\]+ \[{{\text{e}}^{{\text{bx}}}}\])

($\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{e}}^{\text{x}}}} \right) = {{\text{e}}^{\text{x}}}$)

⟹f “(x) = $\dfrac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{x}}^2}}}$ = ab ( a\[{{\text{e}}^{{\text{ax}}}}\] + \[{{\text{e}}^{{\text{bx}}}}\]b)

⟹f “(0) = ab(${{\text{e}}^0}$a+${{\text{e}}^0}$b) (${{\text{e}}^0}$=1)

⟹f”(0) = ab (a+b)

Note: In order to solve this type of question the key is to carefully differentiate the equation.

Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable.

The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.

Last updated date: 30th Sep 2023

•

Total views: 361.2k

•

Views today: 6.61k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE