Question

Suppose $f\left( x \right) = {e^{ax}} + {e^{bx}}$, where $a \ne b$, and that for all x. Then the product ab is equal to
A. 25
B.9
C. -15
D. -9

Answer 1

Hint: Find the derivative and double derivative of f(x) and substitute those values in the equation to get the product of ‘a’ and ‘b’.


Complete step-by-step Solution:
We are given that $f\left( x \right) = {e^{ax}} + {e^{bx}}$, a is not equal to b $a \ne b$ and.
Find the derivative of the function f(x).
$
  y = f\left( x \right) = {e^{ax}} + {e^{bx}} \\
  \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{e^{ax}} + {e^{bx}}} \right)}}{{dx}} \\
  f'\left( x \right) = \dfrac{{d\left( {{e^{ax}}} \right)}}{{dx}} + \dfrac{{d\left( {{e^{bx}}} \right)}}{{dx}} \\
  f'\left( x \right) = a\left( {{e^{ax}}} \right) + b\left( {{e^{bx}}} \right) \\
  \left( {\because \dfrac{{d\left( {{e^{ax}}} \right)}}{{dx}} = a\left( {{e^{ax}}} \right)} \right) \\
  f'\left( x \right) = a{e^{ax}} + b{e^{bx}} \\
 $
Find the double derivative of f(x) or derivative of f’(x).

Substitute the values of f’(x) and f”(x) in
$f'\left( x \right) = a{e^{ax}} + b{e^{bx}}$


${e^{ax}}\& {e^{bx}}$ Cannot be Zero so
$
  \left( {{a^2} - 2a - 15} \right) = 0\& \& \left( {{b^2} - 2b - 15} \right) = 0 \\
  \left( {{a^2} - 5a + 3a - 15} \right) = 0\& \& \left( {{b^2} - 5b + 3b - 15} \right) = 0 \\
  \left( {a\left( {a - 5} \right) + 3\left( {a - 5} \right)} \right) = 0\& \& \left( {b\left( {b - 5} \right) + 3\left( {b - 5} \right)} \right) = 0 \\
  \left( {a + 3} \right)\left( {a - 5} \right) = 0\& \& \left( {b + 3} \right)\left( {b - 5} \right) = 0 \\
  a = - 3,5\& \& b = - 3,5 \\
 $
Given that a and b are not equal. So, if a=-3, b=5 and if a=5, b=-3
In both the cases, product of a and b (ab) is -15.
Therefore, from among the options given in the question option c is correct which is the product if a and b is -15.

Additional Information: Differentiating an expression with respect to x means thinking of the expression as a function of x, so that x is the only variable present in it, and other terms are either consonants or functions of x themselves and then differentiating of course. Let's say you have an equation in terms of x, y=f(x) and you want to know "How much does y change when I change x a little bit?” To do this, you take the derivative of the expression, and x is the variable that you are differentiating. That is, you are differentiating with respect to the variable x. It just means that you care about changing x a little bit to see what happens with y.


Note: Differentiation of ${e^x}$ is ${e^x}$.The value of ‘e’ is 2.718281828 which is a famous irrational number and also called Euler’s number. Differentiation of a function ${x^n} = n.{x^{n - 1}}$