Question

If the function $ f\left( x \right) = {x^2} + \dfrac{a}{x} $ has a local minimum at $ x = 2 $ , then the value of a is
A. $ 8 $
B. $ 16 $
C. $ 18 $
D. none of these

Answer 1

Hint: In order to find the value of $ a $ , find the first derivative of the given function with respect to variable $ x $ . Now equation the derivative equal to zero and put $ x = 2 $ and solve the equation for $ a $ to get the required answer.

Complete step-by-step answer:
We are given a function in variable $ x $ as
 $ f\left( x \right) = {x^2} + \dfrac{a}{x} $ which is having a local minimum at $ x = 2 $
Before proceeding towards the solution, lets understand what are maxima and minima of any function. So, maxima and minima are the points on the graph of the function where the graph flattens or in other words we can say when the graph changes its nature from increasing to decreasing or vice-versa.
The slope of the graph at these flatten points is equal to zero.
To find the local minima or maxima we first find the derivative of the given function with respect to the variable as the derivative of any function is the slope of the graph . Then putting the slope equals to zero we obtain the local maxima and minima.
Here in our question, we first find the first derivative of function with respect to variable $ x $
 $ f'\left( x \right) = \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + \dfrac{a}{x}} \right) $
Rewriting the above equation by splitting the derivative into the terms inside the bracket, we get
 $ f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{a}{x}} \right) $
Now using the rules of derivative $ \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} $ , we get
 $ f'\left( x \right) = 2x - \dfrac{a}{{{x^2}}} $
Since, According to the question we have $ x = 2 $ as local minima, so this value will satisfy the above equation and make $ f'\left( x \right) = 0 $
Putting $ x = 2 $ and $ f'\left( x \right) = 0 $ ,and solving the equation for $ a $ , we get
 $
  0 = 2\left( 2 \right) - \dfrac{a}{{{{\left( 2 \right)}^2}}} \\
  0 = \dfrac{{16 - a}}{4} \\
  0 = 16 - a \\
  a = 16 \;
  $
Therefore, the value of $ a = 16 $ . So the correct option is (B)
So, the correct answer is “Option B”.

Note: 1.We should remember that the local maximum or local minimum values are not same as relative maximum or minimum
2. A function can have more than one local minima or maxima but only one global maxima and minima as they are the highest and lowest values respectively that the function can achieve.
3. Derivative always splits into each and every term of the function.
4. $ \dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} $ is used here.