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Question: If \(f\left( x \right) = {x^2} - 3x\), then the options at which \(f\left( x \right) = f'\left( x \r...

If f(x)=x2−3xf\left( x \right) = {x^2} - 3x, then the options at which f(x)=f′(x)f\left( x \right) = f'\left( x \right) are:
(A) 1,31,3
(B) 1,−31, - 3
(C) −1,3 - 1,3
(D) None of these

Explanation

Solution

Hint : In the given problem, we are provided with a function in variable x and a relation between the function and its derivative. So, we differentiate the function with respect to x and form an equation in variable x. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease.

Complete step-by-step answer :
So, we have, f(x)=x2−3xf\left( x \right) = {x^2} - 3x.
Differentiating both sides of the function with respect to x, we get,
⇒ddx[f(x)]=d(x2−3x)dx\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{{d\left( {{x^2} - 3x} \right)}}{{dx}}
Now, using the power rule of differentiation d(xn)dx=nx(n−1)\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{\left( {n - 1} \right)}}, we get,
⇒f′(x)=(2x−3)\Rightarrow f'\left( x \right) = \left( {2x - 3} \right)
Now, we are given that f(x)=f′(x)f\left( x \right) = f'\left( x \right).
So, we get, x2−3x=(2x−3){x^2} - 3x = \left( {2x - 3} \right).
So, we get a quadratic equation in x. We shift all the terms to the left side of the equation to simplify.
⇒x2−3x−2x+3=0\Rightarrow {x^2} - 3x - 2x + 3 = 0
Adding the like terms, we get,
⇒x2−5x+3=0\Rightarrow {x^2} - 5x + 3 = 0
Now, we solve the quadratic equation x2−5x+3=0{x^2} - 5x + 3 = 0 with the help of quadratic formula. The quadratic formula can be employed for solving an equation only if we compare the given equation with standard form of quadratic equation.
Comparing with standard quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0
Here,a=1a = 1, b=−5b = - 5 andc=3c = 3.
Now, using the quadratic formula, we get the roots of the equation as:
x=(−b)±b2−4ac2ax = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Substituting the values of a, b, and c in the quadratic formula, we get,
x=−(−5)±(−5)2−4×1×(3)2×(1)x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{( - 5)}^2} - 4 \times 1 \times \left( 3 \right)} }}{{2 \times \left( 1 \right)}}
⇒x=5±25−122\Rightarrow x = \dfrac{{5 \pm \sqrt {25 - 12} }}{2}
⇒x=5±13\Rightarrow x = 5 \pm \sqrt {13}
So, the values of x at which f(x)=f′(x)f\left( x \right) = f'\left( x \right) are: x=5+13x = 5 + \sqrt {13} and x=5−13x = 5 - \sqrt {13} .
Therefore, the option (D) is the correct answer.
So, the correct answer is “Option D”.

Note : Quadratic equations are the polynomial equations with degree of the variable or unknown as 2. Quadratic formula is the easiest and most efficient formula to calculate the roots of an equation. Algebraic identities and simplification rules are also of vital importance in such questions. Power rules must be remembered to find the derivative of the expression.