Question

Let \[a\; \in \;R\] and \[f:R \to R\] be given by \[f\left( x \right) = {x^5}-5x + a\], then
(A) \[f\left( x \right)\] has three real roots if \[a > 4\]
(B) \[f\left( x \right)\] has only one real root if \[a > 4\]
(C) \[f\left( x \right)\] has three real roots if \[a < -4\]
(D) \[f\left( x \right)\] has three real roots if\[-{\text{ }}4 < a < 4\].

Answer 1

Hint: Here we will first find the derivative of the given function and find the critical points and then find its double derivative to find the maxima and minima of the given function and then we will put in the values of maxima and minima in the functions to find the maximum and minimum values of the function and then check each of the given options whether they are true of false.

Complete step-by-step answer:
The given function is:-
\[f\left( x \right) = {x^5}-5x + a\]
Differentiating both the sides we get:-
\[\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {{x^5}-5x + a} \right)\]
Simplifying it further we get:-
\[f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^5}} \right) - 5\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( a \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n\left( {{x^{n - 1}}} \right)\]
Also, the derivative of a constant term is zero.
Applying these formulas we get:-
\[f'\left( x \right) = 5{x^4} - 5 + 0\]
Simplifying it further we get:-
\[f'\left( x \right) = 5{x^4} - 5\]
Now we will put \[f'\left( x \right) = 0\] in order to find the critical points.
Hence putting \[f'\left( x \right) = 0\] we get:-
\[5{x^4} - 5 = 0\]
Dividing the equation by 5 as common we get:-
\[\dfrac{{5{x^4}}}{5} - \dfrac{5}{5} = \dfrac{0}{5}\]
Simplifying it we get:-
\[{x^4} - 1 = 0\]
Now factoring it we get;-
\[\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right) = 0\]
Solving for x we get:-
\[
  {x^2} + 1 = 0;{x^2} - 1 = 0 \\
   \Rightarrow {x^2} = - 1;{x^2} = 1 \\
 \]
Now since the value of \[{x^2}\] can never be negative
Hence, \[{x^2} = - 1\] is rejected.
This implies,
\[
  {x^2} = 1 \\
   \Rightarrow x = \pm 1 \\
 \]
Now again differentiating \[f'\left( x \right)\] we get:-
\[\dfrac{d}{{dx}}\left[ {f'\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {5{x^4} - 5} \right)\]
Simplifying it further we get:-
\[f''\left( x \right) = 5\dfrac{d}{{dx}}\left( {{x^4}} \right) - \dfrac{d}{{dx}}\left( 5 \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n\left( {{x^{n - 1}}} \right)\]
Also, the derivative of a constant term is zero.
Applying these formulas we get:-
\[f''\left( x \right) = 5\left( {4{x^3}} \right) - 0\]
Simplifying it further we get:-
\[f''\left( x \right) = 20{x^3}\]
Now when \[x = 1\]
\[
  f''\left( x \right) = 20{\left( 1 \right)^3} \\
   \Rightarrow f''\left( x \right) = 20 > 0 \\
 \]
Hence \[x = 1\] is the local minima.
Now when \[x = - 1\]
\[
  f''\left( x \right) = 20{\left( { - 1} \right)^3} \\
   \Rightarrow f''\left( x \right) = - 20 < 0 \\
 \]
Hence \[x = - 1\] is the local maxima.
Now we will find the value of function f(x) at \[x = 1\] and \[x = - 1\]
Therefore, when \[x = 1\]
\[f\left( 1 \right) = {\left( 1 \right)^5}-5\left( 1 \right) + a\]
Solving it further we get:-
\[
  f\left( 1 \right) = 1 - 5 + a \\
   \Rightarrow f\left( 1 \right) = a - 4 \\
 \]
When \[x = - 1\]
\[f\left( { - 1} \right) = {\left( { - 1} \right)^5}-5\left( { - 1} \right) + a\]
Solving it further we get:-
\[
  f\left( 1 \right) = - 1 + 5 + a \\
   \Rightarrow f\left( 1 \right) = a + 4 \\
 \]
               

Now we will check each of the options.
(a) When \[a > 4\]
\[f( - 1) > 0\;{\text{and}}\;f(1) > 0\]
Hence, there will be only one real root.
Hence it is incorrect.
(b) When \[a > 4\]
\[f( - 1) > 0\;{\text{and}}\;f(1) > 0\]
Hence, there will be only one real root.
Hence it is correct.
(c) When \[a < - 4\]
\[f( - 1) < 0\;{\text{and}} f(1) > 0\]
Hence, there will be one real root.
Hence it is incorrect.
(d) When \[ - 4 < a < 4\]
\[f( - 1) > 0\;{\text{and}}\;f(1) < 0\]
Hence, there will be three real roots.
Hence it is correct.

So, the correct answer is “Option B” and “Option D”.

Note: Students should keep in mind that if the double derivative of a function is greater than zero at any critical point then it is the local minima and the value of the function at that point is the minimum value.
Also, if the double derivative of a function is less than zero at any critical point then it is the local maxima and the value of the function at that point is the maximum value.
Also, if
 \[x \to \infty ,\;f(x) \to \infty \] and \[x \to - \infty \;,\;f(x) \to - \infty \] then the function has at least one root.