Question

Let the given function be f(x) as $f\left( x \right)=-{{x}^{2}}$. Then, the correct statements is/are:
(a) ${f}'\left( 1 \right)< 0$
(b) $f\left( 2 \right)< 0$
(c) ${f}'\left( x \right)\ne 0$ for any $x\in \left( 1,3 \right)$
(d) ${f}'\left( x \right)\ne 0$ for some $x\in \left( 1,3 \right)$

Answer 1

Hint: First, before proceeding for this, we must know that for these kinds of questions, we must proceed by options while checking which one is true or false with given conditions. Then, starting with the first option and finding the derivative and substituting the value gives the final condition. Then, similarly proceeding with all options, we get the final result.

Complete step-by-step solution:
In this question, we are supposed to find the statements which are correct for the function given as $f\left( x \right)=-{{x}^{2}}$.
So, before proceeding for this, we must know that for these kinds of questions, we must proceed by options while checking which one is true or false with given conditions.
So, starting with the first option which needs the derivate of the given function, then the derivate of the given function is given by:
$\begin{align}
  & {f}'\left( x \right)=\dfrac{d}{dx}\left( -{{x}^{2}} \right) \\
 & \Rightarrow {f}'\left( x \right)=\left( -2x \right) \\
\end{align}$
Now, for the first option, by substituting the value of x as 1, we get:
$\begin{align}
  & {f}'\left( 1 \right)=-2\left( 1 \right) \\
 & \Rightarrow {f}'\left( 1 \right)=-2 \\
\end{align}$
So, we get the value of ${f}'\left( 1 \right)$as -2 which is less than zero.
Hence, the (a) option is correct.
Then proceeding with the second option, we need to find the value of f(2).
So, by substituting the value of x as 2 in the given function in the question, we have:
$\begin{align}
  & f\left( 2 \right)=-{{2}^{2}} \\
 & \Rightarrow f\left( 2 \right)=-4 \\
\end{align}$
So, we get the value of function f(2) as -4 which is less than zero.
Hence, the (b) option is also correct.
Now, for the options (c) and (d), we need to find the value of derivate of the function for the range of x from 1 to 3.
As we have already found the value of derivative in option (a) at x=1 which is -2 and that is not equal to zero.
Similarly, for x=2, we get the value of derivate of the function as:
$\begin{align}
  & {f}'\left( 2 \right)=-2\left( 2 \right) \\
 & \Rightarrow {f}'\left( 2 \right)=-4 \\
\end{align}$
So, we get the value of ${f}'\left( 2 \right)$ as -4 which is not equal to zero.
Also, for x=3, we get the value of derivate of the function as:
$\begin{align}
  & {f}'\left( 3 \right)=-2\left( 3 \right) \\
 & \Rightarrow {f}'\left( 3 \right)=-6 \\
\end{align}$
So, we get the value of ${f}'\left( 3 \right)$ as -6 which is not equal to zero.
So, we can say that ${f}'\left( x \right)\ne 0$ for any value at which $x\in \left( 1,3 \right)$.
Also, we can see clearly that for option (d) it says ${f}'\left( x \right)\ne 0$ for some $x\in \left( 1,3 \right)$, but we have seen that ${f}'\left( x \right)\ne 0$ can have any value for the range (1, 3) and not for some values only.
Hence, option (c) is also correct and consequently, option (d) is not correct.
So, finally, we get the answer as options (a), (b), and (c) is correct.

Note: Now, to solve these types of the questions we need to know some of the basics of the differentiation beforehand so that we can solve the question easily. So, the formula required for the differentiation in this question is as:
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ where n is any number.