GATE DA 2025 | Question: 4

Question

​​​​​Let $f(x)=\frac{e^{x}-e^{-x}}{2}, x \in \mathbb{R}$. Let $f^{(k)}(a)$ denote the $k^{t h}$ derivative of $f$ evaluated at $a$. What is the value of $f^{(10)}(0)$? (Note: ! denotes factorial)

• A. $0$
• B. $1$
• C. $\frac{1}{10!}$
• D. $\frac{2}{10!}$

Comments

you can observe that f(0)=f"(0) ( after putting x=0 in f(x) ,for every even derivative, we get value 0)
and f'(0)=f"'(0) (after putting x=0 in f'(x), for every odd derivative we get value 1)
so, as f^10(0) is even derivative...Ans is option A] 0

Answer 1

Suppose $f(x) =\frac{1}{2}(e^x - e^{-x})$, then $f'(x) = \frac{1}{2}(e^x + e^{-x})$ , $f''(x) = \frac{1}{2}(e^x - e^{-x})$.

So $f^k(x) =\begin{cases}\frac{1}{2}(e^x + e^{-x}), & \text{if k is odd}\\\frac{1}{2}(e^x - e^{-x}), & \text{if k is even}\end{cases}$.

so $f^{10} (0) =\frac{1}{2}(e^0 - e^{0}) = \frac{1}{2}(1-1) = 0 $.

The answer is option A