ISI2025-MCS-PCA | Question: 15

Question

Suppose $f$ is a real-valued function defined on the set of real numbers. Let
$$f(x)=\frac{\sin x}{x} \text { for } x \neq 0, \quad f(0)=a$$
where $a$ is a real number. Let $f$ be differentiable at $x=0$. Let $f^{\prime}(0)=b$. Then $a+b$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

a=limit x->0 sin(x)/x      ---> a=1
b=limit x->0 (x*cos(x)-sinx)/x^2      --->b=0

a+b=1

Answer 2

Answer: C