GATE 2025 DA Q.24: solution and answer

Official answer: A, B, CSource: Official answer key, GATE 2025 DA (IIT Roorkee)

1. Sum, difference and product rules: if $f$ and $g$ are differentiable at $c$, then so are $f\pm g$ and $f\cdot g$. So (A) and (B) are always differentiable at $c$.

2. Quotient rule: $\frac{f}{g}$ is differentiable at $c$ when $f$ and $g$ are differentiable at $c$ and $g(c)\ne 0$. Since $g$ takes values in $(1,\infty )$, $g(c)>1$, so $g(c)\ne 0$ and $g$ never vanishes anywhere. Hence (C) is always differentiable at $c$.

3. Chain rule: $f\circ g$ is differentiable at $c$ when $g$ is differentiable at $c$ AND $f$ is differentiable at the point $g(c)$. We are told only that $f$ is differentiable at $c$, not at $g(c)$. Likewise $g\circ f$ needs $g$ differentiable at $f(c)$, which is not given. So (D) can fail, and we need one counterexample.

4. Counterexample for (D): take $c=0$, $f(x)=|x-2|$ and $g(x)=2+\frac{x}{1+x^2}$. Here $f$ is differentiable at $0$ because $f(x)=2-x$ for $x<2$. Also $g$ is differentiable everywhere (the denominator never vanishes), and since $\left|\frac{x}{1+x^2}\right|\le \frac{1}{2}$, we get $g(x)\in [1.5,2.5]\subset (1,\infty )$. So both functions meet the hypotheses at $c=0$.

5. Compute the two compositions. First, $g(0)=2$, so $(f\circ g)(x)=\left|\frac{x}{1+x^2}\right|=\frac{|x|}{1+x^2}$. Its right-hand slope at $0$ is $+1$ and its left-hand slope is $-1$, so it is NOT differentiable at $0$.

6. Second, near $x=0$ we have $f(x)=2-x$, so $(g\circ f)(x)=g(2-x)$, which is differentiable at $0$ because $g$ is differentiable everywhere.

7. A function that is not differentiable at $c$, plus one that is differentiable at $c$, is not differentiable at $c$ (otherwise the difference of the sum and the second function would be differentiable). Hence $f\circ g+g\circ f$ is not differentiable at $0$, and (D) is not always differentiable.

8. Therefore the functions that are always differentiable at $c$ are (A), (B) and (C).