GATE CS Calculus Previous Year Questions

GATE CS 2025 Set 1 | Question 31

NAT · 1 mark

Consider the given function \(f(x)\).

\[ f(x) = \begin{cases} ax + b & \text{for } x \lt 1 \\ x^{3} + x^{2} + 1 & \text{for } x \ge 1 \end{cases} \]

If the function is differentiable everywhere, the value of \(b\) must be ________. (rounded off to one decimal place)

Your answer

Explanation

Answer: -2.1 to -1.9

A function that is differentiable at \(x = 1\) must be continuous there, and its left and right derivatives must be equal.

Continuity at \(x = 1\): \(a + b = 1 + 1 + 1 = 3\). Equal slopes: the left piece has slope \(a\), and the right piece has slope \(3x^{2} + 2x = 5\) at \(x = 1\), so \(a = 5\).

Hence \(b = 3 - 5 = -2\), that is, \(-2.0\).

Answer: −2.0

GATE CS 2025 Set 2 | Question 12

MCQ · 1 mark · −0.33

The value of \(x\) such that \(x \gt 1\), satisfying the equation \(\int_{1}^{x} t \ln t \, dt = \frac{1}{4}\) is

\(\sqrt{e}\) \(e\) \(e^{2}\) \(e - 1\)

Explanation

Answer: (A)

Integrate by parts with \(u = \ln t\) and \(dv = t \, dt\):

\[\int t \ln t \, dt = \frac{t^{2}}{2}\ln t - \frac{t^{2}}{4}.\]

So \(\int_{1}^{x} t \ln t \, dt = \frac{x^{2}}{2}\ln x - \frac{x^{2}}{4} + \frac{1}{4}\). Setting this equal to \(\frac{1}{4}\) gives \(\frac{x^{2}}{2}\ln x = \frac{x^{2}}{4}\). Since \(x \neq 0\), \(\ln x = \frac{1}{2}\), so \(x = \sqrt{e}\).

The other options fail this last check: \(e\), \(e^{2}\) and \(e - 1\) give \(\ln x = 1\), \(2\) and about \(0.54\), not \(\frac{1}{2}\).

Answer: A

GATE CS 2024 Set 1 | Question 11

MCQ · 1 mark · −0.33

Let \(f: \mathbb{R} \to \mathbb{R}\) be a function such that \(f(x) = \max\{x, x^{3}\}\), \(x \in \mathbb{R}\), where \(\mathbb{R}\) is the set of all real numbers. The set of all points where \(f(x)\) is NOT differentiable is

\(\{-1, 1, 2\}\) \(\{-2, -1, 1\}\) \(\{0, 1\}\) \(\{-1, 0, 1\}\)

Explanation

Answer: (D)

The curves \(y = x\) and \(y = x^{3}\) cross where \(x^{3} = x\), that is at \(x = -1, 0, 1\). Between these points the larger one switches:

for \(x \lt -1\) and \(0 \lt x \lt 1\): \(f(x) = x\), slope 1;
for \(-1 \lt x \lt 0\) and \(x \gt 1\): \(f(x) = x^{3}\), slope \(3x^{2}\).

At each crossing the slopes on the two sides differ: at \(x = \pm 1\) they are 1 and 3, and at \(x = 0\) they are 0 and 1. So the graph has a corner at each of these points, and \(f\) is not differentiable exactly on \(\{-1, 0, 1\}\). Elsewhere \(f\) equals a single polynomial near each point, so it is differentiable.

Answer: D

GATE CS 2024 Set 2 | Question 16

MCQ · 1 mark · −0.33

Let \(f(x)\) be a continuous function from \(\mathbb{R}\) to \(\mathbb{R}\) such that

\[f(x) = 1 - f(2 - x)\]

Which one of the following options is the CORRECT value of \(\int_{0}^{2} f(x)\,dx\) ?

0 1 2 \(-1\)

Explanation

Answer: (B)

Let \(I = \int_{0}^{2} f(x)\,dx\). The substitution \(x \to 2 - x\) maps \([0, 2]\) onto itself, so also \(I = \int_{0}^{2} f(2 - x)\,dx\). Adding the two forms and using \(f(x) + f(2 - x) = 1\):

\[2I = \int_{0}^{2} \big[f(x) + f(2 - x)\big]\,dx = \int_{0}^{2} 1\,dx = 2,\]

so \(I = 1\). Geometrically, the graph is symmetric about the point \((1, \frac{1}{2})\), so the parts above and below the line \(y = \frac{1}{2}\) cancel and the integral is \(2 \times \frac{1}{2} = 1\). For example, \(f(x) = \frac{x}{2}\) satisfies the condition, and its integral from 0 to 2 is 1.

Answer: B