Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The number of points in the interval $[2, 4]$, at which the function $f(x) = \left\lfloor x^2 - x - \frac{1}{2} \right\rfloor$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
Show answer and solution
Answer: 10
Key idea: Discontinuity of greatest integer function composition
Step 1: Recognize that $f(x) = \lfloor x^2 - x - \frac{1}{2} \rfloor$ is discontinuous when $g(x) = x^2 - x - \frac{1}{2}$ takes integer values.
Step 2: Find the range of $g(x)$ on $[2, 4]$: $g(2) = 1.5$ and $g(4) = 11.5$. Since $g'(x) = 2x - 1 > 0$ on this interval, $g$ is strictly increasing.
Step 3: The range of $g(x)$ is $[1.5, 11.5]$, so the integer values are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ (10 integers).
Step 4: For each integer $n$, solve $x^2 - x - \frac{1}{2} = n$ to get $x = \frac{1 + \sqrt{4n + 3}}{2}$.
Step 5: Verify that all 10 solutions lie in $[2, 4]$. Since $g$ is continuous and strictly increasing, each integer value is attained exactly once.
Step 6: Therefore, there are **10 points of discontinuity**.
If $y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$, then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:
(1) 3
(2) -1
(3) 1
(4) 2
Show answer and solution
Answer: Option (3)
Key idea: Differentiation of inverse trigonometric functions and simplification using trigonometric identities
Simplify first term: $\tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) = \tan^{-1}\left(\frac{3}{4}\right) - x$ using $\tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}(a) - \tan^{-1}(b)$
Simplify second term: $2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) = \sin^{-1}(x)$ using substitution $x = \sin\theta$ and half-angle formula
Combine: $y = \tan^{-1}\left(\frac{3}{4}\right) - x + \sin^{-1}(x)$
Let $f(x) = \begin{cases} e^{x-1} & , x < 0 \\ x^2 - 5x + 6 & , x \ge 0 \end{cases}$ and $g(x) = f(|x|) + |f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Continuity and differentiability of piecewise functions involving absolute values
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
(1) $3\left(\frac{e+1}{e}\right)$
(2) $3\left(\frac{e-1}{e}\right)$
(3) $\frac{3-e}{e}$
(4) $3e$
Show answer and solution
Answer: Option (2)
Key idea: Functional equations and optimization using derivatives
Substitute $x=y=0$ in the functional equation to get $f(0)=0$
Verify that $f(x)=cx$ satisfies the functional equation, and use $f'(0)=3$ to get $f(x)=3x$
Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_{1}^{x} f(t)dt + (1-x)(\log_{e}x - 1) + e$. Then the value of $f(f(1))$ is :
(1) $(1 + e^{e})$
(2) $(1 + e)$
(3) $(1 + e + e^{e})$
(4) $1 + 2e$
Show answer and solution
Answer: Option (1)
Key idea: Solving functional equations involving integrals using differentiation and first-order linear differential equations
Substitute $x=1$ in the given equation: $f(1) = 0 + 0 + e = e$
Differentiate both sides with respect to $x$ to get: $f'(x) = f(x) - \log_e x + \frac{1}{x}$
Rearrange to form a linear ODE: $f'(x) - f(x) = -\log_e x + \frac{1}{x}$
Solve using integrating factor $e^{-x}$ and integration by parts to obtain: $f(x) = e^x + \log_e x$
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbb{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
(1) $20$
(2) $40$
(3) $-20$
(4) $-40$
Show answer and solution
Answer: Option (2)
Key idea: Integration of differential equations with boundary conditions
Given $f''(x) = g''(x)$, integrate to get $f'(x) - g'(x) = C_1$
Use $f'(1) = 4$ and $g'(1) = 2$ to find $C_1 = f'(1) - g'(1) = 2$
Integrate again: $f(x) - g(x) = 2x + C_2$
Use $f(2) = 3$ and $g(2) = 9$ to find $C_2$: $3 - 9 = 4 + C_2$, so $C_2 = -10$
Let $f(x) = \begin{cases} x^3 + 8; & x < 0,\\ x^2 - 4; & x \ge 0, \end{cases}$ and $g(x) = \begin{cases} (x-8)^{\frac{1}{3}}; & x < 0,\\ (x+4)^{\frac{1}{2}}; & x \ge 0. \end{cases}$ Then the number of points, where the function $gof$ is discontinuous, is _________.
Show answer and solution
Answer: 3
Key idea: Composition of piecewise functions and continuity
For the function $f(x) = e^{\sin|x|} - |x|, x \in \mathbf{R}$, consider the following statements:
**Statement I:** $f$ is differentiable for all $x \in \mathbf{R}$.
**Statement II:** $f$ is increasing in $\left(-\pi, -\frac{\pi}{2}\right)$.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Differentiability of functions involving absolute values and checking monotonicity using derivatives
Let $f(x)=\begin{cases}\frac{1}{3} & , x\leq\pi/2\\ \frac{b(1-\sin x)}{(\pi-2x)^2} & , x>\pi/2\end{cases}$ If $f$ is continuous at $x=\pi/2$, then the value of $\int_{0}^{3b-6}|x^2+2x-3|dx$ is:
(1) $5$
(2) $2$
(3) $3$
(4) $4$
Show answer and solution
Answer: Option (4)
Key idea: Continuity and definite integration with absolute value
Use continuity at $x=\frac{\pi}{2}$: $\lim_{x\to\frac{\pi}{2}^+} \frac{b(1-\sin x)}{(\pi-2x)^2} = \frac{1}{3}$
Substitute $t = x - \frac{\pi}{2}$ and use $1-\cos t \approx \frac{t^2}{2}$ to get $\frac{b}{8} = \frac{1}{3}$, so $b = \frac{8}{3}$
Calculate $3b - 6 = 3 \cdot \frac{8}{3} - 6 = 2$
Factor $x^2 + 2x - 3 = (x+3)(x-1)$, which is negative on $[0,1)$ and positive on $(1,2]$
