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 critical points of the function $f(x) = \begin{cases} \sqrt{\frac{|\sin x|}{x}}, & x \neq 0 \\ 1, & x = 0 \end{cases}$ in the interval $(-2\pi, 2\pi)$ is equal to:
(1) 1
(2) 3
(3) 5
(4)
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Answer: Option (2)
Key idea: Critical points of piecewise functions involving absolute values
1. Check continuity: $\lim_{x \to 0} \sqrt{\frac{|\sin x|}{x}} = 1 = f(0)$, so $f$ is continuous everywhere
2. For $x \neq 0$, write $f(x) = \sqrt{\frac{|\sin x|}{x}}$ and note that $|\sin x| = 0$ at $x = \pm\pi$ in the given interval
3. At $x = \pm\pi$: Since $\sin(\pm\pi) = 0$, we have $f(\pm\pi) = 0$ and the derivative doesn't exist (cusps)
4. Check $x = 0$: Using L'Hôpital or Taylor series, $f'(0^+) = f'(0^-) = 0$, so $x = 0$ is a critical point
5. For other points, differentiate in regions where $\sin x$ has constant sign. Critical points occur when $x\cos x = \sin x$ (i.e., $\tan x = x$) or $x\cos x = -\sin x$
6. The equation $\tan x = x$ has no solutions in $(-2\pi, 2\pi) \setminus \{0\}$ within the valid regions
Let $f(x)$ be a polynomial of degree 5, and have extrema at $x = 1$ and $x = -1$. If $\lim_{x\to 0}\left(\frac{f(x)}{x^3}\right) = -5$, then $f(2) - f(-2)$ is equal to :
(1) $0$
(2) $50$
(3) $92$
(4) $112$
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Answer: Option (4)
Key idea: Polynomial functions with extrema conditions and limit evaluation
Step 1: Since $f(x)$ has degree 5 and $\lim_{x\to 0}\frac{f(x)}{x^3} = -5$ exists, we know $f(0) = 0$ and $f(x) = x^3(Dx^2 + Ex + F)$ where $F = -5$. Thus $f(x) = Dx^5 + Ex^4 - 5x^3$.
Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t \, dt$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + 12 f'\left(-\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to _________
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Answer: 5
Key idea: Differential Equations with Leibniz Rule
Differentiate the given integral equation using Leibniz rule: $f'(x) = -\tan x - f(x)\tan x = -\tan x(1+f(x))$
Find initial condition: $f(0) = 0$ by substituting $x=0$ in the original equation
Differentiate $f'(x)$ to get: $f''(x) = -\sec^2 x - f'(x)\tan x - f(x)\sec^2 x$
Simplify to obtain the differential equation: $f''(x) + f(x) = -1$
Solve with initial conditions $f(0)=0, f'(0)=0$ to get: $f(x) = \cos x - 1$
