Q33 / 54 · JEE Advanced 2026 (Paper 2) · Mathematics
Passage: Consider the curve $C_1$ given by $y = e^{-x}$ for $x \in [0, 10\pi]$, and the curve $C_2$ given by $y = e^{-x}(\sin x + \cos x)$ for $x \in [0, 10\pi]$. Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$. Suppose that $\alpha_1, \alpha_2, \ldots, \alpha_n \in [0, 10\pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that $\alpha_1 < \alpha_2 < \cdots < \alpha_n$.
Question: The value of $n$ is ___________.
Question: The value of $n$ is ___________.
Correct answer: 11
Solution
Equating the equations of the curves $C_1$ and $C_2$:
$e^{-x} = e^{-x}(\sin x + \cos x)$
Since $e^{-x} \neq 0$ for all real $x$, dividing both sides by $e^{-x}$ gives:
$\sin x + \cos x = 1$
Multiplying both sides by $\dfrac{1}{\sqrt{2}}$:
$\dfrac{1}{\sqrt{2}}\sin x + \dfrac{1}{\sqrt{2}}\cos x = \dfrac{1}{\sqrt{2}}$
$\sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
The general solution is given by:
$x + \dfrac{\pi}{4} = 2m\pi + \dfrac{\pi}{4}$ or $x + \dfrac{\pi}{4} = 2m\pi + \dfrac{3\pi}{4}$ for integer $m$
This simplifies to:
$x = 2m\pi$ or $x = 2m\pi + \dfrac{\pi}{2}$
Given the interval $x \in [0, 10\pi]$, the valid values for $x$ are:
For $x = 2m\pi$: $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$ (6 solutions)
For $x = 2m\pi + \dfrac{\pi}{2}$: $x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2}, \dfrac{13\pi}{2}, \dfrac{17\pi}{2}$ (5 solutions)
The total number of points of intersection is $n = 6 + 5 = 11$.
Answer: $11$
$e^{-x} = e^{-x}(\sin x + \cos x)$
Since $e^{-x} \neq 0$ for all real $x$, dividing both sides by $e^{-x}$ gives:
$\sin x + \cos x = 1$
Multiplying both sides by $\dfrac{1}{\sqrt{2}}$:
$\dfrac{1}{\sqrt{2}}\sin x + \dfrac{1}{\sqrt{2}}\cos x = \dfrac{1}{\sqrt{2}}$
$\sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$
The general solution is given by:
$x + \dfrac{\pi}{4} = 2m\pi + \dfrac{\pi}{4}$ or $x + \dfrac{\pi}{4} = 2m\pi + \dfrac{3\pi}{4}$ for integer $m$
This simplifies to:
$x = 2m\pi$ or $x = 2m\pi + \dfrac{\pi}{2}$
Given the interval $x \in [0, 10\pi]$, the valid values for $x$ are:
For $x = 2m\pi$: $x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi$ (6 solutions)
For $x = 2m\pi + \dfrac{\pi}{2}$: $x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2}, \dfrac{13\pi}{2}, \dfrac{17\pi}{2}$ (5 solutions)
The total number of points of intersection is $n = 6 + 5 = 11$.
Answer: $11$
Hence, the correct answer is 11.