JEE Advanced 2026 — Paper 2 — MATH-IV solution

JEE Advanced2026Paper 2MATH-IV

Question Stem (Q1-Q2).

Consider the curve $C_{1}$ given by

$y = e^{- x}$ for $x \in [0, 10 π]$ ,

and the curve $C_{2}$ given by

$y = e^{- x} (sin x + cos x)$ for $x \in [0, 10 π]$ .

Let $n$ be the total number of points of intersection of the curves $C_{1}$ and $C_{2}$ . Suppose that $α_{1}, α_{2}, \dots, α_{n} \in [0, 10 π]$ are the $x$ -coordinates of the points of intersection of the curves $C_{1}$ and $C_{2}$ such that $α_{1} < α_{2} < \dots < α_{n}$ .

Q1. The value of $n$ is ___________.

Your answer

Solution

Setting $e^{- x} = e^{- x} (sin x + cos x)$ and dividing by $e^{- x}$ (always nonzero) yields $sin x + cos x = 1$ .

Rewrite as $2 cos (x - \frac{π}{4}) = 1$ , i.e. $cos (x - \frac{π}{4}) = \frac{1}{2}$ .

So $x - \frac{π}{4} = 2 nπ \pm \frac{π}{4}$ , giving $x = 2 nπ$ or $x = 2 nπ + \frac{π}{2}$ .

The solutions in $[0, 10 π]$ are: $0, \frac{π}{2}, 2 π, \frac{5 π}{2}, 4 π, \frac{9 π}{2}, 6 π, \frac{13 π}{2}, 8 π, \frac{17 π}{2}, 10 π$ .

Counting, $n = 11$ .

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