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Trigonometric Functions & Equations

Section: Chemistry

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Question : 2 of 32

Marks: +1, -0

Consider the curve

C_1

y=e^{-x} \quad \text { for } x \in[0,10 \pi],

C_2

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

n

be the total number of points of intersection of the curves

C_1

C_2

\alpha_1, \alpha_2, \ldots, \alpha_n \in[0,10 \pi]

x

-coordinates of the points of intersection of the curves

C_1

C_2

\alpha_1 < \alpha_2 < \cdots < \alpha_n .

n

\_\_\_\_

[JEE Adv 2026 P2]

Your Answer:

Solution:

Concept:

Find intersection points by equating the

y

-values of the two curves.

Explanation:

Set

e^{-x} = e^{-x}(\sin x + \cos x)

.

Divide both sides by

e^{-x}

(nonzero) to get

\sin x + \cos x = 1

.

Rewrite

\sin x + \cos x = \sqrt{2} \cos(x - \frac{\pi}{4})

.

Thus

\sqrt{2} \cos(x - \frac{\pi}{4}) = 1 \Rightarrow \cos(x - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}

.

General solution:

x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}

,

n \in \mathbb{Z}

.

So

x = 2n\pi

(from minus sign) or

x = 2n\pi + \frac{\pi}{2}

(from plus sign).

In

[0,10\pi]

:

For

x = 2n\pi

:

n = 0,1,2,3,4,5

gives

0,2\pi,4\pi,6\pi,8\pi,10\pi

.

For

x = 2n\pi + \frac{\pi}{2}

:

n = 0,1,2,3,4

gives

\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\frac{13\pi}{2},\frac{17\pi}{2}

.

Total distinct

x

-coordinates:

0,\frac{\pi}{2},2\pi,\frac{5\pi}{2},4\pi,\frac{9\pi}{2},6\pi,\frac{13\pi}{2},8\pi,\frac{17\pi}{2},10\pi

.

This gives

n = 11

points of intersection.

Answer:

11

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