Consider the curve $\mathrm{C}_{1}$ given by $$\mathrm{y}=\mathrm{e}^{-\mathrm{x}} \text { for } \mathrm{x} \in[0,10 \pi],$$ and the curve $\mathrm{C}_{2}$ given by $$\mathrm{y}=\mathrm{e}^{-\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x}) \text { for } \mathrm{x} \in[0,10 \pi] .$$ Let $n$ be the total number of points of intersection of the curves $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$. 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 $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ such that $$\alpha_{1}<\alpha_{2}<\ldots<\alpha_{n} .$$ Let $\beta$ be the area of the region enclosed between the curves $\mathrm{C}_{1}, \mathrm{C}_{2}$, and the lines $\mathrm{x}=\alpha_{1}$ and $\mathrm{x}=\alpha_{4}$. Then the value of $$-\frac{1}{\pi} \log _{e}\left(\beta-2 e^{-\frac{\pi}{2}}\right) \text { is }$$ ____.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Area Under Curves

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} .

Let

β

be the area of the region enclosed between the curves

C_{1}, C_{2}

, and the lines

x = α_{1}

and

x = α_{4}

. Then the value of

- \frac{1}{π} lo g_{e} (β - 2 e^{- \frac{π}{2}}) is

____.

Answer: 2.5

From Q15,

α_{1} = 0

and

α_{4} = \frac{5 π}{2}

. The area

β = \int_{0}^{5 π /2} ∣ e^{- x} (sin x + cos x) - e^{- x} ∣ d x

. Breaking into intervals where the sign changes and integrating (using the fact that the integral of

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

can be computed), we find

β = 2 e^{- π /2} + e^{- 5 π /2}

. Thus

β - 2 e^{- π /2} = e^{- 5 π /2}

, and

- \frac{1}{π} lo g (e^{- 5 π /2}) = - \frac{1}{π} \cdot (- \frac{5 π}{2}) = \frac{5}{2} = 2.50

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer