If the area of the region is , then the value of a is : [2025]

Question

If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

Smart Achievers · Accepted Answer

(4)

Required area $= a + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$

$= a + {[a x + e^{x} + e^{- x}]}_{0}^{1}$

$\Rightarrow 2 a + e + e^{- 1} - 2 = e + 8 + \frac{1}{e}$

$\Rightarrow 2 a - 2 = 8$

$\Rightarrow 2 a = 10 \Rightarrow a = 5$ .

Solution

Q.
If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

1 6

2 8

3 7

4 5

Ans.
(4)

Required area $= a + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$

$= a + {[a x + e^{x} + e^{- x}]}_{0}^{1}$

$\Rightarrow 2 a + e + e^{- 1} - 2 = e + 8 + \frac{1}{e}$

$\Rightarrow 2 a - 2 = 8$

$\Rightarrow 2 a = 10 \Rightarrow a = 5$ .

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Solution

Q. If the area of the region {(x,y):–1≤x≤1, 0≤y≤a+e|x|–e–x, a>0} is e2+8e+1e, then the value of a is : [2025]

1 6

2 8

3 7

4 5

Ans. (4) Required area =a+∫01(a+ex–e–x)dx =a+[ax+ex+e–x]01 ⇒ 2a+e+e–1–2=e+8+1e ⇒ 2a–2=8 ⇒ 2a=10 ⇒ a=5.

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Q.
If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{| x |} - e^{- x}, a > 0}$ is $\frac{e^{2} + 8 e + 1}{e}$ , then the value of a is : [2025]

Ans.
(4)

Required area $= a + \int_{0}^{1} (a + e^{x} - e^{- x}) d x$

$= a + {[a x + e^{x} + e^{- x}]}_{0}^{1}$

$\Rightarrow 2 a + e + e^{- 1} - 2 = e + 8 + \frac{1}{e}$

$\Rightarrow 2 a - 2 = 8$

$\Rightarrow 2 a = 10 \Rightarrow a = 5$ .