Using the principal values of the inverse trigonometric functions

Question 5:- Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of $16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$ is :

(1) 24 π² (2) 18 π²

(3) 31 π² (4) 22 π² [JEE 22 JAN 2025]

Answer :- (4) 22 π²

Explanation:- The principal values of the inverse trigonometric functions

$(\sec^{-1} x)^2 = z \in [0, \pi] - \pi/2$

Since, $\sec^{-1} x + \operatorname{cosec}^{-1} x = \pi/2$

$(\operatorname{cosec}^{-1} x)^2 = \pi/2 - z$

$16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$

= 16[z² + (π/2 – z)²]

= 16[z² + π²/4 + z² – π z]

= 16[2 z² – π z + π²/4]

Let $16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2) = A$

Max Val of $\sec^{-1} x = \pi = z$

Max value of A = 16[(π)² + (π/2 – π)²]

= 16[π² + π²/4] = 16(5π²/4)

= 20 π²

Again, A = 16[2 z² – 2π z + π²/4]

Now d.w.r.t. z

$\dfrac{dA}{dz} = 16[4z - \pi]$

For max and min $\dfrac{dA}{dz} = 0$

So, z = π/4

Again d.w.r.t. z

$\dfrac{d^2A}{dz^2} = 64 >0$

Hence, A is min when z = π/4

min value of A = 16[(π/4)² + (π/2 – π/4)²]

= 16[π²/16 + π²/16]

= 2π²

The sum of maximum and minimum value of $16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$ is

= 20 π² + 2 π²

= 22 π²

Question 2:- Let f : R → R be a twice differentiable function such that f(x + y) = f(x) f(y) for all x, y ∈ R. If f'(0) = 4a and f satisfies f”(x) – 3a f'(x) – f(x) = 0, a > 0, then the area of the region

R = {(x, y) | 0 ≤ y ≤ f(ax), 0 ≤ x ≤ 2} is :

(i) $e^2 - 1$ (ii) $e^4 + 1$

(iii) $e^4 - 1$ (iv) $e^2 + 1$ [JEE 22 JAN 2025]

Solution:- See full solution

Question 3:- Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of ΔPQR is the point (α, β), then 15(α – β) is equal to :

(1) 24 (2) 19

(3) 21 (4) 22 [JEE 22 JAN 2025]

Answer:- See full answer

Question 4:- Let $Z_1, Z_2$ and $Z_3$ be three complex numbers on the circle |Z| = 1 with $\arg(Z_1) = \dfrac{-\pi}{4},\arg(Z_2) = 0$ and $\arg(Z_3) = \dfrac{\pi}{4}$ . If $|Z_1.\bar{Z_2} + Z_2.\bar{Z_3} + Z_3.\bar{Z_1}|^2 = \alpha + \beta\sqrt{2}, \alpha,\beta \in Z$ , then the value of α² + β² is :

(1) 24 (2) 41

(3) 31 (4) 29 [JEE 22 JAN 2025]

Answer : See full Answer

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