Let for (x) = 7 tan^8 x + 7 tan^6 - 3 tan^4 x

Let for (x) = 7 tan^8 x + 7 tan^6 – 3 tan^4 x – 3 tan^2

Question 15 : Let for $f(x) = 7\tan^8 x + 7\tan^6 - 3\tan^4 x - 3\tan^2, I_1 = \displaystyle\int_0^{\pi/4} f(x) dx$ , and $I_2 = \displaystyle\int_0^{\pi/4} x f(x) dx$ . Then $7I_1 + 12 I_2$ is equal to

(i) 2π (ii) π

(iii) 1 (iv) 2

Answer : (iii) 1

Explanation : Let for $f(x) = 7\tan^8 x + 7\tan^6 - 3\tan^4 x - 3\tan^2$

$= 7\tan^6 x(\tan^2 + 1) - 3\tan^ x(\tan^2 x + 1)$

$= (7\tan^6 x - 3\tan^2 x)(\tan^2 x + 1)$

$= (7\tan^6 x - 3\tan^2 x). \sec^2 x$

$I_1 = \displaystyle\int_0^{\pi/4} (7\tan^6 x - 3\tan^2 x). \sec^2 x dx$

Let $\tan x = t$

x = 0, t = 0

x = π/4, t = 1

⇒ $\sec^2x = dt/dx$

⇒ $\sec^2x dx = dt$

Now, $I_1 = \displaystyle\int_0^{1} (7t^6 - 3t^2 ).dt$

⇒ $I_1 = (t^7 - t^3 )_0^1$

⇒ $I_1 = (1 - 1) - (0 - 0) = 0$

⇒ $I_1 = 0$

$I_2 = \displaystyle\int_0^{\pi/4} x(7\tan^6 x - 3\tan^2 x). \sec^2 x dx$

Integration by parts

$I_2 = [x(\tan^7 x - \tan^3 x)]_0^{\pi/4} - \displaystyle \int_0^{\pi/4} (\tan^7 x - \tan^3 x) dx$

⇒ $I_2 = [1(1 - 1) - 0] - \displaystyle \int_0^{\pi/4} \tan^3 x(\tan^2 x - 1)(\tan^2 x + 1) dx$

⇒ $I_2 = - \displaystyle \int_0^{\pi/4} \tan^3 x(\tan^2 x - 1)\sec^2 x dx$

Let $\tan x = t$

x = 0, t = 0

x = π/4, t = 1

⇒ $\sec^2x = dt/dx$

⇒ $\sec^2x dx = dt$

⇒ $I_2 = - \displaystyle \int_0^{1} t^3 (t^2 - 1) dt$

⇒ $I_2 = - \displaystyle \int_0^{1} (t^5 - t^3) dt$

⇒ $I_2 = - [\dfrac{t^6}{6} - \dfrac{t^4}{4}]_0^1$

⇒ $I_2 = - [\dfrac{1}{6} - \dfrac{1}{4}] = \dfrac{1}{12}$

Now,

$7I_1 + 12 I_2 = 0 + 12\times \dfrac{1}{12}$

= 1

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

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 :- See full answer

Question 6:- A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ² denote the mean and variance of X, then the value of 64(μ + σ²) is :

(i) 51 (ii) 48

(iii) 32 (iv) 64 [JEE 22 JAN 2025]

Answer:- See full solution

Question 7:- Let $a_1, a_2, a_3 . . .$ be a G.P. of increasing positive terms. If $a_1.a_5 = 28$ and $a_2 + a_4 = 29$ , the $a_6$ is equal to

(i) 628 (ii) 526

(iii) 784 (iv) 812 [JEE 22 JAN 2025]

Answer:- See full Answer

Question 8:- Let $L_1: \dfrac{x - 1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $L_2 : \dfrac{x - 2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

(i) (-5/3, -7, 1) (ii) (2, 3, 1/3)

(iii) (8/3, -1, 1/3) (iv) (14/3, -3, 22/3) [JEE 22 JAN 2025]

Answer:- See full solution

Question 9:- The product of all solutions of the equation $e^{5(\log_e x)^2 + 3} = x^8, x > 0$ is :

(i) $e^{8/5}$ (ii) $e^{6/5}$

(iii) $e^{2}$ (iv) $e$ [JEE 22 JAN 2025]

Answer :- See full solution

Question 10:- If $\displaystyle \Sigma_{r= 1}^n T_r = \dfrac{(2n - 1)(2n + 1)(2n + 3)(2n + 5)}{64}$ , then $\displaystyle \lim_{n \to \infty} {\Sigma _{r = 1}^n}\dfrac{1}{T_r}$ is equal to

(i) 1 (ii) 0

(iii) 2/3 (iv) 1/3 [JEE 22 JAN 2025]

Answer :- See full Answer

Question 11: From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ‘M’, is :

(1) 14950 (2) 6084

(3) 4356 (4) 5148 [JEE 22 JAN 2025]

Answer : See full Answer

Question 12 : Let x = x(y) be the solution of the differential equation $y^2 dx + (x - \dfrac{1}{y}) dy = 0$ . x(1) = 1 then x(1/2) is :

(i) 1/2 + e (ii) 3/2 + e

(iii) 3 – e (iv) 3 + e

Answer: See full Answer

Question 13: Let the parabola y = x³ + px – 3 meet the coordinate axes at the points P, Q and R. If the circle C with centre at (-1, -1) passes through the points P, Q and R, then the area of ΔPQR is :

(i) 4 (ii) 6

(iii) 7 (iv) 5

Answer :See full Answer

Question 14: A circle C of radius 2 lies in the second quadrant and touches bothe the coordinate axes. Let r be the radius of a circle that has centre at the point (2, 5) and intersects the circle C at exactly two points. If the set of all possible values of r is the interval (α, β), then 3β – 2α is equal to :

(i) 15 (ii) 14

(iii) 12 (iv) 10