JEE Advanced (Matrix Match): Inverse Trigonometric Functions | Chapter-wise Tests for JEE Main & Advanced PDF Download

DIRECTIONS (Q. 1 & 2) : Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q. 1. Match the following

Column I                                                                                         Column II

                                                (p) 1

(B) Sides a, b, c of a triangle ABC are in AP and                           (q)

(C) A line is perpendicular to x + 2y + 2z = 0 and                        (r) 2/3
 passes through (0, 1, 0). The perpendicular
 distance of this line from the origin is

Ans. (A)-(p), (B)-(r), (C)-(q)

Solution. 

t = tan^–1 3 – tan^–1 1 + tan^–15 – tan^–13 + ...... + tan^–1(2n + 1) – tan^–1 (2n – 1) + ...... ∞

(B) ∴ a, b, c are in AP  ⇒ 2b = a + c

(C) Equation of line through (0, 1, 0) and perpendicular to 

For some value of λ, the foot of perpendicular from origin to line is (λ, 2λ + 1, 2λ)
Dr ’s of this ^ from origin are λ , 2λ + 1, 2λ

∴ Foot of perpendicular is 

∴ Required distance

Q. 2. Let (x, y) be such that 

Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I                                                                    Column II 

(A) If a = 1 and b = 0, then (x, y)                                (p) lies on the circle  x2 + y2 = 1 

(B) If a = 1 and b = 1, they (x, y)                                (q) lies on (x2 – 1)(y2 – 1) = 0 

(C) If a = 1 and b = 2, then (x, y)                                (r) lies on y = x 

(D) If a = 2 and b = 2, then (x, y)                                (s) lies on (4x2 – 1) (y2 – 1) = 0

Ans. (A)  → p, (B) → q, (C) → p , (D) → s

Solution. 

⇒ y = cos α, bxy = cos β, αx = cos γ
∴ We get α + β = γ and cos β = bxy
⇒ cos  (γ – α) = bxy
⇒ cos y cos α + sin γ sin α = bxy
⇒ axy + sin γ sin α = bxy ⇒ (a – b) xy = – sin α sin γ
⇒ (a – b)² x²y² = – sin² α sin² γ
= (1– cos² α)  (1– cos² γ)
⇒ (a – b)² x²y² = (1– a²x²) (1– y²) ....(1)

(A) For a = 1, b = 0, equation (1) reduces to
x²y² = (1 – x²) (1– y²) ⇒ x² + y² = 1

(B) For a = 1,  b = 1 equation (1) becomes
(1– x²)  (1– y²) = 0  ⇒ (x² – 1)  (y² – 1) = 0

(C) For a = 1, b = 2 equation (1) reduces to
x²y²  = (1– x²) (1– y²) ⇒ x² + y² = 1

(D) For a = 2,  b = 2 equation (1) reduces to
0 = (1– 4x²)  (1– y²) ⇒ (4x² – 1)  (y² – 1)  = 0

DIRECTIONS (Q. 3) : Following question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q. 3. Match List I with List II and select the correct answer using the code given below the lists :

List I                                                                                                                                           List II

P.                                                        1. 

Q. If cosx + cosy + cosz = 0 = sinx + siny + sinz then possible value of cos          2. √2

R.   cos 2x + sinx sin 2 secx = cosx sin2x secx +                                            3. 1/2
 cos 2x then possible value of secx is

S.                                                                      4. 1
 then possible value of x is

Ans. (b)

Solution. 

(Q) We have cos x + cos y = – cos z
sin x + sin y = – sin z
Squaring and adding we get

(cos x + cos y)² + (sin x + sin y)² = cos² z + sin² z
⇒ 2 + 2 cos (x – y) = 1

(R) We have

= sin2x  secx  (cosx  – sinx)

∴ (R) → (2)