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# Match the following Question of Inverse Trigonometric Functions, Past year Questions JEE Advance JEE Notes | EduRev

## JEE : Match the following Question of Inverse Trigonometric Functions, Past year Questions JEE Advance JEE Notes | EduRev

The document Match the following Question of Inverse Trigonometric Functions, Past year Questions JEE Advance JEE Notes | EduRev is a part of the JEE Course Maths 35 Years JEE Mains & Advance Past year Papers Class 12.
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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 = tanâ€“1 3 â€“ tanâ€“1 1 + tanâ€“15 â€“ tanâ€“13 + ...... + tanâ€“1(2n + 1) â€“ tanâ€“1 (2n â€“ 1) + ...... âˆž

(B) âˆ´ a, b, c are in AP  â‡’ 2b = a + c

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

For some value of Î», the foot of perpendicular from origin to line is (Î», 2Î» + 1, 2Î»)
Dr â€™s of this ^ from origin are Î» , 2Î» + 1, 2Î»

âˆ´ Foot of perpendicular is

âˆ´ 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 Ã— 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 â€“ 1)(y2 â€“ 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 â€“ 1) (y2 â€“ 1) = 0

Ans. (A)  â†’ p, (B) â†’ q, (C) â†’ p , (D) â†’ s

Solution.

â‡’ y = cos Î±, bxy = cos Î², Î±x = cos Î³
âˆ´ We get Î± + Î² = Î³ and cos Î² = bxy
â‡’ cos  (Î³ â€“ Î±) = bxy
â‡’ cos y cos Î± + sin Î³ sin Î± = bxy
â‡’ axy + sin Î³ sin Î± = bxy â‡’ (a â€“ b) xy = â€“ sin Î± sin Î³
â‡’ (a â€“ b)2 x2y2 = â€“ sin2 Î± sin2 Î³
= (1â€“ cos2 Î±)  (1â€“ cos2 Î³)
â‡’ (a â€“ b)2 x2y2 = (1â€“ a2x2) (1â€“ y2) ....(1)

(A) For a = 1, b = 0, equation (1) reduces to
x2y2 = (1 â€“ x2) (1â€“ y2) â‡’ x2 + y2 = 1

(B) For a = 1,  b = 1 equation (1) becomes
(1â€“ x2)  (1â€“ y2) = 0  â‡’ (x2 â€“ 1)  (y2 â€“ 1) = 0

(C) For a = 1, b = 2 equation (1) reduces to
x2y2  = (1â€“ x2) (1â€“ y2) â‡’ x2 + y2 = 1

(D) For a = 2,  b = 2 equation (1) reduces to
0 = (1â€“ 4x2)  (1â€“ y2) â‡’ (4x2 â€“ 1)  (y2 â€“ 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. âˆš2

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 = â€“ cos z
sin x + sin y = â€“ sin z

(cos x + cos y)2 + (sin x + sin y)2 = cos2 z + sin2 z
â‡’ 2 + 2 cos (x â€“ y) = 1

(R) We have

= sin2x  secx  (cosx  â€“ sinx)

âˆ´ (R) â†’ (2)

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