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If θ = sin^-1 x + cos^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which θ lies is given by

1) (π / 2) ≤ θ ≤ 3π / 4 2) 0 < θ < π 3) - π / 4 ≤ θ ≤ 0 4) π / 4 ≤ θ ≤ π / 2 Solution: (2) 0 < θ < π θ = sin-1 x + cos-1 x -... View Article

If tan^-1 x + tan^-1 y + tan^-1 z = π, then x + y + z =

1) xyz 2) 0 3) 1 4) 2xyz Solution: (1) xyz tan-1 x + tan-1 y + tan-1 z = π tan-1 [x + y + z - xyz] / [1 - xy - yz - zx] = π x + y + z - xyz... View Article

The equation 2 cos^-1 x + sin^-1 x = 11π / 6 has

1) No solution 2) Only one solution 3) Two solutions 4) Three solutions Solution: (1) No solution 2 cos-1 x + sin-1 x = 11π / 6 cos-1 x +... View Article

∑_m=1ⁿ tan^-1 [2m] / [m⁴ + m² + 2] =

1) tan-1 [2m] / [n2 + n + 2] 2) tan-1 [n2 + n] / [n2 + n + 2] 3) tan-1 [n2 + n + 2] / [n2 + n] 4) None of these Solution: (2) tan-1 [n2 + n]... View Article

tan [(π / 4) + (1 / 2) cos^-1 (a / b)] + tan^-1 [(π / 4) – (1 / 2) cos^-1 (a / b)] =

1) 2a / b 2) 2b / a 3) a / b 4) b / a Solution: (2) 2b / a tan [(π / 4) + (1 / 2) cos-1 (a / b)] + tan-1 [(π / 4) - (1 / 2) cos-1 (a / b)]... View Article

tan^-1 [√1 + x² + √1 – x²] / [√1 + x² – √1 – x²]

1) (π / 4) + (1 / 2) cos-1 x2 2) (π / 4) + cos-1 x2 3) (π / 4) + (1 / 2) cos-1 x 4) (π / 4) - (1 / 2) cos-1 x2 Solution: (1) (π / 4) + (1 /... View Article

2 tan^-1 [√(a – b / a + b) tan (θ / 2)] =

1) cos-1 [a cos θ + b] / [a + b cos θ] 2) cos-1 [a + b cos θ] / [a cos θ + b] 3) cos-1 [a cos θ / [a + b cos θ]] 4) cos-1 [b cos θ / [a cos θ... View Article

If θ = tan^-1 a, ɸ = tan^-1 b and ab = – 1, then θ – ɸ =

1) 0 2) π / 4 3) π / 2 4) None of these Solution: (3) π / 2 θ = tan-1 a, ɸ = tan-1 b and ab = - 1 tan θ tan ɸ = - 1 tan θ = - cot ɸ θ - ɸ = π / 2... View Article

2 tan^-1 (cos x) = tan^-1 (cosec² x), then x =

1) π / 2 2) π 3) π / 6 4) π / 3 Solution: (4) π / 3 2 tan-1 (cos x) = tan-1 (cosec2 x) tan-1 [2 cos x] / [1 - cos2 x] = tan-1 (1 / sin2 x)... View Article

cot^-1 [√1 – sin x + √1 + sin x] / [√1 – sin x – √1 + sin x] =

1) π - x 2) 2π - x 3) x / 2 4) π - (x / 2) Solution: (4) π - (x / 2) cot-1 [√1 - sin x + √1 + sin x] / [√1 - sin x - √1 + sin x] = cot-1 [√1... View Article

If tan^-1 (x + 2) + tan^-1 (x – 2) – tan^-1 (1 / 2) = 0, then one of the value of x =

1) -1 2) 5 3) 1/2 4) 1 5) -1/2 Solution: (4) 1 tan-1 (x + 2) + tan-1 (x - 2) = tan-1 (1 / 2) tan-1 [(x + 2) + (x - 2)] / [1 - (x + 2) (x -... View Article

If sin^-1 x + sin^-1 2x = π / 3, then the value of x =

1) √3 / 2 √7 2) √3 / √7 3) 0 4) 1 Solution: (1) √3 / 2 √7 sin-1 x + sin-1 2x = π / 3 sin-1 2x = sin-1 (√3 / 2) - sin-1 x sin-1 2x = sin-1... View Article

The solution set of the equation cos^-1 x < sin^-1 x is

1) [- 1, 1] 2) [1 / √2, 1] 3) [0, 1] 4) (1 / √2, 1] Solution: (4) (1 / √2, 1] (π / 2) - sin-1 x < sin-1 x ⇒ sin-1 x > π / 4 (- π / 2)... View Article

x + y = 1, xy = 2, then tan^-1 x + tan^-1 y =

1) π / 4 2) 3π / 4 3) 3π / 2 4) π / 2 Solution: (2) 3π / 4 x + y = 1, xy = 2 (tan-1 x + tan-1 y) = π + tan-1 [x + y] / [1 - xy] (tan-1 x +... View Article

The value of tan^-1 2 + tan^-1 3 =

1) 3π / 4 2) π / 4 3) π / 3 4) tan-1 6 5) tan-1 5 Solution: (1) 3π / 4 Let θ = tan-1 2 + tan-1 3 tan-1 2 > π / 4 and tan-1 3 > π / 4... View Article

If 2 sin^-1 x – cos^-1 x = π / 2, then x =

1) 1 / √2 2) - 1 / √2 3) - √3 / 2 4) √3 / 2 Solution: (4) √3 / 2 2 sin-1 x - cos-1 x = π / 2 ---- (i) sin-1 x + cos-1 x = π / 2 --- (ii)... View Article

05 May 2021: UPSC Exam Comprehensive News Analysis

CNA 5th May 2021:- Download PDF Here TABLE OF CONTENTS A. GS 1 Related B. GS 2 Related POLITY AND GOVERNANCE 1. Supply oxygen or face contempt:... View Article

04 May 2021: PIB Summary & Analysis

May 04th, 2021, PIB:- Download PDF Here TABLE OF CONTENTS 1. India – UK Virtual Summit 2. Ayush-64 3. Moorhen Yoga Mat 1. India – UK Virtual... View Article

The equation of the hyperbola having its eccentricity 2 and the distance between its foci is 8, is

1) (x2 / 12) - (y2 / 4) = 1 2) (x2 / 4) - (y2 / 12) = 1 3) (x2 / 8) - (y2 / 2) = 1 4) (x2 / 16) - (y2 / 9) = 1 Solution: (2) (x2 / 4) - (y2 / 12)... View Article

The locus of the vertices of the family of parabolas y = [a³x² / 3] + (a²x / 2) – 2a is

1) xy = 3/4 2) xy = 35/16 3) xy = 64/105 4) xy = 105/64 Solution: (4) xy = 105/64 y = [a3x2 / 3] + (a2x / 2) - 2a k = [a3h2 / 3] + (a2h / 2)... View Article

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