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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

Axis of a parabola is y = x and vertex and focus are at a distance √2 and 2√2, respectively from the origin. Then, equation of the parabola is

1) (x - y)2 = 8 (x + y - 2) 2) (x + y)2 = 2 (x + y - 2) 3) (x - y)2 = 4 (x + y - 2) 4) (x + y)2 = 2(x - y + 2) Solution: (1) (x - y)2 = 8 (x... View Article

If e₁ is the eccentricity of the ellipse (x² / 16) + (y² / 25) = 1 and e₂ is the eccentricity of the hyperbola passing through the foci of the ellipse and e₁e₂ = 1, then the equation of the hyperbola is

1) (x2 / 9) - (y2 / 16) = 1 2)(x2 / 16) - (y2 / 9) = - 1 3) (x2 / 9) - (y2 / 25) = 1 4) None of these Solution: (2) (x2 / 16) - (y2 / 9) = -... View Article

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Archives

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

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

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

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

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

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

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

05 May 2021: UPSC Exam Comprehensive News Analysis

04 May 2021: PIB Summary & Analysis

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

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

Axis of a parabola is y = x and vertex and focus are at a distance √2 and 2√2, respectively from the origin. Then, equation of the parabola is

If e₁ is the eccentricity of the ellipse (x² / 16) + (y² / 25) = 1 and e₂ is the eccentricity of the hyperbola passing through the foci of the ellipse and e₁e₂ = 1, then the equation of the hyperbola is

If P be the point (1, 0) and Q a point on the locus of y² = 8x. The locus of mid – point of PQ is

The lines x = at² meets the ellipse (x² / a²) + (y² / b²) = 1, in the real points, iff

The curve represented by x = 3 (cos t + sin t) and y = 4(cos t – sin t) is

The parametric representation of a point of the ellipse whose foci are (- 3, 0) and (9, 0) and eccentricity 1/3 is

If (x² / 36) – (y² / k²) = 1 is a hyperbola, then which of the following statements can be correct?

The equation of an ellipse whose eccentricity is 1/ 2 and the vertices are (4, 0) and (10, 0), is

The parametric representation of a point on the ellipse whose foci are ( -1, 0) and (7, 0) and eccentricity 1/2 is

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