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The Angle Between Y 2 X And X 2 Y At The Origin Is

1) 2 tan-1 (3/4) 2) tan-1 (4/3) 3) π/2 4) π/4 Solution: (3) π / 2 y2 = x and... View Article

1) 1 / √a2 – b2 2) 1 / √a2 + b2 3) √a2 + b2 4) √a2 – b2... View Article

1) tan-1 [|(a – b) / (1 + ab)|] 2) tan-1 [|(a – b) / (1 – ab)|] 3) tan-1... View Article

1) 2x + y – 6 = 0 2) 2y + x – 6 = 0 3) x + 3y... View Article

1) 3x + 4y = 16 2) 4x + y = 4 3) x + y = 4 4) 4x... View Article

1) 16 2) 12 3) 8 4) 4 5) 2 Solution: (4) 4 (x2 / a2) + (y2 / b2)... View Article

1) y = – 1 2) – (4/e) 3) x = – 1 4) x = -4/e Solution: (2) –... View Article

1) 3√2 / 8 2) 2√3 / 8 3) 3√2 / 5 4) √3 / 4 Solution: (1) 3v2 /... View Article

1) P(-1) is the minimum and P(1) is the maximum of P 2) P(-1) is not minimum but P(1) is... View Article

1) e 2) -e 3) 1 4) -1 Solution: (3) 1 f (x) = e(x^4 – x^3 + x^2) f’... View Article

1) a < 0, b > 0 2) a = 0, b > 0 3) a < 0, b <... View Article

1) 1 2) 0 3) -(1/2) 4) -1 Solution: (4) -1 f (x) = lim x=-1- f (x) = lim... View Article

For any integer n = 1, the sum Sk=1n k(k+2) is equal to (1) n(n+1)(n+2)/2 (2) n(n+1)(2n+1)/6 (3) n(n+1)(2n+7)/6 (4)... View Article

(1) 2nh (2) n/h (3) nh (4) 2n/h Solution: Given a, a1, a2, ,…,a2n, b are in arithmetic progression. a1,... View Article

(1) 9 (2) 18 (3) 24 (4) 36 Solution: Given AM = 27 HM = 12 We know GM2 =... View Article

(1) (n – 1) (a1 – an) (2) na1an (3) (n – 1)a1an (4) n(a1 – an) Solution: Given a1,... View Article

(1) 1/41 (2) 1/45 (3) 1/49 (4) 1/37 Solution: Given 7th term of HP = 1/10 12th term = 1/25... View Article

(1) 22 (2) 23 (3) 24 (4) 25 Solution: Given a1, a2, a3,…are in HP. So 1/a1, 1/a2, 1/a3 are... View Article

(1) 1/2(1 – v5) (2) 1/2(v5) (3) v5 (4) 1/2(v5 – 1) Solution: Let a, ar, ar2 be the terms... View Article

(1) S2 (2) S2/(2S + 1) (3) 2S/(S2 – 1) (4) S2/(2S – 1) Solution: S = Sn=08 rn =... View Article