If x = sin t and y = sin pt, then the value of (1 – x2) d2y / dx2 – x (dy / dx) + p2y = 1) 0 2) 1 3) –1 4) √2 Solution: (1) 0 x = sin t and y = sin pt (dx / dt) = cos t (dy / dt) = p cos pt dy / dx = p cos pt / cos... View Article
If In = dn (xn log x) / dxn, then In – nIn-1 = 1) n 2) n –1 3) n! 4) (n –1)! Solution: (4) (n –1)! In = dn (xn log x) / dxn In = dn-1 [xn-1 + nxn-1 log x] / dxn-1 In = (n... View Article
Let f (x) and g (x) be two functions having finite non-zero 3rd order derivatives f’’’ (x) and g’’’ (x) for all, x ∈ R. If f (x) g (x) = 1 for all x ∈ R, then (f’’ / f’) – (g’’’ / g’) = 1) 3 [(f’’ / g) - (g’’ / f)] 2) 3 [(f’’ / f) - (g’’ / g)] 3) 3 [(g’’ / g) - (f’’ / g)] 4) 3 [(f’’ / f) - (g’’ / f)] Solution: (2) 3... View Article
If y2 = p (x) is a polynomial of degree 3, then 2 d [y3 (d2y / dx2)] / dx = 1) p’’ (x) + p’ (x) 2) p’’ (x) . p’’ (x) 3) p (x) . p’’’ (x) 4) Constant Solution: (3) p (x) . p’’’ (x) y2 = p (x) 2y (dy / dx) =... View Article
The derivative of tan-1 [(√1 + x2 – 1) / x] with respect to tan-1 [(2x √1 – x2) / (1 – 2x2)] at x = 0 is 1) 1 / 8 2) 1 / 4 3) 1 / 2 4) 1 Solution: (2) 1 / 4 Let y = tan-1 [(√1 + x2 - 1) / x] and z = tan-1 [(2x √1 - x2) / (1 - 2x2)]... View Article
If y = sec-1 (2x) / (1 + x2) + sin-1 [x – 1] / [x + 1], then dy / dx = 1) 1 2) [x - 1] / [x + 1] 3) Does not exist 4) None of these Solution: (3) Does not exist y = sec-1 (2x) / (1 + x2) + sin-1 [x -... View Article
If √1 – x6 + √1 – y6 = a3 (x3 – y3), then dy / dx = 1) (x2 / y2) [(√1 - x6 / √1 - y6)] 2) (y2 / x2) [(√1 - y6 / √1 - x6)] 3) (x2 / y2) [(√1 - y6 / √1 - x6)] 4) None of these Solution:... View Article
d [tan-1 (√1 + x2 + √1 – x2) / (√1 + x2 – √1 – x2)] 1) - x / √1 - x4 2) x / √1 - x4 3) - 1 / 2√1 - x4 4) 1 / 2√1 - x6 Solution: (1) - x / √1 - x4 Let A = (√1 + x2 + √1 - x2) / (√1 +... View Article
If y = (x log x)log log x, then dy / dx = 1) {(x log x)log log x {(1 / x log x) (log x + log log x) + (log log x) [(1 / x) + (1 / x log x)]} 2) {(x log x)log log x log log x = [(2 /... View Article
If y = xx^x^….. ∞, then dy / dx = 1) y2 / [x (1 + y log x)] 2) y2 / [x (1 - y log x)] 3) y / [x (1 + y log x)] 4) y / [x (1 - y log x)] Solution: (2) y2 / [x (1 - y... View Article
If x = sec θ – cos θ and y = secn θ – cosn θ, then 1) (x2 + 4) (dy / dx)2 = n2 (y2 + 4) 2) (x2 + 4) (dy / dx)2 = x2 (y2 + 4) 3) (x2 + 4) (dy / dx)2 = (y2 + 4) 4) None of the above... View Article
If y = f [(2x – 1) / (x2 + 1)] and f’ (x) = sin2 x, then dy / dx = 1) {[6x2 - 2x + 2] / (x2 + 1)2} sin [(2x - 1)2 / (x2 + 1)] 2) {[6x2 - 2x + 2] / (x2 + 1)2} sin2 [(2x - 1) / (x2 + 1)] 3) {[- 2x2 + 2x + 2]... View Article
If u (x, y) = y log x + x log y, then uxuy – uxlog x – uy log y + log x log y = 1) 0 2) –1 3) 1 4) 2 Solution: (3) 1 ux = (y / x) + log y uy = log x + (x / y) (y / x) + log y + log x + (x / y) - [(y / x) +... View Article
If xexy = y + sin2 x, then at x = 0, dy / dx = 1) –1 2) –2 3) 1 4) 2 Solution: (3) 1 xexy = y + sin2 x At x = 0, 0e0 = y + 1 y = 0 1exy + xexy [x (dy / dx) + y] = (dy / dx) +... View Article
Let f, g and h be real-valued suntions defined on the interval [0, 1] by f (x) = ex^2 + e-x^2 and h (x) = x2ex^2 + e-x^2. If a, b and c denote respectively the absolute maximum of f, g and h on [0, 1] then 1) a = b and c ≠b 2) a = c and a ≠b 3) a ≠b and c ≠d 4) a = b = c Solution: (4) a = b = c f (x) = ex^2 + e-x^2 f’ (x) =2x... View Article
If y = cot-1 (cos 2x)½, then the value of dy / dx at x = π / 6 will be 1) (2 / 3)1/2 2) (1 / 3)1/2 3) (3)1/2 4) (6)1/2 Solution: (1) (2 / 3)1/2 y = cot-1 (cos 2x)½ dy / dx = - 1 / [1 + cos 2x] d... View Article
If y = (x / 2) √a2 + x2 + (a2 / 2) log [x + √x2 + a2], then dy / dx = 1) √x2 + a2 2) 1 / [√x2 + a2] 3) 2√x2 + a2 4) 2 / √x2 + a2 Solution: (1) √x2 + a2 y = (x / 2) √a2 + x2 + (a2 / 2) log [x + √x2 +... View Article
Let f (x) = x3. Use mean value theorem to write [f (x + h) – f (x)] / h = f’ (x + θh) with 0 < θ < 1. If x ≠0, then limh→0 θ = 1) –1 2) –0.5 3) 0.5 4) 1 Solution: (3) 0.5 f (x) = x3 f (x + h) = f (x) + hf’ (x) + (h2 / 2) f’’ (x + θ1h), where 0 < θ1... View Article
In which of the following functions, Rolle’s theorem is applicable 1) f (x) = |x| in - 2 ≤ x ≤ 2 2) f (x) = tan x in 0 ≤ x ≤ π 3) f (x) = 1 + (x - 2)⅔ in 1 ≤ x ≤ 3 4) f (x) = x (x - 2)2 in 0 ≤ x ≤ 2... View Article
The value of c in (0, 2) satisfying the mean value theorem for the function f (x) = x (x – 1)2, x ∈ [0, 2] = 1) 3 / 4 2) 4 / 3 3) 1 / 3 4) 2 / 3 5) 5 / 3 Solution: (2) 4 / 3 f (x) = x (x - 1)2, x ∈ [0, 2] f’ (c) = [f (b) - f (a)] / [b... View Article
