The sufficient conditions for the functions f : R → R is to be maximum at x = a will be 1) f’(a) = 0 and f’(a) > 0 2) f’(a) = 0 and f’’(a) = 0 3) f’(a) = 0 and f’’(a) < 0 4) f’(a) > 0 and f’’(a) < 0 Solution:... View Article
Of the given perimeter, the triangle having maximum area is 1) Isosceles triangle 2) Right angle triangle 3) Equilateral 4) None of these Solution: (3) Equilateral AB, the area of triangle... View Article
The area of a rectangle will be maximum for the given perimeter when the rectangle is a 1) Parallelogram 2) Trapezium 3) Square 4) None of these Solution: (3) Square S = 2 (x + y); x and y are adjacent sides. y = (S -... View Article
The necessary condition to be maximum or minimum for the function is 1) f’(x) = 0 and it is sufficient 2) f’’(x) = 0 and it is sufficient 3) f’(x) = 0 but it is not sufficient 4) f’(x) = 0 and f’’(x) = –... View Article
The adjacent sides of a rectangle with a given perimeter as 100 cm and enclosing maximum area are 1) 10 cm and 40 cm 2) 20 cm and 30 cm 3) 25 cm and 25 cm 4) 15 cm and 35 cm Solution: (3) 25 cm and 25 cm 2x + 2y = 100 ⇒ x + y =... View Article
d/dx (ex log sin 2x) = (1) ex(log sin 2x + 2 cot 2x) (2) ex(log cos 2x + 2 cot 2x) (3) ex(log cos 2x + cot 2x) (4) none of these Solution: Let y = ex log... View Article
d/dx tan-1 ( sec x + tan x) = (1) 1 (2) 1/2 (3) cos x (4) sec x Solution: Let y = tan-1 ( sec x + tan x) = tan-1 {(1+sinx)/cos x} = tan-1 {(cos (x/2) + sin... View Article
d/dx log (√(x-a) + √(x-b)) = (1) 1/2 √(x-a) +√(x-b) (2) 1/2 √((x-a)(x-b)) (3) 1/√((x-a)(x-b)) (4) none of these Solution: Let y = log (√(x-a) + √(x-b)) dy/dx... View Article
If y = x + x2 + x3/2! + x4/3! +…then x(dy/dx) = (1) xex (2) y(x+1) (3) x(y+1) (4) y log (1+x) Solution: Given y = x + x2 + x3/2! + x4/3! +.... = x (1 + x + x2/2! + x3/3! +....... View Article
[latex]\frac{d}{dx}e^{x+3\log x}=[/latex] (1) ex.x2(x+3) (2) ex.x(x+3) (3) ex+3/x (4) none of these Solution : Given y = ex+ 3log x = ex.e3 log x = = ex .x3 dy/dx = ex.x3+3x2ex... View Article
If y = log ((1+√x)/(1-√x)), then dy/dx = (1) √x/(1-x) (2) 1/√x(1-x) (3) √x/(1+x) (4) 1/√x(1+x) Solution: Given y = log ((1+√x)/(1-√x)) We know log a/b = log a - log b... View Article
If y = 2log x, then dy/dx is (1) 2log x/log 2 (2) 2log x log 2 (3) 2log x/x (4) 2log x log 2/x Solution: y = 2log x Differentiate w.r.t.x dy/dx = 2log x... View Article
If x2/3 + y2/3 = a2/3, then dy/dx = (1) (y/3)1/3 (2) (-y/x)1/3 (3) (x/y)1/3 (4) (-x/y)1/3 Solution: Given x2/3 + y2/3 = a2/3 y2/3 = a2/3 - x2/3 Differentiate... View Article
If f(x) = √(1+ cos2 (x2), then f’(√π/2) is (1) √π/6 (2) -√(Ï€/6) (3) 1/√6 (4) Ï€/√6 Solution: Given f(x) = √(1+ cos2 (x2) Differentiate w.r.t.x At x = √π/2 =... View Article
If [latex]y=\tan^{-1}\frac{4x}{1+5x^{2}}+\tan^{-1}\frac{2+3x}{3-2x}[/latex] , then dy/dx = (1) 1/(1+25x2) + 2/(1+x2) (2) 5/(1+25x2) + 1/(1+x2) (3) 5/(1+25x2) + 1/(1+25x2) (4) none of these Solution: Given y = tan-1... View Article
If f(x) = mx2 + nx + p, then f’(1) + f’(4) – f’(5) is equal to (1) m (2) -m (3) n (4) -n Solution: Given f(x) = mx2 + nx + p Differentiate w.r.t.x f’(x) = 2mx + n Put x = 1 f’(1) = 2m+n... View Article
If [latex]y=\log \left ( \frac{1+x}{1-x} \right )^{\frac{1}{4}}-\frac{1}{2}\tan^{-1}x[/latex], then dy/dx = (1) x2/(1-x4) (2) 2x2/(1-x4) (3) x2/2(1-x4) (4) none of these Solution: y = log ((1+x)/(1-x))1/4 - ½ tan-1 x = (1/4) log... View Article
If f(x) = logx (log x), then f’(x) at x = e is (1) e (2) 1/e (3) 1 (4) none of these Solution: f(x) = logx (log x) = (log log x)/ log x Use quotient rule of differentiation... View Article
d/dx [ sinn x cos nx] = (1) n sinn-1x cos (n+1)x (2) n sinn-1x cos nx (3) n sinn-1x cos (n-1)x (4) n sinn-1x sin (n+1)x Solution: Let y = sinn x cos nx... View Article
If y = b cos log (x/n)n, then dy/dx = (1) -nb sin log (x/n)n (2) nb sin log (x/n)n (3) (-nb/x) sin log (x/n)n (4) none of these Solution: Given y = b cos log (x/n)n... View Article
