If x2y5 = (x + y)7, then d2ydx2 is equal to
y/x2
x/y
1
0
If x = x = secθ, y = tanθ, then the value of d2ydx2 at θ = π4 is
- 1
2
C.
Given, x = secθ, y = tanθdxdθ = secθtanθ, dydθ = cscθNow, d2ydx2 = ddxdydx = ddxcscθdθdx = - cscθcotθ × 1secθtanθ = - 1tan3θAt θ = π4, d2ydx2θ = π4 = - 1tanπ43 = - 1
If x = f(t) and y =g(t), then the value of d2ydx2 is
f'tg''t - g'tf''tf't3
f'tg''t - g'tf''tf't2
'tf''t - g''tf'tf't2
g'tf''t - g''tf'tf't3
The value of a and b such that the function
fx = - 2sinx, - π ≤ x ≤ - π2asinx + b, - π2 < x < π2cosx, π2 ≤ x ≤ π
is continuous in - π, π, are
- 1, 0
1, 0
1, 1
- 1, 1
The equation of tangent to the curve y2 = ax2 + b at point (2, 3) is y = 4x - 5, then the values of a and b are
3, - 5
6, - 5
6, 15
6, - 15
Let A = cosθ- sinθsinθ- cosθ, then the inverse of A is
cosθ- sinθsinθ- cosθ
- cosθsinθsinθcosθ
sinθ- cosθcosθ- sinθ
- sinθ- cosθ- cosθsinθ
The value of f(4) - f(3) is
∆f2 + ∆2f1 + ∆3f1
∆f3 + ∆2f2 + ∆3f1
∆f2 + ∆2f1 + ∆3f0
None of these
1 + ∆nfa is equal to
f(a + h)
f(a + 2h)
f(a + nh)
f(a + (n - 1)h)
Simplify the Boolean function (x · y) + [(x + y') - y]'.
x + y
xy
If matrix A = abcd, then A- 1 is equal to
ab - bc
1ab - bc
1ab - bcd- b- ca