If fx = x - 3, then f'(3) is
- 1
1
0
does not exist
If fx = xsin1x, x ≠ 00 , x = 0, then at x = 0 the function f(x) is
continuous
differentiable
continuous but not differentiable
None of the above
If Rolle's theorem for f(x) = exsinx - cosx is verified on π4, 5π4, then the value of c is
π3
π2
3π4
π
If the function f(x) defined by
fx = xsin1x, for x ≠ 0k, for x = 0
is continuous at x = 0, then k is equal to
12
If y = emsin-1x and 1 - x2dydx2 = Ay2, then A is equal to
m
- m
m2
- m2
For what value of k, the function defined by
f(x) = log1 + 2xsinx°x2, for x ≠ 0k , for x = 0
is continuous at x = 0 ?
2
π90
90π
If log10x2 - y2x2 + y2 = 2, then dydx is equal to
- 99x101y
99x101y
- 99y101x
99y101x
If g(x) is the inverse function of f(x) and f'x = 11 + x4, then g'(x) is
1 + [g(x)]4
1 - [g(x)]4
1 + [f(x)]4
11 + g(x)4
If the function f(x) = tanπ4 + x1x for x ≠ 0 is = K for x = 0 continuous at x = 0, then K = ?
e
e- 1
e2
e- 2
If x = f(t) and y = g(t) are differentiable functions of t, then d2ydx2 is
f't . g''t - g't . f''tf't3
f't . g''t - g't . f''tf't2
g't . f''t - f't . g''tf't3
g't . f''t + f't . g''tf't3
A.
Given, x = f(t) and y = g(t)On differentiating both sides w.r.t. 't', we getdxdt = f't and dydt = g'tWe know that, dydx = dydtdxdt⇒ dydx = g'tf'tAgain, differentiating both sides w.r.t. 'x', we getd2ydx2 = f't . g''t - g't . f''tf't2 . dtdxd2ydx2 = f't . g''t - g't . f''tf't3