The value of m for which the function f(x) = is differentiable at x = 1, is
0
1
2
does not exist
If f(x) = , is differentiable at x = 0, then (a, b) will be
(- 3, - 1)
(- 3, 1)
(3, 1)
(2, - 1)
Rolle's theorem is not applicable for the function f(x) = , where x [- 1, 1] because
the function f(x) is not continuous in the interval [- 1, 1]
the function f(x) is not differentiable in the interval (- 1, 1)
If the function f(x) = x3 - 6ax2 + 5x satisfies the conditions of Lagrange's Mean Value theorem for the interval [1, 2] and the tangent to the curve y = f(x) at x = 7/4 is parallel to the chord that join the points ofintersection of the curve with the ordinates x = 1 and x = 2 . Then, the value of a is
If 2a + 3b + 6c = 0, then atleast one root of the equation ax2 + bx + c = 0, lies in the interval
(0, 1)
(1, 2)
(2, 3)
None of these
If equation y + , then value of y is
None of these
B.