If the function `f(x)=2x^2 + 3x + 5` satisfies `LMVT` at `x = 2` on the closed interval `[1, a]`, then the value of 'a' is equal to

If the function f(x)=2x^(2)+3x-m log x is monotonic decreasing in the interval (0,1) then the least value of the parameter m is

The fucntion f(x)=x^(3)-6ax^(2)+5x satisfies the conditions of Lagrange's mean value theorem over the interval [1,2] and the value of c (that of LMVT) is 7/4, then find the value of a.

If the function f(x)=logx, x in [1, e] , satisfies all the conditions of LMVT, then the value of c is

The function f(x) = 2x^(3) + 3x^(2) -12 x + 1 decreases in the interval

The least value of the function f(x) = 2 cos x + x in the closed interval [0,pi/2] is

The least value of the function f(x) = 2 cos x + x in the closed interval [0, (pi)/(2)] is:

The function f(x)=x^(2)+2x-5 is increasing in the interval

Consider the function f(x)=8x^2-7x+5 on the interval [-6,6] .the value of c that satisfies the condition of the mean value theorem is