Statement 1: `f(x)=(x^3)/3+(a x^2)/2` + x+5 has positive point of maxima for `a<-2.` Statement 2: `x^2+a x+1=0` has both roots positive for `a<2.`

Let f(x) =x^(3) -x^(2) -x-4 (i) find the possible points of maxima // minima of f(x) x in R (ii) Find the number of critical points of f(x) x in [1,3] (iii) Discuss absolute (global) maxima // minima value of f(x) for x in [-2,2] (iv) Prove that for x in (1,3) the function does not has a Global maximum.

If the function f(x) = x^3 + 3(a-7)x^2 +3(a^2-9)x-1 has a positive point Maximum , then

The values of a and b for which all the extrema of the function, f(x)=a^(2)x^(3)-0.5ax^(2)-2x-b, is positive and the minima is at the point x_(0)=(1)/(3), are

Let f(x) = x^(3)-3(7-a)X^(2)-3(9-a^(2))x+2 The values of parameter a if f(x) has a positive point of local maxima are

Show that the function f(x)=4x^3-18 x^2+27 x-7 has neither maxima nor minima.

Show that the function f(x)=4x^3-18 x^2+27 x-7 has neither maxima nor minima.

Statement 1: f(x)=|x-1|+|x-2|+|x-3| has point of minima at x=3. Statement 2: f(x) is non-differentiable at x=3.

Statement -1: f(x) = x^(3)-3x+1 =0 has one root in the interval [-2,2]. Statement-2: f(-2) and f(2) are of opposite sign.

Statement-1 : f(x) = x^(7) + x^(6) - x^(5) + 3 is an onto function. and Statement -2 : f(x) is a continuous function.

Statement I :The function f(x)=(x^(3)+3x-4)(x^(2)+4x-5) has local extremum at x=1. Statement II :f(x) is continuos and differentiable and f'(1)=0.