Question

Solution

Correct option is

increasing on [1/2, 1]

           

           

Since  for all x, so f’ (x) > 0 if and only if

(x – 1)(2x  – 1) < 0 i.e. –1/2 = min (1, –1/2) < x < max (1, –1/2) = 1

Thus f increases on [–1/2, 1].

SIMILAR QUESTIONS

Q1

If ab > 0 then the minimum value of  

Q2

The curve y = ax3 + bx2 + cx + 8 touches x – axis at P(2, 0) and cuts they – axis at a point Q where its gradient is 3. The value of a, b, c are respectively

Q3

If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, is

Q4

The tangent to the curve 

At the point corresponding to  is

Q5

The points of contact of the vertical tangents to x = 2 – 3 sinθ,  y = 3 + 2 cos θ are

Q6

 the in this interval

Q7

The set of all values of a for which the function

                   

decreases for all real x is

Q8
Q9

The length of a longest interval in which the function 3 sin x – 4Sin3x is increasing is

Q10

The equation e 1 + x – 2 = 0 as