Correct answer is (b) 62
For explanation I would say: We have, f(x) = x^3 – 12x^2 + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x^2 – 24x + 45
3x^2 – 24x + 45 = 0
Or x^2 – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 3^3 – 12*3^2 + 45*3 + 8 = 62.