If f(x) is continuous & differentiable,f(1) = 10 and f’(x) ≥ 3 in 1 ≤ x ≤ 4 thenthe smallest value of f(4) can beA. 10 B. 13C. 14 D. 19

The problem is asking for the smallest possible value of f(4) given that f(x) is continuous and differentiable, f(1) = 10, and f’(x) ≥ 3 for 1 ≤ x ≤ 4.

The function f(x) is differentiable and continuous, and its derivative f’(x) is greater than or equal to 3 in the interval [1, 4]. This means that the function f(x) is increasing at a rate of at least 3 units per unit increase in x.

Since f(1) = 10, and the function increases at a rate of at least 3 units per unit increase in x, the smallest possible value of f(4) would be achieved if the function increased at exactly this minimum rate over the interval from x = 1 to x = 4.

The increase in x over this interval is 4 - 1 = 3 units. If the function increases at a rate of 3 units per unit increase in x, then the increase in f(x) over this interval would be 3 * 3 = 9 units.

Therefore, the smallest possible value of f(4) would be f(1) + 9 = 10 + 9 = 19.

So, the answer is D. 19.