The given limit is: lim(h→0) [f(1+h-h^2) - f(1)] / [f(2+2h+h^2) - f(2)]
Since f(x) is differentiable, we can apply L'Hôpital's Rule. The rule states that if the limit of the ratio of two functions as x approaches a is an indeterminate form (0/0 or ∞/∞), and the derivatives of the numerator and denominator both exist at x = a, then the limit of the ratio is equal to the limit of the ratio of their derivatives as x approaches a.
In our case, let's apply L'Hôpital's Rule. We'll first find the derivatives of the numerator and denominator with respect to h, and then evaluate these derivatives at h = 0.
Derivative of the numerator: [f(1+h-h^2) - f(1)]' = f'(1+h-h^2) - f'(1)
Derivative of the denominator: [f(2+2h+h^2) - f(2)]' = f'(2+2h+h^2) - f'(2)
We are given f'(2) = 6 and f'(1) = 4.
Now, let's evaluate these derivatives at h = 0:
Derivative of the numerator at h = 0: [f'(1+0-0^2) - f'(1)] = f'(1) - f'(1) = 4 - 4 = 0
Derivative of the denominator at h = 0: [f'(2+0+0^2) - f'(2)] = f'(2) - f'(2) = 6 - 6 = 0
Now, let's rewrite the limit using these derivatives:
lim(h→0) [f(1+h-h^2) - f(1)] / [f(2+2h+h^2) - f(2)] = lim(h→0) [0 / 0]
Since we have an indeterminate form (0/0), we can apply L'Hôpital's Rule again:
lim(h→0) [(f'(1+h-h^2) - f'(1))'] / [(f'(2+2h+h^2) - f'(2))'] = lim(h→0) [0 / 0]
Now, let's plug in the given values for f'(1) = 4 and f'(2) = 6:
lim(h→0) [(f'(1+h-h^2) - 4)'] / [(f'(2+2h+h^2) - 6)'] = lim(h→0) [0 / 0]
Now, we apply L'Hôpital's Rule one more time:
lim(h→0) [(f''(1+h-h^2)) / (f''(2+2h+h^2))] = lim(h→0) [0 / 0]
Since f(x) is differentiable and the limit is again an indeterminate form, we can apply L'Hôpital's Rule:
lim(h→0) [(f'''(1+h-h^2)) / (f'''(2+2h+h^2))] = lim(h→0) [0 / 0]
At this point, if the derivatives of f(x) exist and the limit is still an indeterminate form, it suggests that the limit may not have a single definite value. Further analysis of the function f(x) is needed to determine the behavior near these points. It's important to consider the specific properties of f(x) and possibly use more information to evaluate the limit.
The correct answer is option (C): equal to 3
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2\ is :
A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of the function define the rate of change of a function at a point. It is mainly used in fields such as physics, engineering, biology and so on.
The first-order differential equation has a degree equal to 1. All the linear equations in the form of derivatives are in the first order. It has only the first derivative such as dy/dx, where x and y are the two variables and is represented as: dy/dx = f(x, y) = y’
The equation which includes second-order derivative is the second-order differential equation. It is represented as; d/dx(dy/dx) = d2y/dx2 = f”(x) = y”.
Differential equations can be divided into several types namely