Question

The tangent at any point P(x, y) on a curve y = f(x) meets the x-axis at T. If OP = PT, where O being the origin such that f(0) ≠ 0 and f(16) = 4, then $\sqrt{f(2)} + \sqrt{f(4)}$ is equal to:

Answer 1

$4(1+\sqrt{2})$

Answer 2

Let P(x, y) be a point on the curve y = f(x). The slope of the tangent at P is $\frac{dy}{dx} = f'(x)$.

The equation of the tangent at P is $Y - y = \frac{dy}{dx}(X - x)$.

The tangent meets the x-axis at T. So, the y-coordinate of T is 0.

Substituting Y = 0 into the tangent equation:

$0 - y = \frac{dy}{dx}(X - x)$

$-y = \frac{dy}{dx}(X - x)$

$X - x = -y \frac{dx}{dy}$

$X = x - y \frac{dx}{dy}$

So, the coordinates of T are $(x - y \frac{dx}{dy}, 0)$.

The origin O is (0, 0).

The point P is (x, y).

The point T is $(x - y \frac{dx}{dy}, 0)$.

The condition given is $OP = PT$.

$OP^2 = (x-0)^2 + (y-0)^2 = x^2 + y^2$.

$PT^2 = ((x - y \frac{dx}{dy}) - x)^2 + (0 - y)^2 = (-y \frac{dx}{dy})^2 + (-y)^2 = y^2 (\frac{dx}{dy})^2 + y^2$.

Since $OP = PT$, we have $OP^2 = PT^2$.

$x^2 + y^2 = y^2 (\frac{dx}{dy})^2 + y^2$

$x^2 = y^2 (\frac{dx}{dy})^2$

$x^2 = y^2 \left(\frac{1}{dy/dx}\right)^2$

$x^2 \left(\frac{dy}{dx}\right)^2 = y^2$

$\left(\frac{dy}{dx}\right)^2 = \frac{y^2}{x^2}$

$\frac{dy}{dx} = \pm \frac{y}{x}$.

Case 1: $\frac{dy}{dx} = \frac{y}{x}$.

This is a separable differential equation: $\frac{dy}{y} = \frac{dx}{x}$.

Integrating both sides: $\int \frac{dy}{y} = \int \frac{dx}{x}$

$\ln|y| = \ln|x| + \ln|C_1|$, where $C_1$ is a positive constant.

$\ln|y| = \ln|C_1 x|$

$|y| = |C_1 x|$

$y = Cx$, where $C = \pm C_1$ is a non-zero constant.

If $y=Cx$, then $f(x)=Cx$. The condition $f(0) \neq 0$ means $C \cdot 0 \neq 0$, which is $0 \neq 0$. This is a contradiction. So $y=Cx$ is not the solution.

Case 2: $\frac{dy}{dx} = -\frac{y}{x}$.

This is a separable differential equation: $\frac{dy}{y} = -\frac{dx}{x}$.

Integrating both sides: $\int \frac{dy}{y} = \int -\frac{dx}{x}$

$\ln|y| = -\ln|x| + \ln|C_2|$, where $C_2$ is a positive constant.

$\ln|y| = \ln|x^{-1}| + \ln|C_2|$

$\ln|y| = \ln|\frac{C_2}{x}|$

$|y| = |\frac{C_2}{x}|$

$y = \frac{C}{x}$, where $C = \pm C_2$ is a non-zero constant.

If $C=0$, then $y=0$, which means $f(x)=0$ for $x \neq 0$. If we define $f(0)=0$, then $f(0)=0$, contradicting $f(0) \neq 0$. So $C \neq 0$.

For $y = \frac{C}{x}$, $f(0)$ is undefined. However, the condition $f(0) \neq 0$ is given. This suggests that the curve is defined at $x=0$ and $f(0) \neq 0$. The solution $y = C/x$ doesn't satisfy this unless the domain excludes $x=0$. Let's re-examine the tangent meeting the x-axis at T. For $y=C/x$, the tangent at $P(x,y)$ meets the x-axis at $T(2x, 0)$. The condition $OP=PT$ gives $\sqrt{x^2+y^2} = \sqrt{(2x-x)^2+(0-y)^2} = \sqrt{x^2+y^2}$, which is satisfied for any point $(x,y)$ on the curve $y=C/x$.

Let's assume the form of the function is $f(x) = \frac{C}{x}$.

We are given $f(16) = 4$.

$f(16) = \frac{C}{16} = 4$.

$C = 16 \times 4 = 64$.

So the function is $f(x) = \frac{64}{x}$.

Now we need to interpret the condition $f(0) \neq 0$. If the domain of $f(x) = \frac{64}{x}$ includes $x=0$, then $f(0)$ is undefined, which cannot be "not equal to 0". It's possible there is a misunderstanding of the problem statement or the domain of the function.

However, if we strictly follow the derivation from the geometric condition, we get $y=Cx$ or $y=C/x$. The condition $f(0) \neq 0$ rules out $y=Cx$. Thus, $y=C/x$ is the only remaining possibility based on the geometric condition. The condition $f(0) \neq 0$ might imply that the constant $C$ is non-zero, which is already the case for $y=C/x$ to be a non-trivial curve satisfying the property.

Let's proceed with $f(x) = \frac{64}{x}$.

We need to find $\sqrt{f(2)} + \sqrt{f(4)}$.

$f(2) = \frac{64}{2} = 32$.

$f(4) = \frac{64}{4} = 16$.

$\sqrt{f(2)} = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.

$\sqrt{f(4)} = \sqrt{16} = 4$.

$\sqrt{f(2)} + \sqrt{f(4)} = 4\sqrt{2} + 4 = 4(1 + \sqrt{2})$.

Let's check if there's any alternative interpretation.

The derivation $x^2 = y^2 (dx/dy)^2$ implies $x (dy/dx) = \pm y$.

If $y > 0$ and $x > 0$, then $dy/dx = y/x$ or $dy/dx = -y/x$.

If $f(16)=4$, then at $x=16$, $y=4$. Since $16>0$ and $4>0$, the curve is in the first quadrant at $x=16$.

If the curve remains in the first quadrant for $x>0$, then $y>0$ for $x>0$.

If $y>0$ and $x>0$, then $dy/dx = y/x$ gives $y=Cx$ with $C>0$. $f(16)=4 \implies 16C=4 \implies C=1/4$. $f(x)=x/4$. $f(0)=0$, rejected.

If $y>0$ and $x>0$, then $dy/dx = -y/x$ gives $y=C/x$ with $C>0$. $f(16)=4 \implies C/16=4 \implies C=64$. $f(x)=64/x$. For $x>0$, $f(x)>0$. This is consistent.

The condition $f(0) \neq 0$ remains problematic for the function $f(x)=64/x$ if the domain is assumed to include 0. However, if we must choose between $y=Cx$ and $y=C/x$ based on the geometric property and the condition $f(16)=4$, the form $y=C/x$ is the only one that yields a non-zero constant C and satisfies $f(16)=4$. The condition $f(0) \neq 0$ then serves to exclude the form $y=Cx$. It is likely that the domain of the function is intended to be $x \neq 0$, or specifically $x > 0$ given $f(16)=4$. In this context, $f(0) \neq 0$ could be a loose way of saying that the function is not identically zero and not of the form $y=Cx$.

Assuming $f(x) = 64/x$ is the correct function for $x>0$.

We need $\sqrt{f(2)} + \sqrt{f(4)}$. Since $2>0$ and $4>0$, these values are well-defined.

$f(2) = 64/2 = 32$. $\sqrt{f(2)} = \sqrt{32} = 4\sqrt{2}$.

$f(4) = 64/4 = 16$. $\sqrt{f(4)} = \sqrt{16} = 4$.

Sum = $4\sqrt{2} + 4 = 4(1 + \sqrt{2})$.

The final answer is $4(1+\sqrt{2})$.

Explanation of the solution:

1. Find the equation of the tangent at point P(x, y) on the curve y = f(x).
2. Find the coordinates of the point T where the tangent meets the x-axis.
3. Use the given condition OP = PT to set up an equation involving x, y, and dy/dx.
4. Solve the resulting differential equation to find the form of the function y = f(x). The differential equation is $x^2 (dy/dx)^2 = y^2$, leading to $dy/dx = \pm y/x$.
5. Integrate $dy/dx = y/x$ to get $y = Cx$. Use the condition $f(0) \neq 0$ to rule out this solution.
6. Integrate $dy/dx = -y/x$ to get $y = C/x$. This form is consistent with the geometric property.
7. Use the condition $f(16) = 4$ to find the value of the constant C in $y = C/x$. We get $C=64$.
8. The function is $f(x) = 64/x$.
9. Calculate $f(2)$ and $f(4)$ using this function.
10. Calculate $\sqrt{f(2)} + \sqrt{f(4)}$.

The final answer is $4(1+\sqrt{2})$.