Let y = f(x) be a thrice differentiable function in (-5, 5). Let the tangents to the curve y = f(x)

Q: Wait, do I need to find the function $f(x)$ itself to solve this?

Absolutely not! That is the beauty of this problem. We are given information about the derivative, and the integral is structured in a way that we only need the values of the derivative at the boundaries. We never need to know the explicit form of $f(x)$.

Q: Why do we change the limits of integration when we substitute $z = f'(t)$?

This is a fundamental rule of definite integrals. When you change the variable of integration, you are essentially changing the 'domain' of your calculation. If you don't update the limits to reflect the new variable $z$, your result will be incorrect. Always map your $t$-limits to $z$-limits!

Q: How do I handle the $\sqrt{3}$ in the denominator during the final calculation?

Rationalization is your best friend here. When you encounter a term like $\frac{10}{9\sqrt{3}}$, multiply the numerator and denominator by $\sqrt{3}$ to get $\frac{10\sqrt{3}}{27}$. This makes it much easier to combine terms and match the form $\alpha + \beta\sqrt{3}$.

Question

Let $y = f(x)$ be a thrice differentiable function in $(-5, 5)$. Let the tangents to the curve $y = f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively with positive $x$-axis. If $27 \int_{1}^{3} ((f'(t))^2 + 1) f''(t) dt = \alpha + \beta\sqrt{3}$ where $\alpha, \beta$ are integers, then the value of $\alpha + \beta$ equals

Options:

(A) -14

(B) 26

(C) -16

(D) 36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a smooth, continuous curve on a graph, defined by the function $y = f(x)$. The problem states this function is thrice differentiable, ensuring the curve is perfectly smooth with no sharp corners or breaks.

We focus on two specific points on this curve at $x = 1$ and $x = 3$. At these points, we draw tangents which serve as the keys to solving the problem.

The tangent at $x = 1$ makes an angle of $\frac{\pi}{6}$ with the positive $x$-axis, and the tangent at $x = 3$ makes an angle of $\frac{\pi}{4}$. Geometrically, the slope of any line is the tangent of the angle it makes with the $x$-axis, given by $	an(	heta)$.

In calculus, the slope of the tangent line to a curve at any point $x$ is exactly the first derivative, $f'(x)$. Therefore, we establish the bridge: $f'(x) = 	an(	heta)$.

At $x = 1$, we have:
$$f'(1) = 	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$$

At $x = 3$, we have:
$$f'(3) = 	an\left(\frac{\pi}{4}ight) = 1$$

### The Integral's Secret

Now, consider the integral that initially seems intimidating:
$$I = \int_{1}^{3} ((f'(t))^2 + 1) f''(t) dt$$

Look closely at the structure. We have a function of $f'(t)$ multiplied by $f''(t)$, which is a classic setup for the **substitution method**.

If we let $z = f'(t)$, then differentiating both sides with respect to $t$ gives us $dz = f''(t) dt$. This is perfect, as the $f''(t) dt$ term in our integral is exactly $dz$.

We must now update the limits of integration. When $t = 1$, $z = f'(1) = \frac{1}{\sqrt{3}}$. When $t = 3$, $z = f'(3) = 1$.

Our integral transforms into:
$$I = \int_{\frac{1}{\sqrt{3}}}^{1} (z^2 + 1) dz$$

### The Final Calculation

We integrate term by term: $\int z^2 dz = \frac{z^3}{3}$ and $\int 1 dz = z$. Thus, we evaluate:
$$I = \left[ \frac{z^3}{3} + z ight]_{\frac{1}{\sqrt{3}}}^{1}$$

Evaluating at the upper limit $z = 1$:
$$\frac{1^3}{3} + 1 = \frac{1}{3} + 1 = \frac{4}{3}$$

Evaluating at the lower limit $z = \frac{1}{\sqrt{3}}$:
$$\frac{(\frac{1}{\sqrt{3}})^3}{3} + \frac{1}{\sqrt{3}} = \frac{1}{9\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{1 + 9}{9\sqrt{3}} = \frac{10}{9\sqrt{3}}$$

Subtracting the lower limit from the upper limit, we obtain:
$$I = \frac{4}{3} - \frac{10}{9\sqrt{3}}$$

The problem asks for $27I$. Multiplying our result by $27$:
$$27I = 27\left(\frac{4}{3}ight) - 27\left(\frac{10}{9\sqrt{3}}ight) = 36 - \frac{30}{\sqrt{3}}$$

Rationalizing the second term, $\frac{30}{\sqrt{3}} = 10\sqrt{3}$. Thus:
$$27I = 36 - 10\sqrt{3}$$

Comparing this to $\alpha + \beta\sqrt{3}$, we find $\alpha = 36$ and $\beta = -10$. The final answer, $\alpha + \beta$, is:
**36 - 10 = 26**

Select Answer:

Visualized Solution

Visualizing the Curve

Tangent at $x = 1$

Tangent at $x = 3$

Relating Slope to Derivative

Calculating $f^{'} (1)$

Calculating $f^{'} (3)$

The Given Integral

Applying Substitution

Updating Integration Limits

The Transformed Integral

Performing the Integration

Evaluating at Upper Limit

Evaluating at Lower Limit

Final Value of Integral $I$

Finding $α, β$ and Final Answer

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Integral's Secret

The Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Curve

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Relating Slope to Derivative

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(1)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(3)