Question

Let $f$ be a differentiable function in $\left(0,\frac{\pi}{2}\right)$. If $\int_{\cos x}^{1} t^2 f(t) dt = \sin^3 x + \cos x - 1$, then $\left|f^{'}\left(\frac{1}{\sqrt{5}}\right)\right|$ is equal to

• A. $\frac{5}{2}$
• B. $\frac{5\sqrt{5}}{2}$
• C. $\frac{2}{5\sqrt{5}}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

Answer: (B)

$\frac{5\sqrt{5}}{2}$

Answer 2

The problem requires us to find the absolute value of the derivative of a function $f$ at a specific point, given an integral equation involving $f(t)$.

1. Differentiate the given integral equation using Leibniz's Rule.

The given equation is $\int_{\cos x}^{1} t^2 f(t) dt = \sin^3 x + \cos x - 1$.

Let $G(x) = \int_{\cos x}^{1} t^2 f(t) dt$. According to Leibniz's Rule, if $G(x) = \int_{a(x)}^{b(x)} h(t) dt$, then $G'(x) = h(b(x))b'(x) - h(a(x))a'(x)$.

Here, $h(t) = t^2 f(t)$, $a(x) = \cos x$, and $b(x) = 1$. So, $a'(x) = -\sin x$ and $b'(x) = 0$.

Differentiating the left side:

$$\frac{d}{dx} \left( \int_{\cos x}^{1} t^2 f(t) dt \right) = (1)^2 f(1) \cdot (0) - (\cos x)^2 f(\cos x) \cdot (-\sin x)$$ $$= 0 - (-\sin x \cos^2 x f(\cos x)) = \sin x \cos^2 x f(\cos x)$$

Differentiating the right side:

$$\frac{d}{dx} (\sin^3 x + \cos x - 1) = 3 \sin^2 x \cos x - \sin x$$

Equating the derivatives:

$$\sin x \cos^2 x f(\cos x) = 3 \sin^2 x \cos x - \sin x$$

2. Solve for $f(\cos x)$ and then find $f(t)$.

Since $x \in \left(0, \frac{\pi}{2}\right)$, $\sin x \neq 0$. We can divide both sides by $\sin x$:

$$\cos^2 x f(\cos x) = 3 \sin x \cos x - 1$$ $$f(\cos x) = \frac{3 \sin x \cos x - 1}{\cos^2 x}$$

Let $t = \cos x$. Since $x \in \left(0, \frac{\pi}{2}\right)$, $t \in (0, 1)$. Also, for $x \in \left(0, \frac{\pi}{2}\right)$, $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - t^2}$. Substitute $t$ and $\sqrt{1-t^2}$ into the expression for $f(\cos x)$:

$$f(t) = \frac{3t \sqrt{1 - t^2} - 1}{t^2}$$

This can be rewritten as:

$$f(t) = \frac{3t \sqrt{1 - t^2}}{t^2} - \frac{1}{t^2} = 3 \frac{\sqrt{1 - t^2}}{t} - \frac{1}{t^2}$$

3. Differentiate $f(t)$ to find $f'(t)$.

We differentiate $f(t)$ term by term: For the first term, $3 \frac{\sqrt{1 - t^2}}{t}$: Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, where $u = \sqrt{1 - t^2}$ and $v = t$. $u' = \frac{1}{2\sqrt{1 - t^2}}(-2t) = \frac{-t}{\sqrt{1 - t^2}}$ and $v' = 1$.

$$\frac{d}{dt} \left( \frac{\sqrt{1 - t^2}}{t} \right) = \frac{\left(\frac{-t}{\sqrt{1 - t^2}}\right) \cdot t - \sqrt{1 - t^2} \cdot 1}{t^2}$$ $$= \frac{\frac{-t^2}{\sqrt{1 - t^2}} - \frac{1 - t^2}{\sqrt{1 - t^2}}}{t^2} = \frac{\frac{-t^2 - (1 - t^2)}{\sqrt{1 - t^2}}}{t^2} = \frac{-1}{t^2 \sqrt{1 - t^2}}$$

So, the derivative of the first term is $3 \left( \frac{-1}{t^2 \sqrt{1 - t^2}} \right) = -\frac{3}{t^2 \sqrt{1 - t^2}}$.

For the second term, $-\frac{1}{t^2} = -t^{-2}$:

$$\frac{d}{dt} (-t^{-2}) = -(-2)t^{-3} = \frac{2}{t^3}$$

Combining these, we get $f'(t)$:

$$f'(t) = -\frac{3}{t^2 \sqrt{1 - t^2}} + \frac{2}{t^3}$$

4. Substitute $t = \frac{1}{\sqrt{5}}$ into $f'(t)$ and calculate the result.

We need to evaluate $f'\left(\frac{1}{\sqrt{5}}\right)$. First, calculate the necessary values for $t = \frac{1}{\sqrt{5}}$: $t^2 = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5}$. $t^3 = \left(\frac{1}{\sqrt{5}}\right)^3 = \frac{1}{5\sqrt{5}}$. $\sqrt{1 - t^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.

Now substitute these values into $f'(t)$:

$$f'\left(\frac{1}{\sqrt{5}}\right) = -\frac{3}{\left(\frac{1}{5}\right) \left(\frac{2}{\sqrt{5}}\right)} + \frac{2}{\left(\frac{1}{5\sqrt{5}}\right)}$$ $$= -\frac{3}{\frac{2}{5\sqrt{5}}} + 2 \cdot 5\sqrt{5}$$ $$= -3 \cdot \frac{5\sqrt{5}}{2} + 10\sqrt{5}$$ $$= -\frac{15\sqrt{5}}{2} + \frac{20\sqrt{5}}{2}$$ $$= \frac{5\sqrt{5}}{2}$$

5. Find the absolute value.

The question asks for $\left|f^{'}\left(\frac{1}{\sqrt{5}}\right)\right|$.

$$\left|f^{'}\left(\frac{1}{\sqrt{5}}\right)\right| = \left| \frac{5\sqrt{5}}{2} \right| = \frac{5\sqrt{5}}{2}$$

The final answer is $\boxed{\frac{5\sqrt{5}}{2}}$.