Question

If $\int \left(3t^2\sin\left(\frac{1}{t}\right) - t\cos\left(\frac{1}{t}\right)\right) dt = f(t)\sin\left(\frac{1}{t}\right) + c$ then $f(2)$ is equal to

• A. $-2$
• B. $2$
• C. $4$
• D. $8$
• E. $16$

Hint:

When an integral result is given in product form: - Differentiate RHS using product rule - Match coefficients with the integrand to find unknown functions

Answer 1

The Correct Option is D

Concept: We use the idea of reverse differentiation. If \[ \int g(t)\,dt = F(t) + c \] then \[ g(t) = \frac{d}{dt}[F(t)] \] Here, we assume the result is of the form: \[ F(t) = f(t)\sin\left(\frac{1}{t}\right) \] and differentiate using product rule.

Step 1: Differentiate $f(t)\sin\left(\frac{1}{t}\right)$ using product rule.
\[ \frac{d}{dt}\left[f(t)\sin\left(\frac{1}{t}\right)\right] = f'(t)\sin\left(\frac{1}{t}\right) + f(t)\cos\left(\frac{1}{t}\right)\cdot \frac{d}{dt}\left(\frac{1}{t}\right) \] \[ = f'(t)\sin\left(\frac{1}{t}\right) - \frac{f(t)}{t^2}\cos\left(\frac{1}{t}\right) \]

Step 2: Compare with the given integrand.
\[ 3t^2\sin\left(\frac{1}{t}\right) - t\cos\left(\frac{1}{t}\right) \] Comparing coefficients: \[ f'(t) = 3t^2 \] \[ \frac{f(t)}{t^2} = t \]

Step 3: Solve for $f(t)$.
From $\frac{f(t)}{t^2} = t$: \[ f(t) = t^3 \]

Step 4: Evaluate $f(2)$.
\[ f(2) = 2^3 = 8 \]