Class 12 Pre-Board: Three Short Calculus Questions

This lesson works through three short-answer questions from a Class 12 pre-board model paper: an inverse trigonometry value, a differentiation proof, and an integration using a standard exponential rule.

Evaluating $\tan\!\left(\tan^{-1}(-1) + \tfrac{\pi}{3}\right)$

We use the identity $\tan^{-1}(-x) = -\tan^{-1}(x)$, so $\tan^{-1}(-1) = -\tan^{-1}(1) = -\tfrac{\pi}{4}$.

The expression becomes:

$\tan\!\left(-\tfrac{\pi}{4} + \tfrac{\pi}{3}\right) = \tan\!\left(\tfrac{-3\pi + 4\pi}{12}\right) = \tan\tfrac{\pi}{12}.$

Since $\tfrac{\pi}{12} = 15^\circ$, the value is:

$\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}.$

Check with the difference formula

We can confirm the same answer using $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = \tfrac{\pi}{3}$ and $B = \tfrac{\pi}{4}$:

$\tan\!\left(\tfrac{\pi}{3} - \tfrac{\pi}{4}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3}\cdot 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1},$

which matches $\tan\tfrac{\pi}{12}$.

Proving $\tfrac{dy}{dx} - \sec x = 0$ for $y = \log\tan\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right)$

Differentiate using the chain rule, with $\tfrac{d}{dx}\log u = \tfrac{1}{u}\,u'$ and $\tfrac{d}{dx}\tan u = \sec^2 u\,\,u'$:

$\frac{dy}{dx} = \frac{1}{\tan\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right)} \cdot \sec^2\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right) \cdot \frac{1}{2}.$

Writing $\tan = \tfrac{\sin}{\cos}$ and $\sec^2 = \tfrac{1}{\cos^2}$, the cosine terms simplify:

$\frac{dy}{dx} = \frac{1}{2}\cdot\frac{1}{\sin\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right)\cos\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right)}.$

Using $2\sin\theta\cos\theta = \sin 2\theta$ with $\theta = \tfrac{\pi}{4} + \tfrac{x}{2}$:

$\frac{dy}{dx} = \frac{1}{\sin\!\left(2\theta\right)} = \frac{1}{\sin\!\left(\tfrac{\pi}{2} + x\right)}.$

Since $\sin\!\left(\tfrac{\pi}{2} + x\right) = \cos x$:

$\frac{dy}{dx} = \frac{1}{\cos x} = \sec x.$

Therefore $\tfrac{dy}{dx} - \sec x = 0$, as required.

Integrating $\displaystyle\int \frac{x - 3}{(x - 1)^3}\,e^x\,dx$

We use the standard result $\displaystyle\int \big(f(x) + f'(x)\big)e^x\,dx = f(x)\,e^x + C$.

Splitting the fraction

Write the numerator $x - 3$ as $(x - 1) + 1 - 3$, so:

$\frac{x - 3}{(x - 1)^3} = \frac{x - 1}{(x - 1)^3} + \frac{1 - 3}{(x - 1)^3} = \frac{1}{(x - 1)^2} - \frac{2}{(x - 1)^3}.$

Matching to $f + f'$

Take $f(x) = \dfrac{1}{(x - 1)^2}$. Then its derivative is:

$f'(x) = -\frac{2}{(x - 1)^3},$

which is exactly the second term. So the integrand is $\big(f(x) + f'(x)\big)e^x$, and:

$\int \frac{x - 3}{(x - 1)^3}\,e^x\,dx = \frac{e^x}{(x - 1)^2} + C.$

Key takeaways

• $\tan^{-1}(-x) = -\tan^{-1}(x)$, and $\tan\tfrac{\pi}{12} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$.
• For $y = \log\tan\!\left(\tfrac{\pi}{4} + \tfrac{x}{2}\right)$, the chain rule plus the double-angle identity give $\tfrac{dy}{dx} = \sec x$.
• When an integrand has the form $\big(f(x) + f'(x)\big)e^x$, the integral is simply $f(x)\,e^x + C$.