12th Maths Public Exam (July 2021)
Detailed Solutions for Question 25 and Question 42(b)
Question 25: Limits using L'Hôpital's Rule
First, we substitute \(x = \infty\) into the expression:
$$ \lim_{x\rightarrow\infty} e^x = \infty $$ $$ \lim_{x\rightarrow\infty} x^m = \infty $$This results in the indeterminate form \( \frac{\infty}{\infty} \). Therefore, we can apply L'Hôpital's Rule.
We differentiate the numerator and the denominator with respect to \(x\).
First Differentiation:
$$ \lim_{x\rightarrow\infty}\frac{\frac{d}{dx}(e^{x})}{\frac{d}{dx}(x^{m})} = \lim_{x\rightarrow\infty}\frac{e^{x}}{m x^{m-1}} $$This is still \( \frac{\infty}{\infty} \). We differentiate again.
Second Differentiation:
$$ \lim_{x\rightarrow\infty}\frac{e^{x}}{m(m-1) x^{m-2}} $$Continuing this process \(m\) times:
Since \(m\) is a finite integer, after differentiating \(m\) times, the term \(x^m\) in the denominator will become a constant (factorial).
$$ \lim_{x\rightarrow\infty}\frac{e^{x}}{m(m-1)(m-2)...1} $$The denominator becomes \(m!\) (a constant), while the numerator remains \(e^x\).
$$ = \lim_{x\rightarrow\infty}\frac{e^{x}}{m!} $$Since \(m!\) is a constant and \(e^\infty \rightarrow \infty\):
$$ = \frac{\infty}{m!} = \infty $$Question 42(b): Definite Integrals
Multiply the numerator and denominator inside the square root by \((1-x)\) to remove the root from the numerator.
$$ I = \int_{0}^{1}\sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}}dx $$ $$ I = \int_{0}^{1}\frac{1-x}{\sqrt{1-x^2}}dx $$Separate the integral into two parts:
$$ I = \int_{0}^{1}\frac{1}{\sqrt{1-x^2}}dx - \int_{0}^{1}\frac{x}{\sqrt{1-x^2}}dx $$Let's call them \(I_1\) and \(I_2\).
We know the standard formula: \(\int \frac{1}{\sqrt{1-x^2}}dx = \sin^{-1}(x)\)
$$ I_1 = \left[ \sin^{-1}(x) \right]_{0}^{1} $$ $$ I_1 = \sin^{-1}(1) - \sin^{-1}(0) $$ $$ I_1 = \frac{\pi}{2} - 0 = \frac{\pi}{2} $$Use Substitution Method:
- Put \( t = 1-x^2 \)
- Differentiation: \( dt = -2x dx \Rightarrow x dx = -\frac{dt}{2} \)
- Limits change: When \(x=0, t=1\). When \(x=1, t=0\).
However, we can also solve it directly using the form \(\int \frac{f'(x)}{\sqrt{f(x)}}dx = 2\sqrt{f(x)}\):
Rewrite the integral to match the derivative of \((1-x^2)\), which is \(-2x\):
$$ \int \frac{-x}{\sqrt{1-x^2}}dx = \frac{1}{2} \int \frac{-2x}{\sqrt{1-x^2}}dx $$ $$ = \frac{1}{2} \left[ 2\sqrt{1-x^2} \right]_0^1 $$ $$ = \left[ \sqrt{1-x^2} \right]_0^1 $$ $$ = \sqrt{1-1} - \sqrt{1-0} $$ $$ = 0 - 1 = -1 $$Substitute the values of \(I_1\) and the second part back into the main equation:
$$ I = I_1 + \int_{0}^{1}\frac{-x}{\sqrt{1-x^2}}dx $$ $$ I = \frac{\pi}{2} - 1 $$