Accessing ML Aggarwal Class 12 ISC Solutions Chapter 6 Indeterminate Forms MCQs can be a valuable tool for students seeking extra practice.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms MCQs

Choose the correct answer from the given four options in questions (1 to 6) :

Question 1.
$$\ {Lt}_{x \rightarrow 0} \frac{\sin ^{-1} x}{x}$$ is equal to
(a) 0
(b) 1
(c) – 1
(d) does not exist
Solution:
(b) 1

$$\ {Lt}_{x \rightarrow 0} \frac{\sin ^{-1} x}{x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}}}{1}$$
[using L’Hopital’s rule]
= $$\frac{1}{\sqrt{1-0^2}}$$ = 1

Question 2.
$$\ {Lt}_{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x}$$ is eqaul to
(a) 1
(b) $$\frac{2}{3}$$
(c) $$\frac{3}{2}$$
(d) does not exist
Solution:
(c) $$\frac{3}{2}$$

$$\ {Lt}_{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{3 \sec ^2 3 x}{2 \cos 2 x}$$
[using L’Hopital’s rule]
= $$\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$$

Question 3.
$$\ {Lt}_{x \rightarrow 1} \frac{x^3+3 x-4}{2 x^2+x-3}$$ is equals to
(a) $$\frac{3}{2}$$
(b) $$\frac{6}{5}$$
(c) $$\frac{5}{6}$$
(d) none of these
Solution:
(b) $$\frac{6}{5}$$

$$\ {Lt}_{x \rightarrow 1} \frac{x^3+3 x-4}{2 x^2+x-3}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 1} \frac{3 x^2+3}{4 x+1}$$ [usingh L’Hopital’s ruel]
= $$\frac{3 \times 1^2+3}{4 \times 1+1}=\frac{6}{5}$$.

Question 4.
$$\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}$$, n > 1, is equal to
(a) none
(b) n (n – 1)
(c) $$\frac{n(n-1)}{2}$$
(d) does not exist
Solution:
(c) $$\frac{n(n-1)}{2}$$

$$\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}-n}{2 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{n(n-1)(1+x)^{n-2}}{2}$$
= $$\frac{n(n-1)}{2}$$ (1 + 0)n-2
= $$\frac{n(n-1)}{2}$$

Question 5.
$$\ {Lt}_{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\tan ^{-1} \frac{1}{x}}$$ is equal to
(a) 1
(b) 2
(c) $$\frac{1}{2}$$
(d) does not exist
Solution:
(a) 1

= 1 × (0 + 1) = 1.

Question 6.
$$\underset{x \rightarrow \infty}{\mathbf{L t}}$$ x sin-1 $$\frac{3}{x}$$ is equal to
(a) $$\frac{1}{3}$$
(b) 3
(c) 1
(d) does not exist
Solution:
(b) 3

$$\underset{x \rightarrow \infty}{\mathbf{L t}}$$ x sin-1 $$\frac{3}{x}$$ [0 . ∞ form]
= $$\ {Lt}_{x \rightarrow \infty} \frac{\sin ^{-1} \frac{3}{x}}{\frac{1}{x}}$$
= $$\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1-\left(\frac{3}{x}\right)^2}} \cdot\left(-\frac{3}{x^2}\right)}{-\frac{1}{x^2}}$$
= $$\ {Lt}_{x \rightarrow \infty} \frac{3 x}{\sqrt{x^2-9}}$$
= $$\ {Lt}_{x \rightarrow \infty} \frac{3 x}{x \sqrt{1-\frac{9}{x^2}}}$$
= $$\frac{3}{\sqrt{1-0}}$$ = 3.