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.