Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 13 - Limits And Derivatives

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Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 13 Limits And Derivatives are provided here with simple step-by-step explanations. These solutions for Limits And Derivatives are extremely popular among class 11 Science students for Maths Limits And Derivatives Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 11 Science Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 239:

Question 1:

Evaluate :

\lim_{x \to 3} \frac{x^{2} - 9}{x - 3}

Answer:

$\begin{array}{rcl} \lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & = & \lim_{x \to 3} \frac{x^{2} - 3^{2}}{x - 3} \\ = & \lim_{x \to 3} \frac{(x - 3) (x + 3)}{(x - 3)} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to 3} (x + 3) \\ = & 3 + 3 \\ = & 6 \end{array}$

Page No 239:

Question 2:

Evaluate :

\lim_{x \to \frac{1}{2}} \frac{4 x^{2} - 1}{2 x - 1}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{1}{2}} \frac{4 x^{2} - 1}{2 x - 1} & = & \lim_{x \to \frac{1}{2}} \frac{{(2 x)}^{2} - 1^{2}}{2 x - 1} \\ = & \lim_{x \to \frac{1}{2}} \frac{(2 x - 1) (2 x - 1)}{(2 x - 1)} \\ = & \lim_{x \to \frac{1}{2}} (2 x - 1) \\ = & 2 \times \frac{1}{2} + 1 \\ = & 1 + 1 \\ = & 2 \end{array}$

Page No 239:

Question 3:

Evaluate :

\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}

Answer:

Use the formula $\lim_{h \to 0} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1} .$
$∵ h \overset{}{\to} 0 \Rightarrow x + h \overset{}{\to} x Also, h = x + h - x$
$\begin{array}{rcl} ∴ \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} & = & \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{x + h - h} \\ = & \lim_{h \to 0} \frac{{(x + h)}^{\frac{1}{2}} - x^{\frac{1}{2}}}{(x + h) - h} \\ = & \frac{1}{2} x^{\frac{1}{2} - 1} [∴ \lim_{h \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ = & \frac{1}{2} x^{- \frac{1}{2}} \\ = & \frac{1}{2 \sqrt{x}} \end{array}$

Page No 239:

Question 4:

Evaluate :
$\lim_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x} & = & \lim_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{(x + 2) - 2} \\ = & \frac{1}{3} \times 2^{\frac{1}{3} - 1} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ = & \frac{1}{3} \times 2^{\frac{1 - 3}{3}} \\ = & \frac{1}{3} \times 2^{\frac{- 2}{3}} \\ = & \frac{1}{3 {(2)}^{\frac{2}{3}}} \end{array}$

Page No 239:

Question 5:

Evaluate :

\lim_{x \to 1} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}

Answer:

Divide the numerator and denominator by x.
$\begin{array}{rcl} \lim_{x \to 0} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1} & = & \lim_{x \to 0} \frac{\frac{{(1 + x)}^{6} - 1}{x}}{\frac{{(1 + x)}^{2} - 1}{x}} \\ = & \lim_{x \to 0} \frac{\frac{{(1 + x)}^{6} - 1}{(1 + x) - 1}}{\frac{{(1 + x)}^{2} - 1}{(1 + x) - 1}} \\ = & \lim_{x \to 0} \frac{\frac{{(1 + x)}^{6} - 1^{6}}{(1 + x) - 1}}{\frac{{(1 + x)}^{2} - 1^{2}}{(1 + x) - 1}} [∵ \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \\ = & \frac{6 {(1)}^{6 - 1}}{2 {(1)}^{2 - 1}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{a - a} = n a^{n - 1}] \\ = & \frac{6 \times 1}{2 \times 1} \\ = & \frac{6}{2} \\ = & 3 \end{array}$

Page No 239:

Question 6:

Evaluate :

\lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{x - a}

Answer:

$\begin{array}{rcl} \lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{x - a} & = & \lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{x - a + 2 - 2} \\ = & \lim_{x \to a} \frac{{(2 + x)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(2 + x) - (a + 2)} \\ = & \frac{5}{2} {(a + 2)}^{\frac{5 - 1}{2}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1} and (2 + x) \to (a + 2) \sin c e x \overset{}{\to a}] \\ = & \frac{5}{2} {(a + 2)}^{\frac{5 - 2}{2}} \\ = & \frac{5}{2} {(a + 2)}^{\frac{3}{2}} \end{array}$

Page No 240:

Question 7:

Evaluate :

\lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1}

Answer:

$\lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1} = \lim_{x \to 1} \frac{{[{(\sqrt{x})}^{2}]}^{4} - \sqrt{x}}{\sqrt{x} - 1} = \lim_{x \to 1} \frac{{(\sqrt{x})}^{8} - \sqrt{x}}{\sqrt{x} - 1} = \lim_{x \to 1} \frac{\sqrt{x} [{(\sqrt{x})}^{7} - 1]}{\sqrt{x} - 1} = \lim_{x \to 1} \frac{\sqrt{x} (x^{\frac{7}{2}} - 1)}{x^{\frac{1}{2}} - 1} = \lim_{x \to 1} \frac{x^{\frac{7}{2}} - 1}{x^{\frac{1}{2}} - 1} \times \lim_{x \to 1} \sqrt{x} [∵ \lim_{x \to a} f (x) \times g (x) = \lim_{x \to a} f (x) \times \lim_{x \to a} g (x)] = \lim_{x \to 1} \frac{x^{\frac{7}{2}} - 1}{x^{\frac{1}{2}} - 1} \times 1 = \lim_{x \to 1} \frac{\frac{x^{\frac{7}{2}} - 1^{\frac{7}{2}}}{x - 1}}{\frac{x^{\frac{1}{2}} - 1^{\frac{1}{2}}}{x - 1}} = \frac{\lim_{x \to 1} \frac{x^{\frac{7}{2}} - 1^{\frac{7}{2}}}{x - 1}}{\lim_{x \to 1} \frac{x^{\frac{1}{2}} - 1^{\frac{1}{2}}}{x - 1}} [∵ \lim_{x \to 1} \frac{f (x)}{g (x)} = \frac{\lim_{x \to 1} f (x)}{\lim_{x \to 1} g (x)}] = \frac{\frac{7}{2} \times {(1)}^{\frac{7}{2} - 1}}{\frac{1}{2} \times {(1)}^{\frac{1}{2} - 1}} = \frac{\frac{7}{2}}{\frac{1}{2}} = 7$

Page No 240:

Question 8:

Evaluate :

\lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}}

Answer:

Multiply the numerator and denominator by $(\sqrt{3 x} - 2 + \sqrt{x} - 2) .$
$\begin{array}{rcl} \lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}} & = & \lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}} \times \frac{\sqrt{3 x - 2} - \sqrt{x + 2}}{\sqrt{3 x - 2} - \sqrt{x + 2}} \\ = & \lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{{(\sqrt{3 x - 2})}^{2} - {(\sqrt{x + 2})}^{2}} [∵ (a - b) (a + b) = a^{2} - b^{2}] \\ = & \lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{(3 x - 2) - (x + 2)} \\ = & \lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{3 x - 2 - x + 2} \\ = & \lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2 x - 4} \\ = & \lim_{x \to 2} \frac{(x - 2) (x + 2) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2 (x - 2)} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to 2} \frac{(x + 2) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2} \\ = & \frac{(x + 2) (\sqrt{6 - 2} + \sqrt{2 + 2})}{2} \\ = & \frac{4 (\sqrt{4} + \sqrt{4})}{2} \\ = & 2 \times (2 + 2) \\ = & 8 \end{array}$

Page No 240:

Question 9:

Evaluate :

\lim_{x \to \sqrt{2}} \frac{x^{4} - 4}{x^{2} + 3 \sqrt{2 x} - 8}

Answer:

$\begin{array}{rcl} \lim_{x \to \sqrt{2}} \frac{x^{4} - 4}{x^{2} + 3 \sqrt{2 x} - 8} & = & \lim_{x \to \sqrt{2}} \frac{{(x^{2})}^{2} - {(2)}^{2}}{x^{2} + 4 \sqrt{2} x - \sqrt{2} x - 8} \\ = & \lim_{x \to \sqrt{2}} \frac{(x^{2} - 2) - (x^{2} + 2)}{x (x + 4 \sqrt{2}) - \sqrt{2} (x + 4 \sqrt{2})} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to \sqrt{2}} \frac{[x^{2} - {(\sqrt{2})}^{2}] - (x^{2} + 2)}{(x - \sqrt{2}) - (x + 4 \sqrt{2})} \\ = & \lim_{x \to \sqrt{2}} \frac{(x - \sqrt{2}) - (x + \sqrt{2}) (x^{2} + 2)}{(x - \sqrt{2}) - (x + 4 \sqrt{2})} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2}) - (x^{2} + 2)}{(x + 4 \sqrt{2})} \\ = & \frac{(\sqrt{2} + \sqrt{2}) - (x + 2)}{(\sqrt{2} + 4 \sqrt{2})} \\ = & \frac{2 \sqrt{2} \times 4}{5 \sqrt{2}} \\ = & \frac{8}{5} \end{array}$

Page No 240:

Question 10:

Evaluate :

\lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}

Answer:

$\begin{array}{rcl} \lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2} & = & \lim_{x \to 1} \frac{x^{7} - x^{5} - x^{5} + 1}{x^{3} - x^{2} - 2 x^{2} + 2} \\ = & \lim_{x \to 1} \frac{x^{5} (x^{2} - 1) - 1 (x^{5} - 1)}{x^{2} (x - 1) - 2 (x^{2} - 1)} \end{array}$
Divide the numerator and denominator by (x – 1).
$\begin{array}{rcl} \lim_{x \to 1} \frac{\frac{x^{5} (x^{2} - 1)}{x - 1} - \frac{(x^{5} - 1)}{x - 1}}{\frac{x^{2} (x - 1)}{x - 1} - \frac{2 (x^{2} - 1)}{x - 1}} & = & \lim_{x \to 1} \frac{\frac{x^{5} (x^{2} - 1) (x + 1)}{(x - 1)} - \frac{x^{5} - 1^{2}}{x - 1}}{\frac{x^{2} - 2 (x - 1) (x + 1)}{(x - 1)}} \\ = & \lim_{x \to 1} \frac{x^{5} (x + 1) - (\frac{x^{5} - 15}{x - 1})}{x^{2} - 2 (x + 1)} \\ = & \frac{\lim_{x \to 1} x^{5} (x + 1) - \lim_{x \to 1} \frac{(x^{5} - 1^{5})}{x - 1}}{\lim_{x \to 1} (x^{2} - 2 x - 2)} \\ = & \frac{2 - \lim_{x \to 1} \frac{x^{5} - 1^{5}}{x - 1}}{- 3} \\ = & \frac{2 - 5 {(1)}^{5 - 1}}{- 3} [∵ \lim_{x \to 1} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ = & \frac{- 3}{- 3} \\ = & 1 \end{array}$

Page No 240:

Question 11:

Evaluate :

\lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}}

Answer:

Multiplying the numerator and denominator by $(\sqrt{1 + x^{3}} - \sqrt{1 + x^{3}}) .$
$\begin{array}{rcl} \lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}} & = & \lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}} \times \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{\sqrt{1 + x^{3}} + \sqrt{1 + x^{3}}} \\ = & \lim_{x \to 0} \frac{{(\sqrt{1 + x^{3}})}^{2} - {(\sqrt{1 - x^{3}})}^{2}}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 + x^{3}})} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to 0} \frac{1 + x^{3} - 1 + x^{3}}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})} \\ = & \lim_{x \to 0} \frac{2 x^{3}}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})} \\ = & \lim_{x \to 0} \frac{2 x^{3}}{(\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})} \\ = & 0 \end{array}$

Page No 240:

Question 12:

Evaluate :

\lim_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243}

Answer:

Divide the numerator and denominator by (x + 3).
$\begin{array}{rcl} \lim_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243} & = & \lim_{x \to - 3} \frac{\frac{(x^{3} + 27)}{x + 3}}{\frac{(x^{5} + 243)}{x + 3}} \\ = & \frac{\lim_{x \to - 3} (\frac{x^{3} + 27}{x + 3})}{\lim_{x \to - 3} (\frac{x^{5} + 243}{x + 3})} [∵ \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \\ = & \frac{\lim_{x \to - 3} \frac{x^{3} + {(- 3)}^{3}}{x - (3)}}{\lim_{x \to - 3} \frac{x^{5} - {(- 3)}^{5}}{x - (- 3)}} \\ = & \frac{3 {(- 3)}^{3 - 1}}{5 {(- 3)}^{5 - 1}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ = & \frac{3 \times {(- 3)}^{2}}{5 \times {(- 3)}^{4}} \\ = & \frac{1}{15} \end{array}$

Page No 240:

Question 13:

Evaluate :

\lim_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1} & = & \lim_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{{(2 x)}^{2} - 1^{2}} \\ = & \lim_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{(2 x - 1) - (2 x - 1)} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to \frac{1}{2}} \frac{(8 x - 3) (2 x - 1) - (4 x^{2} - 1)}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{16 x^{2} + 8 x - 6 x - 3 - 4 x^{2} - 1}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{12 x^{2} + 2 x - 4}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{2 (6 x^{2} + x - 2)}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{2 (6 x^{2} + 4 x - 3 x - 2)}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{2 [2 x (3 x + 2) - 1 (3 x + 2)]}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{2 (2 x - 1) (3 x + 2)}{(2 x - 1) (2 x + 1)} \\ = & \lim_{x \to \frac{1}{2}} \frac{2 (3 x + 2)}{(2 x - 1)} \\ = & \frac{2 (3 \times \frac{1}{2} + 2)}{(2 \times \frac{1}{2} + 1)} \\ = & \frac{2 (\frac{3 + 4}{2})}{1 + 1} \\ = & \frac{7}{2} \end{array}$

Page No 240:

Question 14:

Evaluate :
Find 'n', if $\lim_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80, n \in N$

Answer:

$\lim_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80, n \in N$
$∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1} ∵ n {(2)}^{n - 1} = 80 \Rightarrow n \times 2^{n - 1} = 5 \times 16 \Rightarrow n \times 2^{n - 1} = 5 \times 2^{4} \Rightarrow n \times 2^{n - 1} = 5 \times 2^{5 - 1} \Rightarrow n = 5$

Page No 240:

Question 15:

Evaluate : $\lim_{x \to a} \frac{\sin 3 x}{\sin 7 x}$

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\sin 3 x}{\sin 7 x} & = & \frac{\lim_{x \to 0} \sin 3 x}{\lim_{x \to 0} \sin 7 x} [∵ \lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{\lim_{x \to 0} f (x)}{\lim_{x \to 0} g (x)}] \\ = & \frac{\lim_{x \to 0} \frac{\sin 3 x}{3 x} \times 3 x}{\lim_{x \to 0} \frac{\sin 7 x}{7 x} \times 7 x} \\ = & \frac{\lim_{x \to 0} \frac{\sin 3 x}{3 x} \cdot \lim_{x \to 0} 3 x}{\lim_{x \to 0} \frac{\sin 7 x}{7 x} \cdot \lim_{x \to 0} 7 x} \\ = & \frac{1 \times \lim_{x \to 0} 3 x}{1 \times \lim_{x \to 0} 7 x} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & \frac{\lim_{x \to 0} 3 x}{\lim_{x \to 0} 7 x} \\ = & \lim_{x \to 0} \frac{3 x}{7 x} \\ = & \lim_{x \to 0} \frac{3}{7} \\ = & \frac{3}{7} \end{array}$

Disclaimer: In the question, the limit is x → 0; not x → a.

Page No 240:

Question 16:

Evaluate :

\lim_{x \to 0} \frac{\sin^{2} 2 x}{\sin^{2} 4 x}

Answer:

$\lim_{x \to 0} \frac{\sin^{2} 2 x}{\sin^{2} 4 x} = \lim_{x \to 0} \frac{\sin^{2} 2 x}{\sin^{2} 2 (2 x)} = \lim_{x \to 0} \frac{\sin^{2} 2 x}{{(2 \sin 2 x \cos 2 x)}^{2}} [∵ \sin 2 x = 2 \sin x \cos x] = \lim_{x \to 0} \frac{\sin^{2} 2 x}{4 \sin^{2} 2 x \cos^{2} 2 x} = \lim_{x \to 0} \frac{1}{4 \cos^{2} 2 x} = \frac{1}{4 \times 1^{2}} [∵ \cos 0 = 1] = \frac{1}{4}$

Page No 240:

Question 17:

Evaluate :

\lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}} & = & \lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} x}{x^{2}} [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ = & \lim_{x \to 0} \frac{2 \sin^{2} x}{x^{2}} \\ = & 2 \lim_{x \to 0} {(\frac{\sin^{2} x}{x^{2}})}^{2} \\ = & 2 \times 1^{2} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & 2 \end{array}$

Page No 240:

Question 18:

Evaluate :

\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} & = & \lim_{x \to 0} \frac{2 \sin x - 2 \sin x \cos x}{x^{3}} [∵ \sin 2 x = 2 \sin x \cos x] \\ = & \lim_{x \to 0} \frac{2 \sin x - 2 \sin x (1 - \cos x)}{x^{3}} \\ = & \lim_{x \to 0} \frac{2 \sin x}{x} \times \lim_{x \to 0} \frac{1 - \cos x}{x^{3}} [∵ \lim_{x \to a} f (x) g (x) = \lim_{x \to a} g (x)] \\ = & 2 \times 1 \times \lim_{x \to 0} \frac{1 - \cos x}{x^{2}} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & 2 \lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} \frac{x}{2}}{x^{2}} [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ = & 2 \lim_{x \to 0} \frac{2 \sin^{2} \frac{x}{2}}{x^{2}} \\ = & 4 \lim_{x \to 0} \frac{{(\sin \frac{x}{2})}^{2}}{\frac{x^{2}}{4} \times 4} \\ = & \frac{4}{4} \lim_{x \to 0} {(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} \\ = & 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Page No 240:

Question 19:

Evaluate :

\lim_{x \to 0} \frac{1 - \cos m x}{1 - \cos n x}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{1 - \cos m x}{1 - \cos n x} & = & \lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} \frac{m x}{2}}{1 - 1 + 2 \sin^{2} \frac{n x}{2}} [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ = & \lim_{x \to 0} \frac{\sin^{2} \frac{m x}{2}}{\sin^{2} \frac{n x}{2}} \\ = & \lim_{x \to 0} \frac{\frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}} \times {(\frac{m x}{2})}^{2}}{\frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}} \times {(\frac{n x}{2})}^{2}} \\ = & \lim_{x \to 0} \frac{\frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}} \times {(m)}^{2}}{\frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}} \times n^{2}} \\ = & \frac{m^{2}}{n^{2}} \lim_{x \to 0} \frac{\frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}}}{\frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}}} \\ = & \frac{m^{2}}{n^{2}} \frac{\lim_{x \to 0} \frac{\sin^{2} \frac{m x}{2}}{{(\frac{m x}{2})}^{2}}}{\lim_{x \to 0} \frac{\sin^{2} \frac{n x}{2}}{{(\frac{n x}{2})}^{2}}} [∵ \lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \\ = & \frac{m^{2}}{n^{2}} \times \frac{1}{1} \\ = & \frac{m^{2}}{n^{2}} \end{array}$

Page No 240:

Question 20:

Evaluate :

\lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} \frac{π}{3} - x}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)} & = & \lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - 1 + 2 \sin^{2} 3 x}}{\sqrt{2} (\frac{π}{3} - x)} [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ = & \lim_{x \to \frac{π}{3}} \frac{\sqrt{2} \sin 3 x}{\sqrt{2} (\frac{π}{3} - x)} \\ = & \lim_{x \to \frac{π}{3}} \frac{\sin (π - 3 x)}{(\frac{π}{3} - x)} [∵ \sin (π - x) = \sin x] \\ = & \lim_{x \to \frac{π}{3}} \frac{3 \sin (π - 3 x)}{(π - 3 x)} \\ = & 3 \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1 and ∵ x \to \frac{π}{3} so π - 3 x \to 0] \end{array}$

Disclaimer:- In the question, the denominator is $\sqrt{2} (\frac{π}{3} - x)$ and not $\sqrt{2} \frac{π}{3} - x .$

Page No 240:

Question 21:

Evaluate :

\lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{x - \frac{π}{4}}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{x - \frac{π}{4}} & = & \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin x \times \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \times \cos x)}{x - \frac{π}{4}} \\ = & \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin x \cos \frac{π}{4} - \sin \frac{π}{4} \cos x)}{x - \frac{π}{4}} \\ = & \sqrt{2} \lim_{x \to \frac{π}{4}} \frac{\sin (x - \frac{π}{4})}{(x - \frac{π}{4})} [∵ \sin (A - B) = \sin A \cos B - \sin B \cos A] \\ = & \sqrt{2} \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & \sqrt{2} \end{array}$

Page No 240:

Question 22:

Evaluate :

\lim_{x \to \frac{π}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{π}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}} & = & \lim_{x \to \frac{π}{6}} \frac{2 (\frac{\sqrt{3}}{2} \times \sin x - \frac{1}{2} \cos x)}{x - \frac{π}{6}} \\ = & \lim_{x \to \frac{π}{6}} \frac{2 (\sin x \times \cos \frac{π}{6} - \sin \frac{π}{6} \cos x)}{x - \frac{π}{6}} \\ = & \lim_{x \to \frac{π}{6}} \frac{2 \sin (x - \frac{π}{6})}{x - \frac{π}{6}} [∵ \sin (A - B) = \sin A \cos B - \sin B \cos A] \\ = & 2 \times 1 \\ = & 2 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Page No 240:

Question 23:

Evaluate :

\lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x} & = & \lim_{x \to 0} \frac{\frac{\sin 2 x + 3 x}{2 x} \times 2 x}{\frac{2 x + \tan 3 x}{3 x} \times 3 x} \\ = & \lim_{x \to 0} \frac{(\frac{\sin 2 x}{2 x} + \frac{3}{2})}{(\frac{2}{3} + \frac{\tan 3 x}{3 x})} \times \frac{2}{3} \\ = & \frac{2}{3} \frac{\lim_{x \to 0} (\frac{\sin 2 x}{2 x} + \frac{3}{2})}{\lim_{x \to 0} (\frac{2}{3} + \frac{\tan 3 x}{3 x})} \\ = & \frac{2}{3} \frac{(1 + \frac{2}{3})}{(\frac{2}{3} + 1)} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1 and \lim_{x \to 0} \frac{\tan x}{x} = 1] \\ = & \frac{2}{3} \frac{(\frac{2 + 3}{2})}{(\frac{2 + 3}{3})} \\ = & \frac{2}{3} \frac{\frac{5}{2}}{\frac{5}{3}} \\ = & \frac{2}{3} \times \frac{3}{2} \\ = & 1 \end{array}$

Page No 240:

Question 24:

Evaluate :

\lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}

Answer:

$\begin{array}{rcl} \lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} & = & \lim_{x \to a} \frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{\sqrt{x} - \sqrt{a}} [∵ \sin C - \sin D = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2}] \\ = & \lim_{x \to a} \frac{2 \cos (\frac{x - a}{2}) \sin (\frac{x - a}{2})}{(\sqrt{x} - \sqrt{a}) \times (\sqrt{x} + \sqrt{a})} \times (\sqrt{x} + \sqrt{a}) \\ = & \lim_{x \to a} \frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{x - a} \times (\sqrt{x} + \sqrt{a}) [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = & \lim_{x \to a} 2 \cos (\frac{x + a}{2}) \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2} \times 2} (\sqrt{x} + \sqrt{a}) \\ = & \lim_{x \to a} \cos (\frac{x + a}{2}) \times (\sqrt{x} + \sqrt{a}) \lim_{x \to a} \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} [∵ \lim_{x \to a} f (x) g (x) = \lim_{x \to a} f (x) + \lim_{x \to a} g (x)] \\ = & \cos (\frac{a + a}{2}) (\sqrt{a} + \sqrt{a}) \times \lim_{x \to a} \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} \\ = & \cos a \times 2 \sqrt{a} \times \lim_{x \to a} \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} \\ = & 2 \sqrt{a} \cos a \times 1 [∵ \lim_{x \to a} \frac{\sin x}{x} = 1] \\ = & 2 \sqrt{a} \cos a \end{array}$

Page No 240:

Question 25:

Evaluate :

\lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2}

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2} & = & \lim_{x \to \frac{π}{6}} \frac{{cosec}^{2} x - 1 - 3}{cosec x - 2} [∵ {cosec}^{2} x = 1 + \cot^{2} x] \\ = & \lim_{x \to \frac{π}{6}} \frac{{cosec}^{2} x - 4}{cosec x - 2} \\ = & \lim_{x \to \frac{π}{6}} \frac{{cosec}^{2} x - 2^{2}}{cosec x - 2} \\ = & \lim_{x \to \frac{π}{6}} \frac{(cosec x - 2) (cosec x + 2)}{cosec x - 2} \\ = & \lim_{x \to \frac{π}{6}} (cosec x - 2) \\ = & cosec \frac{π}{6} + 2 \\ = & 2 + 2 \\ = & 4 \end{array}$

Page No 240:

Question 26:

Evaluate :

\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^{2} x}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^{2} x} & = & \lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos^{2} \frac{x}{2} - 1}}{\sin^{2} x} [∵ \cos 2 x = 2 \cos^{2} x - 1] \\ = & \lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos^{2} \frac{x}{2}}}{\sin^{2} x} \\ = & \lim_{x \to 0} \frac{\sqrt{2} (1 - \cos \frac{x}{2})}{\sin^{2} x} \\ = & \lim_{x \to 0} \frac{\sqrt{2} [1 - 1 + 2 \sin^{2} (\frac{x}{4})]}{\sin^{2} x} [∵ \cos 2 x = 1 - 2 \sin^{2} x] \\ = & \lim_{x \to 0} 2 \sqrt{2} \frac{\sin^{2} \frac{x}{4}}{\sin^{2} x} \times \frac{{(\frac{x}{4})}^{2}}{{(\frac{x}{4})}^{2}} \\ = & \lim_{x \to 0} 2 \sqrt{2} \frac{\sin^{2} \frac{x}{4}}{{(\frac{x}{4})}^{2}} \times \frac{{(\frac{x}{4})}^{2}}{\sin^{2} x} \\ = & 2 \sqrt{2} \lim_{x \to 0} \frac{\sin^{2} \frac{x}{4}}{{(\frac{x}{4})}^{2}} \times \frac{1}{\frac{16 \sin^{2} x}{x^{2}}} \\ = & 2 \sqrt{2} \lim_{x \to 0} \frac{\sin^{2} \frac{x}{4}}{{(\frac{x}{4})}^{2}} \times \lim_{x \to 0} \frac{1}{\frac{16 \sin^{2} x}{x^{2}}} \\ = & \frac{2 \sqrt{2}}{16} \lim_{x \to 0} \frac{\sin^{2} \frac{x}{4}}{{(\frac{x}{4})}^{2}} \times \lim_{x \to 0} \frac{1}{\frac{\sin^{2} x}{x^{2}}} [∵ \lim_{x \to 0} f (x) g (x) = \lim_{x \to 0} f (x) \times \lim_{x \to a} g (x)] \\ = & \frac{1}{4 \sqrt{2}} \lim_{x \to 0} {(\frac{\sin \frac{x}{4}}{\frac{x}{4}})}^{2} \times \lim_{x \to 0} \frac{1}{{(\frac{\sin x}{x})}^{2}} \\ = & \frac{1}{4 \sqrt{2}} \times 1 \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & \frac{1}{4 \sqrt{2}} \end{array}$

Page No 240:

Question 27:

Evaluate :

\lim_{x \to 0} \frac{\sin x - 2 \sin 3 x + \sin 5 x}{x}

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\sin x - 2 \sin 3 x + \sin 5 x}{x} & = & \lim_{x \to 0} (\frac{\sin x}{x} - \frac{2 \sin 3 x}{x} + \frac{\sin 5 x}{x}) \\ = & \lim_{x \to 0} (\frac{\sin x}{x} - \frac{6 \sin 3 x}{3 x} + \frac{5 x \sin 5 x}{5 x}) \\ = & 1 - 6 \times 1 + 5 \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & 1 - 6 + 5 \\ = & 0 \end{array}$

Page No 240:

Question 28:

Evaluate :

If

\lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = \lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}},

then find the value of k.

Answer:

$\lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = \lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \Rightarrow \lim_{x \to 1} \frac{x^{4} - 1^{4}}{x - 1} = \lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \Rightarrow 4 {(1)}^{4 - 1} = \lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \Rightarrow 4 = \lim_{x \to k} \frac{\frac{x^{3} - k^{3}}{x - k}}{\frac{x^{2} - k^{2}}{x - k}} \Rightarrow 4 = \frac{\lim_{x \to k} \frac{x^{3} - k^{3}}{x - k}}{\lim_{x \to k} \frac{x^{2} - k^{2}}{x - k}} [∵ \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \Rightarrow 4 = \frac{\lim_{x \to k} \frac{(x - k) (x^{2} + x k + k^{2})}{x - k}}{\lim_{x \to k} \frac{(x - k) (x + k)}{x - k}} \Rightarrow 4 = \frac{\lim_{x \to k} (x^{2} + x k + k^{2})}{\lim_{x \to k} (x + k)} \Rightarrow 4 = \frac{k^{2} + k^{2} + k^{2}}{k + k} \Rightarrow 4 = \frac{3 k^{2}}{2 k} \Rightarrow \frac{3 k}{2} = 4 \Rightarrow k = \frac{8}{3}$

Page No 240:

Question 29:

Differentiate the functions w. r. to x:

$\frac{x^{4} + x^{3} + x^{2} + 1}{x}$

Answer:

$\begin{matrix} \frac{\frac{d (x^{4} + x^{3} + x^{2} + 1)}{x}}{d x} = \frac{d (x^{3} + x^{2} + x + \frac{1}{x})}{d x} \\ = \frac{d x^{3}}{d x} + \frac{d x^{2}}{d x} + \frac{d x}{d x} + \frac{d (\frac{1}{x})}{d x} \\ = 3 x^{2} + 2 x + 1 - \frac{1}{x^{2}} \\ = \frac{3 x^{4} + 2 x^{3} + x^{2} - 1}{x^{2}} \end{matrix}$

Page No 240:

Question 30:

Differentiate the functions w. r. to x:

${(x + \frac{1}{x})}^{3}$

Answer:

Let y = ${(x + \frac{1}{x})}^{3}$
$\begin{matrix} ∴ \frac{d y}{d x} = 3 {(x + \frac{1}{x})}^{3 - 1} \frac{d (x + \frac{1}{x})}{d x} (By chain rule) \\ = 3 {(x + \frac{1}{x})}^{2} [\frac{d x}{d x} + \frac{d (\frac{1}{x})}{d x}] \\ = 3 {(x + \frac{1}{x})}^{2} (1 - \frac{1}{x^{2}}) \\ = 3 \frac{{(x + 1)}^{2}}{x^{2}} \times \frac{x^{2} - 1}{x^{2}} \\ = 3 \frac{{(x + 1)}^{2} (x - 1) (x + 1)}{x^{4}} \\ = \frac{3 (x - 1) {(x + 1)}^{3}}{x^{4}} \end{matrix}$

Page No 240:

Question 31:

Differentiate the functions w. r. to x:

(3x + 5) (1 + tanx)

Answer:

Let y = (3x + 5) (1 + tan x)
$\begin{matrix} ∴ \frac{d y}{d x} = \frac{d}{d x} [(3 x + 5) (1 + tan x)] \\ = (3 x + 5) \frac{d}{d x} (1 + tan x) + (1 + tan x) \frac{d}{d x} (3 x + 5) [By product rule] \\ = (3 x + 5) {sec}^{2} x + (1 + tan x) \times 3 \\ = 3 x {sec}^{2} x + 5 {sec}^{2} x + 3tan x + 3 \end{matrix}$

Page No 241:

Question 32:

(sec x – 1) (sec x + 1)

Answer:

Let y = (sec x – 1) (sec x + 1)

= sec²x – 1
= tan²x

Differentiate y with respect to x.
â€‹
â€‹ $\begin{matrix} \frac{d y}{d x} = \frac{d ({tan}^{2} x)}{d x} \\ = 2 tan x \frac{d (tan x)}{d x} [By chain rule] \\ =2tan x {sec}^{2} x \end{matrix}$

Page No 241:

Question 33:

Differentiate the functions w. r. to x:

$\frac{3 x + 4}{5 x^{2} - 7 x + 9}$

Answer:

Let y = $\frac{3 x + 4}{5 x^{2} - 7 x + 9}$
Use the quotient rule to differentiate y with respect to x.
$\begin{matrix} \frac{d y}{d x} = \frac{(5 x^{2} - 7 x + 9) \frac{d (3 x + 4)}{d x} - (3 x + 4) \frac{d (5 x^{2} - 7 x + 9)}{d x}}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{3 (5 x^{2} - 7 x + 9) - (3 x + 4) (10 x - 7)}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{15 x^{2} - 21 x + 27 - 30 x^{2} + 21 x - 40 x + 28}{{(5 x^{2} - 7 x + 9)}^{2}} \\ = \frac{- 15 x^{2} - 40 x + 55}{{(5 x^{2} - 7 x + 9)}^{2}} \end{matrix}$

Page No 241:

Question 34:

Differentiate the functions w. r. to x:

$\frac{x^{5} - \cos x}{\sin x}$

Answer:

Let y = $\frac{x^{5} - \cos x}{\sin x}$
Use the quotient rule to differentiate y with respect to x.
$\begin{matrix} \frac{d y}{d x} = \frac{sin x \frac{d (x^{5} – cos x)}{d x} - (x^{5} - cos x) \frac{d (sin x)}{d x}}{{sin}^{2} x} \\ = \frac{sin x (5 x^{4} + sin x) - (x^{5} - cos x) {cos}^{} x}{{sin}^{2} x} \\ = \frac{5 x^{4} \sin x + \sin^{2} x - x^{5} - \cos x + \cos^{2} x}{} \\ = \frac{5 x^{4} sin x - x^{5} cos x + 1}{{sin}^{2} x} [∵ {Sin}^{2} x + {cas}^{2} x = 1] \end{matrix}$

Page No 241:

Question 35:

Differentiate the functions w. r. to x:

$\frac{x^{2} \cos \frac{π}{4}}{\sin x}$

Answer:

Let y = $\frac{x^{2} \cos \frac{π}{4}}{\sin x}$
$\begin{array}{l} \Rightarrow y = \frac{x^{2} \times (\frac{1}{\sqrt{2}})}{sin x} \\ \Rightarrow y = \frac{1}{\sqrt{2}} \frac{x^{2}}{sin x} \end{array}$
Differentiate y with respect to x using quotient rule.
$\begin{matrix} \frac{d y}{d x} = \frac{1}{\sqrt{2}} \frac{sin x \frac{d x^{2}}{d x} - x^{2} \frac{d (sin x)}{d x}}{{sin}^{2} x} \\ = \frac{1}{\sqrt{2}} (\frac{2 x sin x - x^{2} cos x}{{sin}^{2} x}) \\ = \frac{x}{\sqrt{2}} (\frac{2 sin x}{{sin}^{2} x} - \frac{x cos x}{{sin}^{2} x}) \\ = \frac{x}{\sqrt{2}} (2 cosec x - x cot x cosec x) \\ = \frac{x cosec x}{\sqrt{2}} (2 - x cot x) \end{matrix}$

Page No 241:

Question 36:

(ax² + cotx) (p + q cosx)

Answer:

Let y = (ax² + cot x) (p + q cos x)
Differentiate y with respect to x using product rule.
â€‹ $\begin{matrix} \frac{d y}{d x} = (a x^{2} + cot x) . \frac{d (p + q cos x)}{d x} + (p + q cos x) \frac{d (a x^{2} + cot x)}{d x} \\ = (a x^{2} + cot x) (- q sin x) + (p + q cos x) (2 a x - {cosec}^{2} x) \\ = - q sin x (a x^{2} + cot x) + (p + q cos x) (2 a x - {cosec}^{2} x) \end{matrix}$

Page No 241:

Question 37:

$\frac{a + b \sin x}{c + d \cos x}$

Answer:

$Let y = \frac{a + b sin x}{c + d cos x}$
Differentiate y with respect to x using quotient rule.
â€‹ $\begin{matrix} \frac{d y}{d x} = \frac{(c + d cos x) \frac{d (a + b sin x)}{d x} - (a + b sin x) \frac{d (c + d cos x)}{d x}}{{(c + d cos x)}^{2}} \\ = \frac{(c + d cos x) (b cos x) - (a + b sin x) (- d sin x)}{{(c + d cos x)}^{2}} \\ = \frac{b c cos x + b d {cos}^{2} x + a d sin x + b d {sin}^{2} x}{{(c + d cos x)}^{2}} \\ = \frac{b c cos x + a d sin x + b d ({cos}^{2} x + {sin}^{2} x)}{{(c + d cos x)}^{2}} \\ = \frac{b c cos + a d sin x + b d}{{(c + d cos x)}^{2}} [∵ {cos}^{2} x + {sin}^{2} x = 1] \end{matrix}$

Page No 241:

Question 38:

(sin x + cosx)²

Answer:

Let y = (sin x + cos x)²
= sin²x + cos²x + 2 sin x cos x
= 1 + sin 2x
Differentiate y with respect to x
$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d}{d x} (1 + \sin 2 x) \\ = & \cos 2 x \times \frac{d (2 x)}{d x} \end{array}$ [By chain rule]
= 2cos 2x

Page No 241:

Question 39:

(2x – 7)² (3x + 5)³

Answer:

Let y = (2x – 7)²(3x + 5)³
Differentiate y with respect to x using product rule.
$\frac{d y}{d x} = {(2 x - 7)}^{2} \frac{d}{d x} {(3 x + 5)}^{3} + {(3 x + 5)}^{3} \frac{d y}{d x} {(2 x - 7)}^{2}$
$= {(2 x - 7)}^{2} \times 3 {(3 x + 5)}^{2} \frac{d}{d x} + {(3 x + 5)}^{3} \times 2 (2 x - 7) \frac{d (2 x - 7)}{d x}$
= (2x – 7)²× 3(3x + 5)²× 3 + (3x + 5)³× 2(2x × 7) × 2
= 9(2x – 7)² (3x + 5)²+ 4(2x – 7) (3x + 5)³
= (2x – 7) (3x + 5)²[9(2x – 7) + 4(3x + 5)]
= (2x – 7) (3x + 5)²[18x – 63 + 12x + 20]
= (2x – 7) (3x + 5)²[18x – 63 + 12x + 20]
= (2x – 7) (3x + 5)²(30x – 43)

Page No 241:

Question 40:

x² sinx + cos2x

Answer:

Let y = x²sin x + cos 2x
Differentiate y with respect to x using sum and product rule.
$\frac{d y}{d x} = x^{2} \frac{d (\sin x)}{d x} + \sin x \frac{d x^{2}}{d x} + \frac{d (\cos 2 x)}{d x}$
$= x^{2} \cos x + \sin x 2 x + (- \sin 2 x) \frac{d (2 x)}{d x}$ (By chain rule)
= x²cos x + 2x sin x – 2sin 2x

Page No 241:

Question 41:

sin³x cos³x

Answer:

Let y = sin³x cos³x
Differentiate y with respect to x using product rule.
$\frac{d y}{d x} = \sin^{3} x \frac{d (\cos^{3} x)}{d x} + \cos^{3} x \frac{d (\sin^{3} x)}{d x} = \sin^{3} x 3 \cos^{2} x \frac{d (\cos x)}{d x} + \cos^{3} x 3 \sin^{2} x \frac{d (\sin x)}{d x}$
= sin³x 3cos²x (– sin x) + cos³x 3sin²x (Note: ignore it cos x)
= –3 sin⁴x cos²x + 3cos⁴x sin²x
= 3 sin²x cos²x (cos²x – sin²x)
$= \frac{3}{4} (4 \sin^{2} x \cos^{2} x) \cos 2 x = \frac{3}{4} {(2 \sin x \cos x)}^{2} \cos 2 x = \frac{3}{4} \sin^{2} 2 x \cos 2 x$

Page No 241:

Question 42:

$\frac{1}{a x^{2} + b x + c}$

Answer:

Let $y = \frac{1}{a x^{2} + b x + c}$
⇒ y = (ax²+ bx + c)^–1
Differentiate y with respect to x using product and chain rule.
$\begin{array}{rcl} \frac{d y}{d x} & = & - {(a x^{2} + b x + c)}^{- 2} \frac{d (a x^{2} + b x + c)}{d x} \\ = & - \frac{1}{{(a x^{2} + b x + c)}^{2}} \times (2 a x + b) \\ = & \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} \end{array}$

Page No 241:

Question 43:

Differentiate of the function with respect to ‘x’
cos (x² + 1)

Answer:

Let f(x) = cos(x²+ 1)
Differentiate f(x) with respect to x using first principle.
$\frac{d f (x)}{d x} = \underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h} = \underset{h \to 0}{l i m} \frac{\cos [{(x + h)}^{2} + 1] - \cos (x^{2} + 1)}{h} = \underset{h \to 0}{l i m} \frac{- 2 \sin \{\frac{{(x + h)}^{2} + 1 + x^{2} + 1}{2}\} \sin \{\frac{{(x + h)}^{2} + 1 - x^{2} - 1}{2}\}}{h}$
$[∵ C osC - C osD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})]$
$= \underset{h \to 0}{l i m} \frac{1}{h} [- 2 \sin \{\frac{{(x + h)}^{2} + x^{2} + 2}{2}\} \sin \{\frac{h^{2} + 2 h x}{2}\}] = \underset{h \to 0}{l i m} - 2 \sin \{\frac{{(x + h)}^{2} + x^{2} + 2}{2}\} \underset{h \to 0}{l i m} \frac{\sin \{\frac{h (h + 2 x)}{2}\}}{h (\frac{h + 2 x}{2})} \times \frac{(h + 2 x)}{2}$
$= - 2 \underset{h \to 0}{l i m} \sin \{\frac{{(x + h)}^{2} + x^{2} + 2}{2}\} \underset{h \to 0}{l i m} \frac{(h + 2 x)}{2} [∵ \underset{h \to 0}{l i m} \frac{\sin x}{x} = 1] = - 2 \sin \{\frac{x^{2} + x^{2} + 2}{2}\} \times \frac{2 x}{2}$
= –2x sin(x² + 1)

Page No 241:

Question 44:

Differentiate of the function with respect to ‘x’
$\frac{a x + b}{c x + d}$

Answer:

Let f(x) = $\frac{a x + b}{c x + d}$
$\Rightarrow f (x + h) = \frac{a (x + h) + b}{c (x + h) + d}$
Differentiate f(x) with respect to x using first principle.
$\frac{d}{d x} f (x) = \underset{h \to 0}{l i m} \frac{1}{h} [f (x + h) - f (x)] = \underset{h \to 0}{l i m} \frac{1}{h} [\frac{a (x + h) + b}{c (x + h) + d} - \frac{a x + b}{c x + d}] = \underset{h \to 0}{l i m} \frac{1}{h} [\frac{(a x + a h + b) (c x + d) - (a x + b) (c x + c h + d)}{\{c (x + h) + d\} (c x + d)}] = \underset{h \to 0}{l i m} \frac{1}{h} [\frac{a c x^{2} + a d x + a c h x + a d h + b c x + b d - a c x^{2} - a c h x - a d x - b c x - b c h - b d}{\{c (x + h) + d\} (c x + d)}] = \underset{h \to 0}{l i m} \frac{1}{h} [\frac{a d h - b c h}{\{c (x + h) + d\} (c x + d)}] = \underset{h \to 0}{l i m} \frac{1}{h} [\frac{a d - b c}{\{c (x + h) + d\} (c x + d)}] = \frac{a d - b c}{{(c x + d)}^{2}}$

Page No 241:

Question 45:

Differentiate of the function with respect to ‘x’
$x^{\frac{2}{3}}$

Answer:

Let f(x) = $x^{\frac{2}{3}}$
∴ f(x + h) = ${(x + h)}^{\frac{2}{3}}$
Differentiate f(x) with respect to x using first principle.
$\begin{array}{rcl} \frac{d}{d x} f (x) & = & \underset{h \to 0}{l i m} \frac{1}{h} [f (x + h) - f (x)] \\ = & \underset{h \to 0}{l i m} \frac{1}{h} [{(x + h)}^{\frac{2}{3}} - x^{\frac{2}{3}}] \\ = & \underset{h \to 0}{l i m} \frac{{(x + h)}^{\frac{2}{3}} = x^{\frac{2}{3}}}{(x + h) - x} \\ = & \frac{2}{3} x^{\frac{2}{3} - 1} [∵ \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = {n a}^{n - 1} and x + h \to x since h \to 0] \\ = & \frac{2}{3} x^{- \frac{1}{3}} \\ = & \frac{2}{3 x^{\frac{1}{3}}} \end{array}$

Page No 241:

Question 46:

Differentiate of the function with respect to ‘x’
x cosx

Answer:

Let f(x) = x cos x
⇒ f(x + h) = (x + h) cos(x + h)
Differentiate f(x) with respect to x using first principle.
$\begin{array}{rcl} \frac{d f (x)}{d x} & = & \underset{h \to 0}{l i m} \frac{1}{h} [(x + h) \cos (x + h) - x \cos x] \\ = & \underset{h \to 0}{l i m} \frac{1}{h} [x \cos (x + h) + h \cos (x + h) - x \cos x] \\ = & \underset{h \to 0}{l i m} \frac{1}{h} [x \{\cos (x + h) - \cos x\} + h \cos (x + h)] \\ = & \underset{h \to 0}{l i m} \frac{1}{h} [x \{\cos (x + h) - \cos x\} + h \cos (x + h)] \end{array}$
                                 $[∵ cosC - cosD = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}]$
   $= - 2 \underset{h \to 0}{l i m} x \sin (x + \frac{h}{2}) \underset{h \to 0}{l i m} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \frac{1}{2} + \underset{h \to 0}{l i m} \cos (x + h) = - \underset{h \to 0}{l i m} x \sin (x + \frac{h}{2}) \times 1 + \underset{h \to 0}{l i m} \cos (x + h) [∵ \underset{h \to 0}{l i m} \frac{\sin x}{x} = 1]$
          = – sin x + cos x
= cos x – sin x

Page No 241:

Question 47:

Evaluate

$\lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y}$

Answer:

$\underset{y \to 0}{l i m} \frac{(x + y) s e c (x + y) - x s e c x}{y} = \underset{y \to 0}{l i m} \frac{\frac{(x + y)}{\cos (x + y)} - \frac{x}{\cos x}}{y} = \underset{y \to 0}{l i m} \frac{(x + y) \cos x - x \cos (x + y)}{y \cos x \cos (x + y)} = \underset{y \to 0}{l i m} \frac{x \cos x = y \cos x - x \cos (x + y)}{y \cos x \cos (x + y)} = \underset{y \to 0}{l i m} \frac{x \{\cos x - \cos (x + y)\} + y \cos x}{y \cos x \cos (x + y)} = \underset{y \to 0}{l i m} \frac{x \{- 2 \sin (\frac{x - x - y}{2}) \sin (\frac{x + x + y}{2})\} + y \cos x}{y \cos x \cos (x + y)} = \underset{y \to 0}{l i m} \frac{2 x \sin (\frac{y}{2}) \sin (\frac{2 x + y}{2}) + y \cos x}{y \cos x \cos (x + y)} = \underset{y \to 0}{l i m} \frac{2 x}{2} \times \frac{\sin (\frac{y}{2})}{(\frac{y}{2})} \times \frac{\sin (\frac{2 x + y}{2})}{\cos x \cos (x + y)} + \frac{l i m}{y \to 0} \frac{1}{\cos (x + y)} = \underset{y \to 0}{l i m} \frac{x \times 1 \times \sin (\frac{2 x + y}{2})}{\cos x \cos (x + y)} + \underset{y \to 0}{l i m} \frac{1}{\cos (x + y)}$
$[∵ \underset{y \to 0}{l i m} \frac{\sin x}{x} = 1]$
$= \frac{x \sin (\frac{2 x}{2})}{\cos^{2} x} + \frac{1}{\cos x}$
= x tanx sexx + secx
= sec x (x tanx + 1)

Page No 241:

Question 48:

Evaluate
$\lim_{x \to 0} \frac{(\sin (α + β) x + \sin (α - β) x + \sin 2 α x)}{\cos 2 β x - \cos 2 α x} \cdot x$

Answer:

$\underset{x \to 0}{l i m} \frac{[\sin (α + β) x + \sin (α - β) x + \sin 2 α x]}{\cos 2 β x - \cos 2 α x} \times x = \underset{x \to 0}{l i m} \frac{2 \sin (\frac{α x + β x + α x - β x}{2}) \cos (\frac{α x + β x - α x + β x}{2}) + \sin 2 α x}{\cos 2 β x - \cos 2 α x} \times x = \underset{x \to 0}{l i m} \frac{2 \sin α x \cos β x + 2 \sin α x \cos α x}{- 2 \sin (\frac{2 β x + 2 α x}{2}) \sin (\frac{2 β x + 2 α x}{2})} \times x$
$[∵ \cos C - \cos D = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})]$
$= \frac{l i m}{x \to 0} \frac{2 \sin α x (\cos β x + \cos α x)}{2 \sin (α x + β x) \sin (α x - β x)} \times x = \frac{l i m}{x \to 0} \frac{x \sin α x [2 \cos (\frac{α + β}{2}) x \cos (\frac{α - β}{2}) x]}{2 \sin (\frac{α + β}{2}) x \cos (\frac{α + β}{2}) x \times 2 \sin (\frac{α - β}{2}) x \cos (\frac{α - β}{2}) x}$
$[∵ \cos C + \cos D = 2 \cos \frac{C + D}{2} \cos \frac{C - D}{2}]$
$= \frac{l i m}{x \to 0} \frac{2 x \sin α x}{4 \sin (\frac{α + β}{2}) x \sin (\frac{α - β}{2}) x} = \frac{l i m}{x \to 0} \frac{2 α x^{2} \frac{\sin α x}{α x}}{4 (\frac{α + β}{2}) x \frac{\sin (\frac{α + β}{2}) x}{(\frac{α + β}{2}) x} (\frac{α - β}{2}) x \frac{\sin (\frac{α - β}{2}) x}{(\frac{α - β}{2}) x}}$
$= \underset{x \to 0}{l i m} \frac{2 α x^{2}}{4 (\frac{α + β}{2}) (\frac{α - β}{2}) x^{2}}$ $[∵ \underset{x \to 0}{l i m} \frac{\sin x}{x} = 1]$
$= \frac{2 α}{4 \frac{(α + β) (α - β)}{4}} = \frac{2 α}{α^{2} - β^{2}}$

Page No 241:

Question 49:

Evaluate
$\lim_{x \to \frac{π}{4}} \frac{\tan^{3} x - \tan x}{\cos x + \frac{π}{4}}$

Answer:

$\underset{x \to \frac{π}{4}}{l i m} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})} = \underset{x \to \frac{π}{4}}{l i m} \frac{- \tan x (1 - \tan^{2} x)}{\cos (x + \frac{π}{4})} = \underset{x \to \frac{π}{4}}{l i m} \frac{- \tan x (1 - \tan x) (1 + \tan x)}{\cos (x + \frac{π}{4})} [∵ a^{2} - b^{2} = (a - b) (a + b)] = \underset{x \to \frac{π}{4}}{l i m} - \tan x (1 + \tan x) \underset{x \to \frac{π}{4}}{l i m} \frac{(1 - \tan x)}{\cos (x + \frac{π}{4})} = - 1 \times (1 + 1) \underset{x \to \frac{π}{4}}{l i m} \frac{(\cos x - \sin x)}{\cos x \cos (x + \frac{π}{4})} = - 2 \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt{2} [\cos \frac{π}{4} \cos x - \sin x \sin \frac{π}{4}]}{\cos x \cos (x + \frac{π}{4})} = - 2 \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt{2} \cos (x + \frac{π}{4})}{\cos x \cos (x + \frac{π}{4})} [∵ \cos (A + B) = cosAcosB - sinAsinB] = - 2 \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt{2}}{\cos x} = \frac{- 2 \sqrt{2}}{\cos \frac{π}{4}} = \frac{- 2 \sqrt{2}}{\frac{1}{\sqrt{2}}} = - 2 \sqrt{2} \times \sqrt{2} = - 4$

Page No 241:

Question 50:

Evaluate
$\lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} \cos \frac{x}{4} - \sin \frac{x}{4}}$

Answer:

$\begin{array}{rcl} \lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} & = & \lim_{x \to π} \frac{1 - 2 \sin \frac{x}{4} \cos \frac{x}{4}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} \\ = & \lim_{x \to π} \frac{\sin^{2} \frac{x}{4} + \cos^{2} \frac{x}{4} - 2 \sin \frac{x}{4} \cos \frac{x}{4}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} \\ = & \lim_{x \to π} \frac{{(\cos \frac{x}{4} - \sin \frac{x}{4})}^{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} [∵ {(a - b)}^{2} = a^{2} + b^{2} - 2 a b] \\ = & \lim_{x \to π} \frac{(\cos \frac{x}{4} - \sin \frac{x}{4})}{\cos \frac{x}{2}} \\ = & \lim_{x \to π} \frac{(\cos \frac{x}{4} - \sin \frac{x}{4})}{\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4}} [∵ \cos 2 x = \cos^{2} x - \sin^{2} x] \\ = & \lim_{x \to π} \frac{(\cos \frac{x}{4} - \sin \frac{x}{4})}{(\cos \frac{x}{4} - \sin \frac{x}{4}) (\cos \frac{x}{4} + \sin \frac{x}{4})} [∵ a^{2} - b^{2} = (a - b) - (a + b)] \\ = & \lim_{x \to π} \frac{1}{\cos \frac{x}{4} + \sin \frac{x}{4}} \\ = & \frac{1}{\cos \frac{π}{4} + \sin \frac{π}{4}} \\ = & \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} \\ = & \frac{\sqrt{2}}{2} \\ = & \frac{1}{\sqrt{2}} \end{array}$

Page No 241:

Question 51:

Evaluate
Show that $\lim_{x \to 4} \frac{|x - 4|}{x - 4}$ does not exists

Answer:

Left hand limit:

$\begin{array}{rcl} LHL & = & \lim_{x \to \frac{π^{-}}{4}} \frac{- (x - 4)}{x - 4} [∵ |x - 4| = - (x - 4) for x < 4] \\ = & - 1 \end{array}$

Right hand limit:

$\begin{array}{rcl} RHL & = & \lim_{x \to \frac{π^{+}}{4}} \frac{(x - 4)}{x - 4} [∵ |x - 4| = (x - 4) for x > 4] \\ = & - 1 \end{array}$

∴ LHL ≠ RHL

Thus, limit does not exist.

Page No 242:

Question 52:

Evaluate
Let $f (x) = \{\begin{array}{cr} \frac{k \cos x}{π - 2 x} & when x \neq \frac{π}{2} \\ 3 & x = \frac{π}{2} \end{array} and if \lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}),$ find the value of k.

Answer:

Left hand limit:

$\begin{array}{rcl} LHL & = & \lim_{x \to \frac{π^{-}}{2}} \frac{k \cos x}{π - 2 x} \\ = & \lim_{h \to 0} \frac{k \cos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \\ = & \lim_{h \to 0} \frac{k \sin h}{π - π + 2 h} [∵ \cos (\frac{π}{2} - x) = \sin x] \\ = & \lim_{h \to 0} \frac{k \sin h}{2 h} \\ = & \frac{k}{2} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Right hand limit:

$\begin{array}{rcl} RHL & = & \lim_{x \to \frac{π^{+}}{2}} \frac{k \cos x}{π - 2 x} \\ = & \lim_{h \to 0} \frac{k \cos (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = & \lim_{h \to 0} \frac{k (- \sin h)}{π - π - 2 h} [∵ \cos (\frac{π}{2} + x) = - \sin x] \\ = & \lim_{h \to 0} \frac{k \sin h}{2 h} \\ = & \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} \\ = & \frac{k}{2} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Since $\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ and $f (\frac{π}{2}) = 3,$

$\Rightarrow \frac{k}{2} = 3 \Rightarrow k = 6$

Hence, the value of k is 6.

Page No 242:

Question 53:

Evaluate
Let $f (x) = \{\begin{array}{l} x + 2 & x ⩽ - 1 \\ c x^{2} & x > - 1 \end{array},$ find 'c' if $\lim_{x \to - 1} f (x)$ exists.

Answer:

$f (x) = \{\begin{matrix} x + 2, & x \leq - 1 \\ c x^{2}, & x > - 1 \end{matrix}$

Left hand limit:

$\begin{array}{rcl} LHL & = & \lim_{x \to - 1^{-}} f (x) \\ = & \lim_{x \to - 1^{-}} f (x + 2) \\ = & \lim_{h \to 0} (- 1 - h + 2) \\ = & \lim_{h \to 0} (1 - h) \\ = & 1 \end{array}$

Right hand limit:

$\begin{array}{rcl} RHL & = & \lim_{x \to - 1^{+}} f (x) \\ = & \lim_{x \to - 1^{+}} c x^{2} \\ = & \lim_{h \to 0^{-}} c {(- 1 + h)}^{2} \\ = & c \end{array}$

If $\begin{array}{rcl} \lim_{x \to - 1} f (x) \end{array}$ exists, then LHL = RHL.

∴ c = 1

Hence, the value of c is 1.

Page No 242:

Question 54:

$\lim_{x \to π} \frac{\sin x}{x - π} is$

(A) 1
(B) 2
(C) −1
(D) −2

Answer:

$\begin{array}{rcl} \lim_{x \to π} \frac{\sin x}{x - π} & = & \lim_{x \to π} \frac{\sin x}{- (π - x)} \\ = & \lim_{x \to π} \frac{\sin x (π - x)}{- (π - x)} [∵ \sin (π - x) = \sin x] \\ = & - \lim_{x \to π} \frac{\sin (π - x)}{(π - x)} \\ = & - 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1 and π - x \to 0 \Rightarrow x - π] \end{array}$

Hence, the correct answer is option c.

Page No 242:

Question 55:

$\lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x} is$

(A) 2

(B) $\frac{3}{2}$

(C) $\frac{- 3}{2}$

(D) 1

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x} & = & \lim_{x \to 0} \frac{x^{2} \cos x}{2 \sin^{2} \frac{x}{2}} [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}] \\ = & \lim_{x \to 0} \frac{x^{2}}{2 \sin^{2} \frac{x}{2}} \times \lim_{x \to 0} \cos x [∵ \lim_{x \to a} f (x) g (x) = \lim_{x \to a} f (x) + \lim_{x \to a} g (x)] \\ = & 2 \lim_{x \to 0} \frac{{(\frac{x}{2})}^{2}}{\sin^{2} \frac{x}{2}} \times 1 \\ = & 2 \lim_{x \to 0} {[\frac{(\frac{x}{2})}{\sin (\frac{x}{2})}]}^{2} \\ = & 2 \times 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & 2 \end{array}$

Hence, the correct answer is option a.

Page No 242:

Question 56:

$\lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x} is$

(A) n

(B) 1

(C) −n

(D) 0

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{x} & = & \lim_{x \to 0} \frac{{(1 + x)}^{n} - 1}{(1 + x) - 1} \\ = & \lim_{(1 + x) \to 1} \frac{{(1 + x)}^{n} - 1^{n}}{(1 + x) - 1} \\ = & n {(1)}^{n - 1} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = x a^{n - 1}] \\ = & n \end{array}$

Hence, the correct answer is option a.

Page No 242:

Question 57:

$\lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1} is$

(A) 1

(B) $\frac{m}{n}$

(C) $- \frac{m}{n}$

(D) $\frac{m^{2}}{n^{2}}$

Answer:

$\begin{array}{rcl} \lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1} & = & \lim_{x \to 1} \frac{\frac{x^{m} - 1}{x - 1}}{\frac{x^{n} - 1}{x - 1}} \\ = & \frac{\lim_{x \to 1} \frac{x^{m} - 1^{n}}{x - 1}}{\lim_{x \to 1} \frac{x^{n} - 1^{n}}{x - 1}} [∵ \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \\ = & \frac{m {(1)}^{m - 1}}{n {(1)}^{n - 1}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n {(a)}^{n - 1}] \\ = & \frac{m}{n} \end{array}$

Hence, the correct answer is option b.

Page No 242:

Question 58:

$\lim_{x \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} is$

(A) $\frac{4}{9}$

(B) $\frac{1}{2}$

(C) $- \frac{1}{2}$

(D) −1

Answer:

$\begin{array}{rcl} \lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} & = & \lim_{θ \to 0} \frac{1 - \cos 2 (2 θ)}{1 - \cos 2 (3 θ)} \\ = & \lim_{θ \to 0} \frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ} \\ = & \lim_{θ \to 0} \frac{\frac{\sin^{2} 2 θ}{{(2 θ)}^{2}} \times {(2 θ)}^{2}}{\frac{\sin^{2} 3 θ}{{(3 θ)}^{2}} \times {(3 θ)}^{2}} \\ = & \frac{4}{9} \lim_{θ \to 0} \frac{\frac{\sin^{2} 2 θ}{{(2 θ)}^{2}}}{\frac{\sin^{2} 3 θ}{{(3 θ)}^{2}}} \\ = & \frac{4}{9} \frac{\lim_{θ \to 0} \frac{\sin^{2} (2 θ)}{{(2 θ)}^{2}}}{\lim_{θ \to 0} \frac{\sin^{2} (3 θ)}{{(3 θ)}^{2}}} [∵ \lim_{x \to a} \frac{f (x)}{g (x)} = \frac{\lim_{x \to a} f (x)}{\lim_{x \to a} g (x)}] \\ = & \frac{4}{9} \times \frac{1}{1} \\ = & \frac{4}{9} \end{array}$

Hence, the correct answer is option c.

Page No 243:

Question 59:

$\lim_{x \to 0} \frac{cosec x - \cot x}{x} is$

(A) $\frac{- 1}{2}$

(B) 1

(C) $\frac{1}{2}$

(D) −1

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{cosec x - \cot x}{x} & = & \lim_{x \to 0} \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} \\ = & \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} \\ = & \lim_{x \to 0} \frac{2 \sin^{2} \frac{x}{2}}{x 2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ = & \lim_{x \to 0} \frac{\tan \frac{x}{2}}{x} \\ = & \lim_{x \to 0} \frac{\tan \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{2} \\ = & 1 \times \frac{1}{2} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1] \\ = & \frac{1}{2} \end{array}$

Hence, the correct answer is option c.

Page No 243:

Question 60:

$\lim_{x \to 0} \frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} is$

(A) 2

(B) 0

(C) 1

(D) −1

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} & = & \lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - \sqrt{1 - x}} \times \frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \\ = & \lim_{x \to 0} \frac{\sin x (\sqrt{1 + x} + \sqrt{1 - x})}{1 + x - 1 + x} \\ = & \lim_{x \to 0} \frac{\sin x (\sqrt{1 + x} + \sqrt{1 - x})}{2 x} \\ = & \frac{1}{2} \lim_{x \to 0} \frac{\sin x}{x} \times \lim_{x \to 0} (\sqrt{1 + x} + \sqrt{1 - x}) [∵ \lim_{x \to a} f (x) g (x) = \lim_{x \to a} f (x) \times \lim_{x \to a} g (x)] \\ = & \frac{1}{2} \times 1 \times (\sqrt{1} + \sqrt{1}) [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = & \frac{1}{2} \times 2 \\ = & 1 \end{array}$

Hence, the correct answer is option c.

Page No 243:

Question 61:

$\lim_{x \to \frac{π}{4}} \frac{\sec^{2} x - 2}{\tan x - 1} is$

(A) 3

(B) 1

(C) 0

(D) $\sqrt{2}$

Answer:

$\begin{array}{rcl} \lim_{x \to \frac{π}{4}} \frac{\sec^{2} x - 2}{\tan x - 1} & = & \lim_{x \to \frac{π}{4}} \frac{1 + \tan^{2} x - 2}{\tan x - 1} \\ = & \lim_{x \to \frac{π}{4}} \frac{\tan^{2} x - 1}{\tan x - 1} \\ = & \lim_{x \to \frac{π}{4}} \frac{(\tan x - 1) (\tan x + 1)}{(\tan x - 1)} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to \frac{π}{4}} (\tan x + 1) \\ = & \tan (\frac{x}{4}) + 1 \\ = & 1 + 1 \\ = & 2 \end{array}$

Page No 243:

Question 62:

$\lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} is$

(A) $\frac{1}{10}$

(B) $\frac{- 1}{10}$

(C) 1

(D) None of these

Answer:

$\begin{array}{rcl} \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} & = & \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + 3 x - 2 x - 3} \\ = & \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{x (2 x + 3) - 1 (2 x + 3)} \\ = & \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{(x - 1) (2 x + 3)} \\ = & \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{[{(\sqrt{x})}^{2} - 1^{2}] (2 x + 3)} \\ = & \lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{(\sqrt{x} - 1) (\sqrt{x} + 1) (2 x + 3)} [∵ a^{2} - b^{2} = (a - b) (a + b)] \\ = & \lim_{x \to 1} \frac{2 x - 3}{(\sqrt{x} + 1) (2 x + 3)} \\ = & \frac{2 - 3}{(1 + 1) (2 + 3)} \\ = & \frac{- 1}{10} \end{array}$

Hence, the correct answer is option b.

Page No 243:

Question 63:

If $f (x) = \{\begin{array}{cc} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{array},$ where [.] denotes the greatest integer function, then $\lim_{x \to 0} f (x)$ is equal to

(A) 1

(B) 0

(C) −1

(D) None of these

Answer:

$f (x) = \{\begin{array}{cc} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{array}$

Left hand limit:

$\begin{array}{rcl} LHL & = & \lim_{x \to 0^{-}} f (x) \\ = & \lim_{x \to 0^{-}} \frac{\sin [x]}{[x]} \\ = & \lim_{h \to 0} \frac{\sin [0 - h]}{[0 - h]} \\ = & \frac{\sin (- 1)}{- 1} \\ = & \frac{- \sin (1)}{1} \\ = & \sin (1) \end{array}$

Right hand limit:

$\begin{array}{rcl} RHL & = & \lim_{x \to 0^{+}} f (x) \\ = & \lim_{x \to 0^{+}} \frac{\sin [x]}{[x]} \\ = & \lim_{h \to 0} \frac{\sin [0 + h]}{[0 + h]} \\ = & \frac{\sin 0}{0} \\ = & undefined \end{array}$

∴ LHL ≠ RHL

Thus, limit does not exist.

Hence, the correct answer is option d.

Page No 243:

Question 64:

$\lim_{x \to 0} \frac{|\sin x|}{x} is$

(A) 1

(B) −1

(C) does not exist

(D) None of these

Answer:

$\lim_{x \to 0} \frac{|\sin x|}{x}$

Left hand limit:

$\begin{array}{rcl} LHL & = & \lim_{x \to 0^{-}} (\frac{- \sin x}{x}) [∵ |\sin x| = - \sin x for x < 0] \\ = & \underset{x \to 0^{-}}{- \lim} \frac{\sin x}{x} \\ = & \underset{h \to 0}{- \lim} \frac{\sin (0 - h)}{0 - h} \\ = & \underset{h \to 0}{- \lim} \frac{\sin h}{h} \\ = & - 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Right hand limit:

$\begin{array}{rcl} RHL & = & \lim_{x \to 0^{+}} (\frac{\sin x}{x}) \\ = & \lim_{h \to 0} \frac{\sin (0 + h)}{0 + h} \\ = & \lim_{h \to 0} \frac{\sin h}{h} \\ = & 1 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] \end{array}$

Since LHL ≠ RHL, the limit does not exists.

Hence, the correct answer is option c.

Page No 243:

Question 65:

Let $f (x) = \{\begin{array}{l} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{array},$ the quadratic equation whose roots are $\lim_{x \to 2^{-}} f (x)$ and $\lim_{x \to 2^{+}} f (x)$ is

(A) x² − 6x + 9 = 0
(B) x² − 7x + 8 = 0
(C) x² − 14x + 49 = 0
(D) x² − 10x + 21 = 0

Answer:

$f (x) = \{\begin{matrix} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{matrix}$

$\begin{array}{rcl} ∴ \lim_{x \to 2^{-}} f (x) & = & \lim_{x \to 2^{-}} (x^{2} - 1) \\ = & \lim_{h \to 0} [{(2 - h)}^{2} - 1] \\ = & \lim_{h \to 0} (4 + h^{2} - 4 h - 1) \\ = & \lim_{h \to 0} (3 - 4 h + h^{2}) \\ = & 3 \end{array}$

$\begin{array}{rcl} Now, \lim_{x \to 2^{+}} f (x) & = & \lim_{x \to 2^{+}} (2 x + 3) \\ = & \lim_{h \to 0} [2 (2 + h) + 3] \\ = & \lim_{h \to 0} (4 + 2 h + 3) \\ = & 7 \end{array}$

Thus, the roots of the required quadratic equation is 3 and 7.

∴ x² – (3 + 7) x + 3 × 7 = 0

⇒ x² – 10x + 21 = 0

Hence, the correct answer is option d.

Page No 244:

Question 66:

$\lim_{x \to 0} \frac{\tan 2 x - x}{3 x - \sin x} is$

(A) 2

(B) $\frac{1}{2}$

(C) $\frac{- 1}{2}$

(D) $\frac{1}{4}$

Answer:

$\begin{array}{rcl} \lim_{x \to 0} \frac{\tan 2 x - x}{3 x - \sin x} & = & \lim_{x \to 0} \frac{x (\frac{\tan 2 x}{x} - 1)}{x (3 - \frac{\sin x}{x})} \\ = & \lim_{x \to 0} \frac{(\frac{\tan 2 x}{2 x} \times 2 - 1)}{(3 - \frac{\sin x}{x})} \\ = & \frac{\lim_{x \to 0} (\frac{2 \tan 2 x}{2 x} - 1)}{\lim_{x \to 0} (3 - \frac{\sin x}{x})} \\ = & \frac{2 \times 1 - 1}{3 - 1} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1 and \lim_{x \to 0} \frac{\tan x}{x} = 1] \\ = & \frac{1}{2} \end{array}$

Hence, the correct answer is option (B).

Page No 244:

Question 67:

Let f(x) = x − [x]; ∈ R, then f ′ $\frac{1}{2}$ is

(A) $\frac{3}{2}$

(B) 1

(C) 0

(D) −1

Answer:

The derivative of greatest integer function is zero i.e., $\frac{d [x]}{d x} = 0$ .
$∴ \frac{d f (x)}{d x} = \frac{d (x - [x])}{d x} \Rightarrow f' (x) = \frac{d x}{d x} - \frac{d [x]}{d x} \Rightarrow f' (x) = 1 - 0 \Rightarrow f' (x) = 1 ∴ f' (\frac{1}{2}) = 1$

Hence, the correct answer is option (B).

Page No 244:

Question 68:

If $y = \sqrt{x} + \frac{1}{\sqrt{x}},$ then $\frac{d y}{d x}$ at x = 1 is
(A) 1

(B) $\frac{1}{2}$

(C) $\frac{1}{\sqrt{2}}$

(D) 0

Answer:

$y = \sqrt{x} + \frac{1}{\sqrt{x}} \Rightarrow y = x^{\frac{1}{2}} + x^{- \frac{1}{2}}$
Differentiate y with respect to x.
$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{1}{2} x^{\frac{1}{2} - 1} + (- \frac{1}{2}) x^{- \frac{1}{2} - 1} (By chain rule) \\ = & \frac{1}{2 \sqrt{x}} - \frac{1}{2 x^{\frac{3}{2}}} \end{array}$

$\begin{array}{rcl} ∴ {(\frac{d y}{d x})}_{x = 1} & = & \frac{1}{2} - \frac{1}{2} \\ = & 0 \end{array}$

Hence, the correct answer is option D.

Page No 244:

Question 69:

If $f (x) = \frac{x - 4}{2 \sqrt{x}},$ then f ′(1) is

(A) $\frac{5}{4}$

(B) $\frac{4}{5}$

(C) 1

(D) 0

Answer:

Use the quotient rule $[\frac{d (\frac{u}{v})}{d x} = \frac{v \frac{d u}{d x} - 4 \frac{d v}{d x}}{v^{2}}]$ to differentiate f(x) with respect to x.
$\begin{array}{rcl} f' (x) & = & \frac{2 \sqrt{x} \frac{d (x - 4)}{d x} - (x - 4) \frac{d (2 \sqrt{x})}{d x}}{{(2 \sqrt{x})}^{2}} \\ = & \frac{2 \sqrt{x} - (x - 4) \times 2 \times \frac{1}{2 \sqrt{x}}}{4 x} \\ = & \frac{2 x - (x - 4)}{4 x \sqrt{x}} \\ = & \frac{2 x - x + 4}{4 x^{\frac{3}{2}}} \\ = & \frac{x + 4}{4 x^{\frac{3}{2}}} \end{array}$
$\begin{array}{rcl} ∴ f' (1) & = & \frac{1 + 4}{4 {(1)}^{\frac{3}{2}}} \\ = & \frac{5}{4} \end{array}$
Hence, the correct answer is option (A).

Page No 244:

Question 70:

If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}},$ then $\frac{d y}{d x}$ is

(A) $\frac{- 4 x}{{(x^{2} - 1)}^{2}}$

(B) $\frac{- 4 x}{x^{2} - 1}$

(C) $\frac{1 - x^{2}}{4 x}$

(D) $\frac{4 x}{x^{2} - 1}$

Answer:

$\begin{array}{rcl} y & = & \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} \\ = & \frac{\frac{x^{2} + 1}{x^{2}}}{\frac{x^{2} - 1}{x^{2}}} \\ = & \frac{x^{2} + 1}{x^{2} - 1} \end{array}$
Differentiate y with respect to x using the quotient rule.
$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{(x^{2} - 1) 2 x - (x^{2} + 1) 2 x}{{(x^{2} - 1)}^{2}} \\ = & \frac{2 x^{3} - 2 x - 2 x^{3} - 2 x}{{(x^{2} - 1)}^{2}} \\ = & \frac{- 4 x}{{(x^{2} - 1)}^{2}} \end{array}$

Hence, the correct answer is option (A).

Page No 244:

Question 71:

If $y = \frac{\sin x + \cos x}{\sin x - \cos x},$ then $\frac{d y}{d x}$ at x = 0 is

(A) −2

(B) 0

(C) $\frac{1}{2}$

(D) does not exist

Answer:

$y = \frac{\sin x + \cos x}{\sin x - \cos x}$
Differentiate y with respect to x using the quotient rule.
$\begin{array}{rcl} ∴ \frac{d y}{d x} & = & \frac{(\sin x - \cos x) \frac{d}{d x} (\sin x + \cos x) - (\sin x + \cos x) \frac{d}{d x} (\sin x - \cos x)}{{(\sin x - \cos x)}^{2}} \\ = & \frac{(\sin x - \cos x) (\cos x - \sin x) - (\sin x + \cos x) (\cos x + \sin x)}{{(\sin x - \cos x)}^{2}} \\ = & \frac{- {(\sin x - \cos x)}^{2} - {(\sin^{2} x + \cos x)}^{2}}{{(\sin x - \cos x)}^{2}} \\ = & - \frac{[\sin^{2} x + \cos^{2} x - 2 \sin x \cos x + \sin^{2} x + \cos^{2} x + 2 \sin x \cos x]}{{(\sin x - \cos x)}^{2}} \\ = & \frac{- (1 + 1)}{{(\sin x - \cos x)}^{2}} \\ = & \frac{- 2}{{(\sin x - \cos x)}^{2}} \end{array}$
$\begin{array}{rcl} ∴ {(\frac{d y}{d x})}_{x = 0} & = & \frac{- 2}{{(\sin x - \cos 0)}^{2}} \\ = & \frac{- 2}{{(0 - 1)}^{2}} \\ = & - 2 \end{array}$

Hence, the correct answer is option (A).

Page No 245:

Question 72:

If $y = \frac{\sin (x + 9)}{\cos x}$ then $\frac{d y}{d x}$ at x = 0 is

(A) cos 9

(B) sin 9

(C) 0

(D) 1

Answer:

Using the quotient rule to differentiate y with respect to x.
$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{\cos x \frac{d [\sin (x + 9)]}{d x} - \sin (x + 9) \frac{d (\cos x)}{d x}}{\cos^{2} x} \\ = & \frac{\cos x \cos (x + 9) - \sin (x + 9) (- \sin x)}{\cos^{2} x} \\ = & \frac{\cos x \cos (x + 9) + \sin x \sin (x + 9)}{\cos^{2} x} \\ = & \frac{\cos (x + 9 - x)}{\cos^{2} x} [∵ \cos x \cos y + \sin x \sin y = \cos (x - y)] \\ = & \frac{\cos 9}{\cos^{2} x} \end{array}$
$\begin{array}{rcl} ∴ {(\frac{d y}{d x})}_{x = 0} & = & \frac{\cos 9}{\cos^{2} 0} \\ = & \frac{\cos 9}{1} \\ = & \cos 9 \end{array}$
Hence, the correct answer is option (A).

Page No 245:

Question 73:

If $f (x) = 1 + x + \frac{x^{2}}{2} + . . . + \frac{x^{100}}{100},$ then f ′(1) is equal to

(A) $\frac{1}{100}$

(B) 100

(C) does not exist

(D) 0

Answer:

$f (x) = 1 + x + \frac{x^{2}}{2} + . . . + \frac{x^{100}}{100} ∴ f' (x) = 0 + 1 + \frac{2 x}{2} + . . . + \frac{100 x^{99}}{100} = 1 + x + . . . + x^{99}$

Put x = 1 in f'(x).

$\begin{array}{rcl} f' (1) & = & 1 + 1 + . . . + 1 (100 times) \\ = & 100 \end{array}$

Hence, the correct answer is option B.

Page No 245:

Question 74:

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant 'a' then f ′(a) is

(A) 1

(B) 0

(C) does not exist

(D) $\frac{1}{2}$

Answer:

$f (x) = \frac{x^{n} - a^{n}}{x - a}$
Using the quotient rule to differentiate f(x) with respect to x.
$\begin{array}{rcl} f' (x) & = & \frac{(x - a) \frac{d}{d x} (x^{n} - a^{n}) - (x^{n} - a^{n}) \frac{d (x - a)}{d x}}{{(x - a)}^{2}} \\ = & \frac{(x - a) n x^{n - 1} - (x^{n} - a^{n}) \times 1}{{(x - a)}^{2}} \\ = & \frac{n x^{n} - n a x^{n - 1} - x^{n} + a^{n}}{{(x - a)}^{2}} \end{array}$
$\begin{array}{rcl} ∴ f' (a) & = & \frac{n a^{n} - n a a^{n - 1} - a^{n} + a^{n}}{{(x - a)}^{2}} \\ = & \frac{n a^{n} - n a^{n}}{0^{2}} \\ = & \frac{0}{0} \end{array}$
Thus, f'(a) does not exist.

Hence, the correct answer is option C.

Page No 245:

Question 75:

If f(x) = x¹⁰⁰ + x⁹⁹ + ... + x + 1, then f ′(1) is equal to

(A) 5050

(B) 5049

(C) 5051

(D) 50051

Answer:

$f (x) = x^{100} + x^{99} + . . . + x + 1 ∴ f' (x) = 100 x^{99} + 99 x^{98} + . . . + 1 + 0$

$\begin{array}{rcl} \Rightarrow & f' (1) = 100 + 99 + . . . + 1 \\ = & \frac{100}{2} [2 \times 100 + (100 - 1) (- 1)] [∵ S_{n} = \frac{n}{2} \{2 a + (n - 1) d\}] \\ = & 50 (200 - 99) \\ = & 50 \times 101 \\ = & 5050 \end{array}$

Hence, the correct answer is option (A).

Page No 245:

Question 76:

If f(x) = 1 − x + x² − x³ ... − x⁹⁹ + x¹⁰⁰, then f ′(1) is equal to

(A) 150

(B) −50

(C) −150

(D) 50

Answer:

$f (x) = 1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100} ∴ f' (x) = 0 - 1 + 2 x - 3 x^{2} + . . . - 99 x^{98} + 100 x^{99}$

$\begin{array}{rcl} \Rightarrow & f' (x) = - 1 + 2 - 3 + . . . - 99 + 100 \\ = & - (1 + 3 + 5 + . . . + 99) + (2 + 4 + . . . + 100) \\ = & - \frac{50}{2} [2 \times 1 + (50 - 1) 2] + \frac{50}{2} [2 \times 2 + (50 - 1) 2] [∵ S_{n} = \frac{n}{2} \{2 a + (n - 1) d\}] \\ = & - 25 (2 + 98) + 25 (4 + 98) \\ = & - 25 \times 100 + 25 \times 102 \\ = & - 2500 + 2550 \\ = & 50 \end{array}$

Hence, the correct answer is option (D).

Page No 245:

Question 77:

Fill in the blank.

If $f (x) = \frac{\tan x}{x - π},$ then $\lim_{x \to π} f (x) =$ _____________

Answer:

$\begin{array}{rcl} \lim_{x \to π} f (x) & = & \lim_{x \to π} \frac{\tan x}{x - π} \\ = & \lim_{x \to π} \frac{- \tan (π - x)}{x - π} [∵ \tan (π - x) = - \tan x] \\ = & \lim_{x \to π} \frac{- \tan (π - x)}{- (π - x)} \\ = & \lim_{x \to π} \frac{\tan (π - x)}{(π - x)} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1 and x \to π \Rightarrow π - x \to 0] \\ = & 1 \end{array}$

Page No 245:

Question 78:

Fill in the blank.

$\lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2,$ then m = _____________

Answer:

$\lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2 \Rightarrow \lim_{x \to 0} \frac{\sin m x}{m x} \times m x \times \frac{1}{\tan \frac{x}{\sqrt{3}}} = 2 \Rightarrow \lim_{x \to 0} \frac{\sin m x}{m x} \times m x \times \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}} \times \frac{x}{\sqrt{3}}} = 2 \Rightarrow \lim_{x \to 0} \frac{\sin m x}{m x} \times \lim_{x \to 0} \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \times \lim_{x \to 0} \frac{m x}{\frac{x}{\sqrt{3}}} = 2 \Rightarrow 1 \times 1 \times \sqrt{3} m = 2 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1 and \lim_{x \to 0} \frac{\tan x}{x} = 1] \Rightarrow m = \frac{2}{\sqrt{3}} \Rightarrow m = \frac{2 \sqrt{3}}{3}$

Page No 245:

Question 79:

Fill in the blank.
If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .,$ then $\frac{d y}{d x} =$ _____________

Answer:

$y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$
$\begin{array}{rcl} ∴ \frac{d y}{d x} & = & 0 + 1 + \frac{2 x}{2!} + \frac{3 x^{2}}{3!} + . . . \\ = & 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . \\ = & y \end{array}$

Page No 245:

Question 80:

Fill in the blank.
$\lim_{x \to 3^{+}} \frac{x}{[x]} =$ _____________

Answer:

$\begin{array}{rcl} \lim_{x \to 3^{+}} \frac{x}{[x]} & = & \lim_{h \to 0} \frac{3 + h}{[3 + h]} \\ = & \frac{3}{3} \\ = & 1 \end{array}$

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