NCERT Solutions For Class 11 Maths Chapter 13

NCERT Solutions Class 11 Maths Limits and Derivatives

Ncert Solutions For Class 11 Maths Chapter 13 PDF Free Download

NCERT Solutions for Class 11 Maths Chapter 13 limits and derivatives have been provided here for students of class 11 who are looking to clear their doubts and improve their knowledge in this chapter. These NCERT maths class 11 solutions for limits and derivatives (chapter 13) will help the students to instantly clear their doubts.

The limits and derivative chapter contains several important topics such as limits of polynomials and rationals functions, logarithmic, trigonometric as well as exponential functions. While solving the questions, students often face doubts and up piling those up. Thus, the class 11 NCERT Solutions for Maths Chapter 13 (Limits and Derivatives) are provided here to help the students. These solutions are extremely easy to understand and are given in stepwise format. The solutions are given for each of the exercises and solutions to all the questions in the NCERT book for class 11 maths are given in a simple but detailed way. The NCERT Solutions for Class 11 Maths Chapter 13 pdf is also provided here which the students can download and access the content offline.

Exercise 13.1

Q1.

Evaluate \(\lim_{x \rightarrow 3} x + 5\) limit:

Soln:

\(\lim_{x \rightarrow 3} x + 5 = 3 + 5 = 8\)

 

Q2.

Evaluate \(\lim_{x \rightarrow \pi}\left ( x – \frac{22}{7} \right )\)

Soln:

\(\lim_{x \rightarrow \pi} ( x – \frac{22}{7}) =( \pi – \frac{22}{7})\)

 

Q3.

Evaluate \(\lim_{r \rightarrow l} \pi r^{2}\)

Soln:

\(\lim_{r \rightarrow l} \pi r^{2} = \pi (l)^{2} = \pi \)

 

Q4.

Evaluate \(\lim_{x + 4}\frac{4x + 6}{x – 3}\)

Soln:

\(\lim_{x + 4}\frac{4x + 6}{x – 3} = \frac{4 (4) + 6}{4 – 3} = \frac{16 + 6}{1} = \frac{22}{1} = 22\)

 

Q5.

Evaluate \(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1}\)

Soln:

\(\lim_{x \rightarrow -1} \frac{x^{12} + x^{7} + 1}{x – 1} = \frac{(-1)^{12} + (-1)^{7} + 1}{-1 – 1 } = \frac{1 – 1 + 1}{-2} = -\frac{1}{2}\)

 

Q6.

Evaluate \(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x}\)

Put x + 3 = y so that \(y \rightarrow 1 \; as \; x \rightarrow 0\)

Accordingly, \(\\ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = \lim_{y \rightarrow 1}\frac{y^{5} – 3}{y – 3} \\ = \lim_{y \rightarrow 1}\frac{y^{5} – 3^{5}}{y – 3} \\ = 5.3^{5 – 1} \\ = 405 \\ ∴ \lim_{x \rightarrow 0}\frac{(x + 3)^{5} – 3}{x} = 405\)

 

Q7.

Evaluate \(\lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4}\)

Soln:

At x = 2, the given rational function has the value of 0/ 0

\(\\ \lim_{x \rightarrow 2}\frac{3x^{2} – x – 10}{x^{2} – 4} = \lim_{x \rightarrow 2}\frac{ \left ( x – 2 \right ) \left ( 3x + 5 \right ) }{\left ( x – 2 \right ) \left ( x + 2 \right ) } \\ = \lim_{x \rightarrow 2} \frac{3x + 5}{x + 2} \\ = \frac{3 (2) + 5}{2 + 2} \\ = \frac{11}{4}\)

 

Q8.

Evaluate \(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3}\)

Soln:

At x = 2, the given rational function has the value of 0/ 0

\(\\ \lim_{x \rightarrow 3}\frac{x^{4} – 81}{2x^{2} – 5x – 3} = \lim_{x \rightarrow 3}\frac{(x- 3)(x + 3)(x^{2} + 9)}{(x – 3)(2x + 1)} \\ = \lim_{x \rightarrow 3}\frac{(x + 3)(x^{2} + 9)}{2x + 1} \\ = \frac{(3 + 3)(3^{2} + 9)}{2(3) + 1} \\ = \frac{6 \times 18}{7} \\ = \frac{108}{7}\)

 

Q9.

Evaluate \(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0)+ 1} = b\)

 

Q10.

Evaluate \(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)

Soln:

\(\lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1}\)

At z = 1, the given function has the value of 0/ 0

Put \(z^{\frac{1}{6}} = x \; so \; that \; z\rightarrow 1 \; as \; x \rightarrow 1 \\ Accordingly, \; \lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}} – 1}{z^{^{\frac{1}{6}}} – 1} = \lim_{x \rightarrow 1} \frac{x^{2} – 1}{x – 1} \\ = \lim_{x \rightarrow 1}\frac{x^{2} – 1^{2}}{x – 1} \\ = 2.1^{2 – 1} \\ = 2 \\ ∴ \lim_{z \rightarrow 1}\frac{z^{\frac{1}{3}} – 1}{z^{\frac{1}{6}} – 1} = 2\)

 

Q11.

Evaluate \(\lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a}, \; a + b + c \neq 0\)

Soln:

\(\\ \lim_{x \rightarrow 1}\frac{ax^{2} + bx + c}{cx^{2} + bx + a} = \frac{a(1)^{2} + b(1) + c}{c(1)^{2} + b(1) + a}\\ = \frac{a + b + c}{a + b + c} \\ = 1\)

 

Q12.

\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)

Soln:

\(\lim_{x \rightarrow 2}\frac{\frac{1}{x} + \frac{1}{2}}{x + 2}\)

At x = -2, the given function has the value of 0/ 0

\(Now, \; \lim_{x \rightarrow 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \rightarrow -2} \frac{\left ( \frac{2 + x}{2x} \right )}{x + 2} \\ = \lim_{x \rightarrow -2}\frac{1}{2x} \\ = \frac{1}{2 (-2) } = \frac{-1}{4}\)

 

Q13.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{\sin ax}{bx}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \lim_{x \rightarrow 0}\frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \frac{sin ax}{ax} \times \frac{ax}{bx} \\ = \lim_{x \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \times \left ( \frac{a}{b} \right ) \\ = \frac{a}{b} \lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right ) \\ = \frac{a}{b} \times 1 \\ = \frac{a}{b}\)

 

Q14.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)

Soln:

\(\lim_{x \rightarrow 0}\frac{\sin ax}{\sin bx}, a, \; b \neq 0\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \lim_{x \rightarrow 0}\frac{sin ax}{sin bx} = \lim_{x \rightarrow 0} \frac { \left (\frac{ \sin ax}{ax} \right ) \times ax}{\left ( \frac{\sin bx}{bx} \right ) \times bx}\\ = \left ( \frac{a}{b} \right ) \times \frac{\lim_{ax \rightarrow 0}\left ( \frac{\sin ax}{ax} \right )}{\lim_{bx \rightarrow 0} \left ( \frac{\sin bx}{bx} \right )} \\ = \left ( \frac{a}{b} \right ) \times \frac{1}{1} \\ = \frac{a}{b}\)

 

Q15.

Evaluate \(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)

Soln:

\(\lim_{x \rightarrow \pi}\frac{\sin (\pi – x)}{\pi (\pi – x)}\)

It seems \(x \rightarrow \pi \Rightarrow \left ( \pi – x \right ) \rightarrow 0\)

\(\\ ∴ \lim_{x \rightarrow \pi} \frac{\sin (\pi – x)}{\pi (\pi – x)} = \frac{1}{\pi}\lim_{\left ( z – x \right ) – 1}\frac{\sin(\pi – x)}{(\pi – x)} \\ = \frac{1}{\pi} \times 1 \\ = \frac{1}{\pi}\)

 

Q16.

Evaluate \(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x}\)

Soln;

\(\lim_{x \rightarrow 0}\frac{\cos x}{\pi – x} = \frac{\cos 0}{\pi – 0} = \frac{1}{\pi}\)

 

Q17.

Evaluate \(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)

Soln:

\(\lim_{x \rightarrow 0} \frac{\cos 2x – 1}{\cos x – 1}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\cos 2x – 1}{\cos x – 1} = \lim_{x \rightarrow 0}\frac{1 – 2 \sin^{2} x – 1}{1 – 2 \sin^{2} \frac{x}{2} – 1} \\ = \lim_{x \rightarrow 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}} = \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin^{2}x}{x^{2}} \right ) \times x^{2}}{\left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )\times \frac{x^{2}}{4}} \\ = 4 \frac{\lim_{x\rightarrow 0}\left ( \frac{\sin^{2}x}{x^{2}} \right )}{\lim_{x \rightarrow 0} \left ( \frac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )}\)

\(\\ = 4 \frac{\left (\lim_{x \rightarrow 0}\frac{\sin x}{x} \right )^{2}}{\left ( \lim_{\frac{x}{2} \rightarrow 0} \frac{sin \frac{x}{2}}{\frac{x}{2}} \right )^{2}} \\ = 4\frac{1^{2}}{1^{2}} \\ = 4\)

 

Q18.

Evaluate \(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)

Soln:

\(\lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x}\)

At x = 0, the given function has the value of 0/ 0

\(\\ Now \\ \lim_{x \rightarrow 0}\frac{ax + x \cos x}{b \sin x} = \frac{1}{b} \lim_{x \rightarrow 0}\frac{x(a + \cos x)}{\sin x}\\ = \frac{1}{b} \lim_{x \rightarrow 0} \left ( \frac{x}{\sin x} \right ) \times \lim_{x \rightarrow 0}(a + \cos x) \\ = \frac{1}{b} \times \frac{1}{\left ( \lim_{x \rightarrow 0} \frac{\sin x}{x} \right )} \times \lim_{x \rightarrow 0}\left ( a + \cos x \right ) \\ = \frac{1}{b} \times \left ( a + \cos 0 \right ) \\ = \frac{a + 1}{b}\)

 

Q19.

Evaluate \(\lim_{x \rightarrow 0} x \sec x\)

Soln:

\(\lim_{x \rightarrow 0} x \sec x = \lim_{x \rightarrow 0}\frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0\)

 

Q20.

Evaluate \(\lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} a, \;b \; \; a + b \neq 0\)

Soln:

At x =0, the given function has the value of 0/ 0

\(\\ Now, \\ \lim_{x \rightarrow 0}\frac{\sin ax + bx}{ax + \sin bx} \\ \lim_{x \rightarrow 0}\frac{\left ( \frac{\sin ax}{ax} \right )ax + bx}{ax + bx\left ( \frac{\sin bx}{bx} \right )} \\ =\)

\(\\ = \frac{\left ( \lim_{ax \rightarrow 0} \frac{\sin ax}{ax} \right ) \times \lim_{x \rightarrow 0}\left ( ax \right ) + \lim_{x \rightarrow 0} bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0} bx\left ( \lim_{bx \rightarrow 0} \frac{\sin bx}{bx} \right )} \\ = \frac{\lim_{x \rightarrow 0}(ax) + \lim_{x \rightarrow 0}bx}{\lim_{x \rightarrow 0}ax + \lim_{x \rightarrow 0}bx} \\ = \frac{\lim_{x \rightarrow 0}(ax + bx)}{\lim_{x \rightarrow 0}(ax + bx)} \\ = \lim_{x \rightarrow 0}(1) \\ = 1\)

 

Q21.

Evaluate \(\lim_{x \rightarrow 0}(\csc x – \cot x)\)

Soln:

At x =0, the given function has the value of ∞ – ∞

\(\\ Now, \\ \lim_{x \rightarrow 0}(\csc x – \cot x) \\ = \lim_{x \rightarrow 0}\left ( \frac{1}{\sin x} – \frac{\cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\left ( \frac{1 – \cos x}{\sin x} \right ) \\ = \lim_{x \rightarrow 0}\frac{\left (\frac{1 – \cos x}{\sin x} \right )}{\left (\frac{\sin x}{x} \right )} \\ = \frac{\lim_{x \rightarrow 0}\frac{1 – \cos x}{x}}{\lim_{x \rightarrow 0}\frac{\sin x}{x}} \\ = \frac{0}{1} \\ = 0\)

 

Q22.

\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)

Soln:

\(\lim_{x \rightarrow \frac{\pi}{x}}\frac{\tan 2x}{x – \frac{\pi}{2}}\)

At \( x = \frac{\pi}{2}\), the given function has the value of 0/ 0

\(\\ Now, \; put \; x – \frac{\pi}{2} = y \; so \; that \; x \rightarrow \frac{\pi}{2}, \; y \rightarrow 0 \\ ∴ \lim_{x \rightarrow \frac{\pi}{2}}\frac{\tan2x}{x – \frac{\pi}{2}} = \lim_{y \rightarrow 0}\frac{tan 2\left ( y + \frac{\pi}{2} \right ) }{y} \\ = \lim_{y \rightarrow 0} \frac{tan \left ( \pi + 2y \right ) }{y} \\ =\lim_{y \rightarrow 0} \frac{\tan 2y}{y} \\ = \lim_{y \rightarrow 0}\frac{\sin 2y}{y \cos 2y} \\ = \lim_{y \rightarrow 0}\left ( \frac{\sin 2y}{2y} \times \frac{2}{\cos 2y} \right ) \\ = \left ( \lim_{2y \rightarrow 0} \frac{\sin 2y}{2y} \right ) \times \lim_{y \rightarrow 0}\left ( \frac{2}{\cos 2y} \right ) \\ =1 \times \frac{2}{\cos 0} \\ = 1 \times \frac{2}{1} \\ = 2\)

 

Q23.

Find \(\\ \lim_{x\rightarrow 0} f(x) \; and \; \lim_{x\rightarrow 0} f(x), \\ where f(x) = \left\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\right.\)

Soln:

Given function \(f(x) =\{\begin{matrix} 2x + 3 & x \leq 0\\ 3(x + 1) & x > 0 \end{matrix}\)

 

\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0}\left [ 2x + 3 \right ] = 2(0) + 3 = 3\)

 

\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} 3(x + 1) = 3(0 + 1) = 3\)

 

\(∴ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f(x) = 3\)

 

\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)

 

\(\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x + 1) = 3(1 + 1) = 6\)

 

\(∴ \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} f(x)\) = 6

 

Q24.

Find \(\lim_{x \rightarrow 1} f(x), \; where \; f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\)

Soln:

Given function is

\(f(x) = \left\{\begin{matrix} x^{2} – 1, & x \leq 1 \\ -x^{2} – 1, & x \geq 1 \end{matrix}\right.\)

\(\lim_{x \rightarrow 1^{-} f(x)} = \lim_{x \rightarrow 1}[x^{2} – 1] = 1^{2} – 1 = 1 – 1 = 0\)

\(\lim_{x \rightarrow 1^{+} f(x)} = \lim_{x \rightarrow 1}[-x^{2} – 1] = -1^{2} – 1 = 1 – 1 = 0\)

We observed that \(\lim_{x \rightarrow 1^{-}} f(x) \neq \lim_{x \rightarrow 1^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 1} f(x)\) doesn’t exist.

 

25.

Evaluate \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

Soln:

Given function \(f(x) = \left\{\begin{matrix} \frac{\left | x \right |}{x}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

When \( \left | x \right | = -x \)

\(\lim_{x \rightarrow 0 ^{-}} f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{\left | x \right |}{x} \right ]\)

= \(\lim_{x \rightarrow 0 \left ( \frac{ -x }{x} \right )}\)

= \(\lim_{x \rightarrow 0 \left ( -1 \right )}\)

= -1

When \( \left | x \right | = x \)

\(\lim_{x \rightarrow 0 ^{+}} f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{\left | x \right |}{x} \right ]\)

= \(\lim_{x \rightarrow 0 \left [ \frac{ -x }{x} \right ]}\)

= \(\lim_{x \rightarrow 0 \left ( 1 \right )}\)

= 1

We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.

 

Q26.

Find \(\lim_{x \rightarrow 0} f(x), \; where \; f(x) = \left\{\begin{matrix} \frac{x}{\left | x \right |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\right.\)

Soln:

f(x) = \(\{\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{matrix}\)

When \(\left | x \right | = -x\)

\(\lim_{x \rightarrow 0^{-}}f(x) = \lim_{x \rightarrow 0^{-}}\left [ \frac{x}{\left | x \right |} \right ]\)

\( = \lim_{x \rightarrow 0}\left [ \frac{x}{ – x } \right ]\)

\( = \lim_{x \rightarrow 0}\left ( – 1 \right )\)

= -1

When \(\left | x \right | = x\)

\(\lim_{x \rightarrow 0^{+}}f(x) = \lim_{x \rightarrow 0^{+}}\left [ \frac{x}{\left | x \right |} \right ]\)

\( = \lim_{x \rightarrow 0}\left [ \frac{x}{ x } \right ]\)

\( = \lim_{x \rightarrow 0}\left ( 1 \right )\)

= 1

We observe that \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

Hence, \(\lim_{x \rightarrow 0 } f(x) \) doesn’t exist.

 

Q27.

Find \(\lim_{x \rightarrow 5} f(x), \; where \; f(x) = \left | x \right | – 5\)

Soln:

Given function is \( f(x) = \left | x \right | – 5\)

When \( x > 0, \left | x \right | = x \)

\(\lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}}[\left | x \right | – 5]\)

= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)

= 5 – 5

= 0

When \( x > 0, \left | x \right | = x \)

\(\lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}}(\left | x \right | – 5)\)

= \( \lim_{x \rightarrow 5} \left ( x – 5\right )\)

= 5 – 5

= 0

\(∴ \lim_{x \rightarrow 5^{-}}f(x) = \lim_{x \rightarrow 5^{+}}f(x) = 0\)

Hence, \(\lim_{x \rightarrow 5 } f(x) = 0 \)

 

Q28.

Suppose \(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\)

And \(\lim_{x \rightarrow 1} f(x) = f(1)\) what can be the values of b and a?

Soln:

Given function

\(f(x) = \left\{\begin{matrix} a + bx, & if \; x < 1 \\ 4, & if \; x = 0, \\ b – ax, & if \; x > 1 \end{matrix}\right.\)

\(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(a + bx) = a + b\)

\(\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1}(b – ax) = b – a \)

f(1) = 4

Given that \(\lim_{x \rightarrow 1} f(x) = f(1)\).

\(∴ \lim_{x \rightarrow 1^{-}}f(x) = \lim_{x \rightarrow 1^{+}}f(x) = \lim_{x \rightarrow 1} f(x) = f(1)\)

=> a+ b = 4 and b – a = 4

Solving both the equations, we get a = 0 and b = 4.

Hence, b and a are 4 and 0

 

Q29.

A function \(f(x) = (x – a_{1})(x – a_{2}) . . . . (x – a_{n})\) is define by fixed real numbers a1, a2, . . . . an.

Find \(\lim_{x \rightarrow a_{1}}f(x)\). For some \(a \neq a_{1}, a_{2}, . . . . , a_{n} \; compute \; \lim_{x \rightarrow a_{n}} f(x)\)

Soln:

Given function = \(f(x) = (x – a_{1})(x – a_{2}) . . . (x – a_{n})\)

\(\lim_{x \rightarrow a_{1}}f(x) = \lim_{x \rightarrow a_{1}}[(x – a_{1})(x – a_{2}) . . . (x – a_{n})]\)

 

\(= [\lim_{x \rightarrow a_{1}}(x – a_{1})][\lim_{x \rightarrow a_{1}}(x – a_{2})] . . . [\lim_{x \rightarrow a_{1}}(x – a_{n})]\)

 

\(= \left ( a_{1} – a_{1} \right ) \left ( a_{1} – a_{2} \right ) . . . . (a_{1} – a_{n}) = 0\)

 

\(∴ \lim_{x \rightarrow a_{1}]}f(x) = 0\)

 

Now, \(\lim_{x \rightarrow a_{1}]}f(x) = \lim_{x \rightarrow a}[\left (x – a_{1} \right )\left ( x – a_{2} \right ) . . . \left ( x – a_{n} \right )]\)

 

= \(= \lim_{x \rightarrow a}[ x – a_{1} ]  [ x – a_{2}] . . . [x – a_{n}]\)

 

\(( a_{1} – a_{1}) ( a_{1} – a_{2} ) . . . . (a_{1} – a_{n})\)
\( ∴ \lim_{x \rightarrow a} f(x) =( a_{1} – a_{1})( a_{1} – a_{2}) . . . . (a_{1} – a_{n}) \)

 

Q30.

If \(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)

\(\lim_{x \rightarrow a} f(x)\) will exist for what values?

Soln:

Given function:

\(f(x) = \left\{\begin{matrix} \left | x \right | + 1, & x \leq 0\\ 0, & x = 0 \\ \left | x \right | – 1, & x > 0 \end{matrix}\right.\)

When a = 0

\(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0^{-}}\left ( \left | x \right | + 1 \right )\)

\(= \lim_{x \rightarrow 0}\left ( – x + 1 \right )\)   [if x < 0, |x| = -x]

= -0 + 1

= 1

\(\lim_{x \rightarrow 0^{+} f(x)} = \lim_{x \rightarrow 0^{+}}\left ( \left | x \right | – 1 \right )\)

\(= \lim_{x \rightarrow 0}\left ( x – 1 \right )\)   [if x > 0, |x| = x]

= 0 – 1

= -1

We have, \(\lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x)\)

\(∴ \lim_{x \rightarrow 0} f(x)\) doesn’t exist.

When a < 0

\(\lim_{x \rightarrow a^{-}}f(x) = \lim_{x \rightarrow a^{-}}(\left | x \right | + 1)\)

\(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)

= -a + 1

\(\lim_{x \rightarrow a^{+}}f(x) = \lim_{x \rightarrow a^{+}}(\left | x \right | + 1)\)

\(= \lim_{x \rightarrow a}(- x + 1) \; \; \; \; [ x < a < 0 \Rightarrow \left | x \right | = – x ]\)

= – a + 1

\(∴ \lim_{x \rightarrow a^{-} } = \lim_{x \rightarrow a^{+} } = – a + 1\)

Hence, at x = a limit of f(x) exist, where a > 0

Thus, \(\lim_{x \rightarrow a} f(x) \; exist \; for \; all \; a \neq 0\)

 

Q31.

Evaluate \(\lim_{x \rightarrow 1} f(x)\)

If f(x) satisfies, \(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\)

Soln:

\(\lim_{x \rightarrow 1} \frac{f(x) – 2}{x^{2} – 1} = \pi\)

\( \Rightarrow \frac{\lim_{x \rightarrow 1} \left ( f(x) – 2 \right )}{ \lim_{x \rightarrow 1}  \left ( x^{2} – 1 \right )} = \pi\)

\(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi \lim_{x \rightarrow 1} (x^{2} – 1)\)

\(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = \pi (1^{2} – 1)\)

\(\Rightarrow \lim_{x \rightarrow 1} (f(x) – 2) = 0 \)

\(\Rightarrow \lim_{x \rightarrow 1} f(x) – \lim_{x \rightarrow 1} 2 = 0 \)

\(\Rightarrow \lim_{x \rightarrow 1} f(x) – 2 = 0 \)

\(∴ \lim_{x \rightarrow 1} f(x) = 2\)

 

Q32.

If \(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\)

For what values of a and b does \(\lim_{x \rightarrow 0} f(x) \; and \; \lim_{x \rightarrow 1} f(x)\) exist?

 

Soln:

Given function

\(f(x) = \left\{\begin{matrix} ax^{2} + b, & x < 0 \\ bx + a, & 0 \leq x \leq 1 \\ bx^{3} + a & x > 1 \end{matrix}\right.\)

\(\lim_{x \rightarrow 0^{-} f(x)} = \lim_{x \rightarrow 0}(ax^{2} + b)\)

= a(0)2 + b

= b

\(\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0}(bx + a)\)

= b(1) + a

= a + b

\(∴ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} f(x)\)

Thus, for integral values of n and m \(\lim_{x \rightarrow 1}f(x)\) exist.

 

 

Exercise-13.2

Q1. Find derivative for x2 – 2 at x = 10

 

Soln:

Let f(x) = x2 – 2

Accordingly,

\(f'(10) = \lim_{h \rightarrow 0} \frac {f(10 + h) – f(10)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {[(10 + h)^{2} – 2] – (10^{2} – 2)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {10^{2} + 2. 10. h + h^{2} – 2 – 10^{2} + 2}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac {20h + h^{2}}{h} \)

 

\(= \lim_{h \rightarrow 0} (20 + h) = (20 + 0) = 0 \)

 

Hence, derivative for x2 – 2 at x = 10 is 20

 

 

Q2. Find derivative

99x at x = 100

 

Soln:

Let f(x) = 99x,

Accordingly,

\(f'(100) = \lim_{h \rightarrow 0} \frac{f(100 + h) – f(100)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{99(100 + h) – 99(100)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{99 \times 100 + 99h – 99 \times 100}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{ 99h }{h}\)

 

\( = \lim_{h \rightarrow 0} (99h) = 99 \)

 

Hence, derivative for 99x at x = 100 is 99

 

 

Q3. Find derivative

x at x = 1

 

Soln:

Let f(x) = x

Accordingly,

\(f'(1) = \lim_{h \rightarrow 0} \frac{f(1 + h) – f(1)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{(1 + h) – 1}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{h}{h}\)

 

\(= \lim_{h \rightarrow 0} (1)\)

 

= 1

Hence, derivative for x at x = 1 is 1

 

 

Q4. Using first principle find derivative

(i) x3 – 27

 

(ii) (x – 1) (x – 2)

 

(iii) \(\frac{1}{x^{2}}\)

 

(iv) \(\frac{x + 1}{x – 1}\)

Soln:

(i) Let f(x) = x3 – 27

From first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\( = \lim_{h \rightarrow 0}\frac{ [(x + h)^{3} – 27] – (x^{3} – 27)} {h} \)

 

\( = \lim_{h \rightarrow 0}\frac{ x^{3} + h^{3} + 3x^{2}h + 3xh^{2} – x^{3}}{h} \)

 

\( = \lim_{h \rightarrow 0}\frac{ h^{3} + 3x^{2}h + 3xh^{2}} {h} \)

 

\( = \lim_{h \rightarrow 0} \left ( h^{2} + 3x^{2} + 3xh \right ) \)

 

= 0 + 3x2 + 0 = 3x2

 

(ii) Let f(x) = (x – 1)(x – 2)

From first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(x + h – 1) (x + h – 2) – (x – 1) (x – 2)} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(x^{2} + hx – 2x + hx + h^{2} – 2h – x – h + 2 ) – (x^{2} – 2x – x + 2)} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{(hx + hx + h^{2} – 2h -h )} {h}\)

 

\(= \lim_{h \rightarrow 0}\frac{( 2hx + h^{2} – 3h )} {h}\)

 

\(= \lim_{h \rightarrow 0} ( 2x + h – 3 )\)

 

\( ( 2x + h – 3 )\)

 

= 2x – 3

 

(iii) Let \(f(x) = \frac{1}{x^{2}}\)

From first principle

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac{\frac{1}{(x + h)^{2}} – \frac{1}{x^{2}}}{h}\)

 

\( = \lim_{h \rightarrow 0}  \frac{1}{h} \left [\frac{ x^{2} – (x + h )^{2}}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0}  \frac{1}{h} \left [\frac{ x^{2} – x^{2} – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{ – h^{2} – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\( = \lim_{h \rightarrow 0} \left [\frac{ – h – 2hx}{x^{2} (x + h )^{2}}\right ]\)

 

\(\frac{0 – 2x}{x^{2}(x + 0)^{2}} = \frac{-2}{x^{3}}\)

 

(iv) Let \(f(x) = \frac{x +1}{x – 1}\)

From first principle,

\( = \lim_{h \rightarrow 0} \frac{\left ( \frac{x + h + 1}{x + h – 1} – \frac{x + 1}{x – 1} \right )}{h}\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x – 1 \right ) \left ( x+ h + 1 \right ) – \left ( x + 1 \right ) \left ( x + h – 1 \right )}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{\left ( x^{2} + hx + x – x – h – 1 \right ) – \left ( x^{2} + hx – x + x + h – 1 \right )}{\left ( x- 1 \right )\left ( x + h + 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \frac {1}{h} \left [ \frac{-2h}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\( = \lim_{h \rightarrow 0} \left [ \frac{-2}{\left ( x – 1 \right ) \left ( x+ h – 1 \right )} \right ]\)

 

\(= \frac{-2}{\left ( x – 1 \right )\left ( x – 1 \right )} = \frac{-2}{\left ( x – 1 \right )^{2}}\)

 

 

Q5. Prove \(f'(1) = 100 f'(0)\)

For the function \(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)

 

Soln:

Given function

\(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1\)

 

\(\frac{d}{dx} f(x) = \frac{d}{dx}\left [ \frac{x^{100}}{100} + \frac{x^{99}}{99} + … + \frac{x^{2}}{2} + x + 1 \right ]\)

 

\(\frac{d}{dx} f(x) = \frac{d}{dx} \left ( \frac{x^{100}}{100} \right ) + \frac{d}{dx}\left ( \frac{x^{99}}{99} \right ) + … + \frac{d}{dx}\left ( \frac{x^{2}}{2} \right ) + \frac{d}{dx}\left ( x \right )+ \frac{d}{dx}\left ( 1 \right )\)

 

Using theorem \(\frac{d}{dx} \left ( x^{n} \right ) = n x^{n – 1}, \) we get

 

\(\frac{d}{dx} f(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + . . . . + \frac{2x}{2} + 1 + 0\)

 

\(= x^{99} + x^{98} + …. + x + 1\)

 

∴ \(\frac{d}{dx} f(x) = x^{99} + x^{98} + …. + x + 1\)

 

At x = 0

 

f'(0) = 1

 

At x = 1,

 

f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100

 

Hence,  f’(1) = 100 x f1 (0)

 

 

Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an

Soln:

Let \(f(x) = x ^{n} +  ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n}\)

 

\(f’(x) = \frac{d}{dx} \left ( x ^{n} + ax ^{n – 1} + a^{2} x ^{n – 2} + . . . + a^{n – 1}x + a^{n} \right ) \)

 

\(= \frac{d}{dx} \left ( x ^{n} \right ) + a\frac{d}{dx} \left ( x ^{n – 1} \right ) + a^{2} \frac{d}{dx} \left ( x ^{n – 2} \right ) + . . . + a^{n – 1}\frac{d}{dx} \left ( x \right ) + a^{n} \frac{d}{dx} \left ( 1 \right )\)

 

Using \(\frac{d}{dx}x^{n} = nx^{n -1}\), we have

 

\(f'(x) = nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1} + a^{n}(0)\)

 

\(= nx^{n – 1} + a(n – 1)x^{n – 2} + a^{2} (n – 2)x^{n -3} + … + a^{ n – 1}\)

 

 

Q7. Find the derivative for

(i) (x – m)(x – n)

 

(ii) (ax2 + b)2

 

(iii) \(\frac{x – a}{x – b}\)

Soln:

(i) Let f(x) = (x – m)(x – n)

 

\(\Rightarrow f(x) = x^{2} – (m + n) x + mn\)

 

\(f'(x) = \frac{d}{dx}(x^{2} – (m + n)x + mn)\)

 

\(= \frac{d}{dx}(x^{2}) – (m + n) \frac{d}{dx}(x) + \frac{d}{dx}(mn)\)

 

Using \(= \frac{d}{dx}(x^{n}) = nx^{n – 1}\), we get

 

\(f'(x) = 2x -(m + n) + 0 = 2x – m – n\)

 

(ii) Let f(x) = (ax2 + b)2

 

\(\Rightarrow f(x) = a^{2}x^{2} + 2abx + b^{2}\)

 

\(f'(x) = \frac{d}{dx} (a^{2}x^{4} + 2abx^{2} + b^{2}) = a^{2}\frac{d}{dx}(x^{4}) + 2ab\frac{d}{dx}(x^{2}) + \frac{d}{dx}(b^{2})\)

 

Using \(\frac{d}{dx}x^{n} = nx^{n – 1}\), we have

 

\(f’\left ( x \right ) = a ^{2}(4x^{3}) + 2ab (2x) + b^{2}(0)\)

 

= 4a2x3 + 4abx

 

= 4ax(ax2 + b)

 

(iii) Let \(f(x) = \frac{(x – a)}{(x – b)}\)

 

\(f'(x) = \frac{d}{dx} \left (\frac{x – a}{x – b} \right )\)

 

Quotient rule,

\(f'(x) = \frac {(x – b)\frac{d}{dx}(x – a) – (x – a)\frac{d}{dx}(x- b)}{(x – b)^{2}}\)

 

\(= \frac {(x – b)(1) – (x – a)(1)}{(x – b)^{2}}\)

 

\(= \frac {x – b – x + a}{(x – b)^{2}}\)

 

\(= \frac { a – b }{(x – b)^{2}}\)

 

 

Q8. If a is constant find derivative of \(\frac {x^{n} – a^{n}}{x – a}\)

Soln:

Let \(\frac {x^{n} – a^{n}}{x – a}\)

 

\(\Rightarrow f'(x) = \frac{d}{dx}\left ( \frac {x^{n} – a^{n}}{x – a} \right )\)

 

By Question rule

\(\Rightarrow f'(x) = \frac{(x – a) \frac{d}{dx}(x^{n} – a^{n}) – (x^{n} – a^{n})\frac{d}{dx}(x – a)}{(x – a)^{2}}\)

 

\(= \frac{(x -a)(nx^{n – 1} – 0) – (x^{n} – a^{n})}{(x – a)^{2}}\)

 

\(= \frac{nx^{n} – anx^{n – 1} – x^{n} + a^{n}}{(x – a)^{2}}\)

 

 

Q9.

(i) \( 2x – \frac {3}{4} \)

 

(ii) (5x3 + 3x – 1) (x – 1)

 

(iii) x-3 (5 + 3x)

 

(iv) x5 (3 – 6x-9)

 

(v) x-4 (3 – 4x-5)

 

(vi) \(= \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)

 

Soln.

(i) Let \( f(x) = 2x – \frac {3}{4} \)

 

\(f'(x) = \frac{d}{dx}\left ( 2x – \frac{3}{4} \right )\)

 

\(= 2\frac{d}{dx}\left ( x \right ) – \frac {d}{dx} \left ( \frac{3}{4} \right )\)

 

= 2 – 0

 

= 2

 

(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)

By Leibnitz product rule,

\(f'(x) = (5x^{3} + 3x – 1) \frac{d}{dx}(x – 1) + (x – 1) \frac{d}{dx}(5x^{3} + 3x – 1 )\)

 

= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)

 

= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)

 

= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3

 

= 20x3 – 15x2 + 6x – 4

 

(iii) Let f(x) = x-3 (5 + 3x)

 

\(f'(x) = x^{-3}\frac{d}{dx}(5 + 3x) + (5 + 3x)\frac{d}{dx}(x^{-3})\)

 

\( = x^{-3}(0 + 3) + (5 + 3x) (-3x^{-3 – 1} )\)

 

\( = x^{-3}( 3) + (5 + 3x) (-3x^{-4} )\)

 

= 3x-3 – 15x-4 – 9x-3

 

= -6x-3 – 15x-4

 

\(= -3 x ^{-3} \left ( 2 + \frac{5}{x} \right )\)

 

\(= \frac{-3 x ^{-3} }{x}\left ( 2x + 5 \right )\)

 

\(= \frac{-3 }{ x ^{4} }\left ( 2x + 5 \right )\)

 

(iv) let f(x) = x5 (3 – 6x-9)

 

From Leibnitz product rule,

\(f'(x) = x^{5} \frac{d}{dx}(3 – 6x^{-9}) + (3 – 6x^{-9}) \frac{d}{dx}(x^{5})\)

 

\(= x^{5} \left \{ 0 – 6 (-9)x^{-9 – 1} \right \} + (3 – 6x^{-9})(5x^{4})\)

 

\(= x^{5} (54x ^{-10}) + 15x^{4} – 30x^{-5}\)

 

= 54x-5 + 15x4 – 30x-5

 

= 24x-5 + 15x4

 

= \(= 15x^{4} + \frac{24}{x^{5}}\)

 

(v) Let f(x) = x-4 (3 – 4x-5)

 

From Leibnitz product rule,

\(f'(x) = x^{-4}\frac{d}{dx}\left ( 3 – 4x^{-5} \right ) + (3 – 4 x^{-5})\frac{d}{dx}(x^{-4})\)

 

\(= x^{-4}\left \{ 0 – 4 (-5)x^{-5 – 1} + (3 – 4x^{-5})(-4)x^{-4 -1} \right \}\)

 

= x-4 (20x-6) + (3 – 4x-5)(-4x-5)

 

= 36x-10 – 12x-5

 

\(= -\frac{12}{x^{5}} + \frac{36}{x^{10}}\)

 

(iv) Let \( f(x) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}\)

\(f'(x) = \frac{d}{dx}\left ( \frac{2}{x + 1} \right ) – \frac{d}{dx} \left ( \frac{x^{2}}{3x – 1} \right )\)

 

From quotient rule,

\(f'(x) = \left [ \frac{\left ( x + 1 \right ) \frac{d}{dx}(2) – 2 \frac{d}{dx} (x + 1) }{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1) \frac{d}{dx} (x^{2}) – x^{2} \frac{d}{dx}\left ( 3x – 1 \right )}{(3x – 1)^{2}} \right ]\)

 

\(= \left [ \frac{(x + 1)(0) – 2(1)}{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1)(2x) – (x^{2})(3)}{(3x – 1)^{2}} \right ]\)

 

\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{6x^{2} – 2x – 3x^{2}}{(3x – 1)^{2}} \right ]\)

 

\(= \frac{-2}{(x + 1)^{2}} – \left [ \frac{3x^{2} – 2x} {(3x – 1)^{2}} \right ]\)

 

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

 

 

Q10. Using first principle find derivative of \(\cos x\)

Soln:

Let \(f(x) = \cos x\).

 

According to first principle

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{\cos (x + h) – \cos x}{h}\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \right ]\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h) – \sin x \sin h }{h} \right ]\)

 

\(= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h)}{h} – \frac{\sin x \sin h }{h} \right ]\)

 

\(=- \cos x \left ( \lim_{h \rightarrow 0} \frac{1 – \cosh}{h} \right ) – \sin x\lim_{h \rightarrow 0}\left ( \frac{sin h}{h} \right )\)

 

= \(= – \cos x (0) – \sin x (1)\)   \(\left [ \lim_{h \rightarrow 0} \frac{1 – cos h}{h} = 0 \; and \; \lim_{h \rightarrow 0}\frac{\sin h }{h} = 1 \right ]\)

 

= – sin x

 

f’(x) = – sin x

 

 

Q11.

(i) sin x cos x

 

(ii) sec x

 

(iii) 5 sec x + 4 cos x

 

(iv) cosec x

 

(v) 3cot x + 5cosec x

 

(vi) 5sin x – 6cos x + 7

 

(vii) 2tan x – 7sec x

Soln:

(i) Let f’(x) = sin x cos x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) -f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{\sin (x + h) \cos(x + h) – \sin x \cos x}{h}\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2\sin (x + h) \cos (x + h) – 2 \sin x cos x \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ \sin 2(x + h) – \sin 2x \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2 \cos \frac{2x + 2h + 2x}{2} . \sin \frac{2x + 2h – 2x}{2} \right ]\)

 

\(= \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos \frac{ 4x + 2h }{2}  \sin \frac{ 2h }{2} \right ]\)

 

\(\lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (2x + h ) \sin h \right ]\)

 

\(\lim_{h \rightarrow 0} \cos (2x + h ) \lim_{h \rightarrow 0} \frac{\sin h}{h}\)

 

= cos(2x + 0). 1

 

= cos 2x

 

(ii) Let f(x) = sec x

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{\sec (x + h) – \sec x}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos(x + h)}{\cos x \cos(x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \right ) \sin \left ( \frac{x – x – h}{2} \right )}{\cos (x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{2x + h}{2} \right ) \sin \left ( \frac{- h}{2} \right )}{\cos (x + h)} \right ]\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\left [ \sin \left ( \frac{2x + h}{2} \right ) \frac{\sin \left ( \frac{h}{2}\right )}{\left (\frac{h}{2}\right )} \right ]}{\cos (x + h)}\)

 

\(= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}. \lim_{h \rightarrow 0}\frac{\sin\left ( \frac{2x + h}{2} \right ) }{\cos (x + h)}\)

 

\(= \frac{1}{\cos x} . 1 . \frac{\sin x}{ \cos x}\)

 

\(= \sec x \tan x\)

 

(iii)

Let f(x) = 5 sec x + 4 cos x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(\lim_{h \rightarrow 0}\frac{5\sec (x + h) + 4 \cos(x + h) – \left [ 5 \ sec x + 4 \ cos x \right ]}{h}\)

 

\(= 5\lim_{h \rightarrow 0}\frac{[\sec(x + h) – \sec x]}{h} + 4 \lim_{h \rightarrow 0}\frac{[\cos(x + h) – \cos x]}{h}\)

 

\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)

 

\(= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{\cos x – \cos (x + h)}{\cos x \cos (x + h)} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h}\left [ \cos x \cos h – \sin x \sin h – \cos x \right ]\)

 

\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{x + x + h}{2} \right )\sin \left ( \frac{x – x – h}{2} \right )}{cos(x + h)} \right ] + 4 \lim_{h\rightarrow 0}\frac{1}{h} [-\cos x (1 – \cos h) – \sin x sin h]\)

 

\(\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{ 2x + h}{2} \right )\sin \left ( \frac{ – h}{2} \right )}{cos(x + h)} \right ] + 4\left [ – \cos x \lim_{h \rightarrow 0} \frac{(1 – \cos h)}{h} – \sin x \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ] \)

 

\(= \frac{5}{\cos x}. \lim_{h \rightarrow 0} \left [ \frac{\sin\left ( \frac{2x + h}{2} \right ). \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}}{\cos (x + h)}\right ] + 4 [(-\cos x).(0) – (\sin x). 1]\)

 

\(= \frac{5}{\cos x}. \left [ \lim_{h \rightarrow 0} \frac{\sin\left ( \frac{2x + h}{2} \right )}{\cos (x + h)}. \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ] + 4 \sin x\)

 

\(= \frac{5}{\cos x}. \frac{\sin x}{\cos x}. 1 – 4 \sin x\)

 

\(= 5 \sec x \tan x – 4 \sin x\)

 

(iv)

Let f(x) = cosec x

 

According to first principle

\(f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}\)

 

\(f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ] \)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin(x + h)}{ \sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ). \sin \left ( \frac{x – x – h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ). \sin \left ( \frac{- h}{2} \right ) }{\sin (x + h) . \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )} }{\sin (x + h) . \sin x}\)

 

\(= \lim_{h \rightarrow 0} \left ( \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). }{\sin (x + h) . \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)

 

\(= \left ( \frac{-\cos x}{\sin x \sin x} \right ).1\)

 

= – cosec x cot x

 

(v)

Let f(x) = 3cot x + 5cosec x

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{3 \cot (x + h) + 5 \csc (x + h) – 3 \cot x – 5 \csc x}{h}\)

 

\(= 3 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] + 5 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \csc x]\) ….. (1)

 

Now,

\( \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] \)

 

= \( \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos (x + h)}{\sin (x + h)} – \frac{\cos x}{\sin x} \right ] \)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos(x + h)\sin x – \cos x \sin (x + h)}{\sin x sin(x + h)} \right ]\)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x – x – h)}{\sin x sin(x + h)} \right ]\)

 

= \(\lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (- h)}{\sin x sin(x + h)} \right ]\)

 

= \(- \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ). \left ( \lim_{h \rightarrow 0} \frac{1}{\sin x. \sin (x + h)} \right )\)

 

= \(- 1 . \frac{1}{\sin x . \sin (x + 0)} = \frac{-1}{\sin^{2}x} = -\csc ^{2}x\) ……. (2)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin (x + h) }{\sin x \sin (x + h)} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ) . \sin \left ( \frac{x – x – h}{2} \right )}{\sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ) . \sin \left ( \frac{- h}{2} \right )}{\sin (x + h) \sin x} \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{ -\cos \left ( \frac{ 2x + h}{2} \right ) . \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}}{\sin (x + h) \sin x}\)

 

\(= \lim_{h \rightarrow 0} \left ( \frac{ -\cos \left ( \frac{ 2x + h}{2} \right )} {\sin (x + h) \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}\)

 

\(= \left ( \frac{- \cos x }{\sin x \sin x} \right ) . 1\)

 

\(= – \csc x \cot x\) ………. (3)

 

From eqn (1), (2), and (3), we obtain

\(f'(x) = – 3 \csc^{2} x – 5 \csc x \cot x\)

 

(vi)

Let f(x) = 5sin x – 6cos x + 7

 

According to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \sin(x + h) – 6 \cos (x + h) + 7 – 5 \sin x + 6 \cos x – 7 \right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \left \{ \sin(x + h) – \sin x \right \} – 6 \left \{ \cos (x + h) – \cos x \right \} \right ]\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sin(x + h) – \sin x \right ] – 6 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \cos (x + h) – \cos x \right ]\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{x + x + h}{2} \right )  \sin \frac{x + h – x}{2} \right ] – 6 \lim_{h \rightarrow 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h}\)

 

\(= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{ 2x + h}{2} \right ) \sin \frac{ h }{2} \right ] – 6 \lim_{h \rightarrow 0} \left [  \frac{ – \cos x (1 – \cos h ) – \sin x \sin h }{h} \right ] \)

 

\(= 5 \lim_{h \rightarrow 0} \left ( \cos\left ( \frac{ 2x + h}{2} \right ) \frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ) – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h )}{h} – \frac {\sin x \sin h }{h} \right ]\)

 

\(= 5 \left [ \lim_{h \rightarrow 0} \cos\left ( \frac{ 2x + h}{2} \right ) \right ] \left [ \lim_{\frac{h}{2} \rightarrow 0}\frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ] – 6 \left [ (- \cos x ) \left ( \lim_{h \rightarrow 0} \frac{ (1 – \cos h )}{h} \right ) – \sin x \lim_{h \rightarrow 0} \left ( \frac {\sin h }{h} \right ) \right ]\)

 

\(= 5 \cos x. 1 – 6 [(-\cos x). (0) – \sin x.1]\)

 

= 5cos x + 6 sin x

 

(vii)

Let f (x) = 2 tan x – 7 sec x

 

Accordingly to first principle,

\(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \tan (x + h) – 7 \sec (x + h) – 2 \tan x + 7 \sec x\right ]\)

 

\(= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \left \{ \tan (x + h) – \tan x \right \} – 7 \left \{ \sec (x + h) – \sec x \right \} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \tan (x + h) – \tan x \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sec (x + h) – \sec x \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h)}{\cos (x + h)} – \frac{\sin x}{\cos x} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h) \cos x – \sin x \cos (x + h) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos (x + h) }{\cos x \cos (x + h)} \right ]\)

 

\(= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{ \sin (x + h – x) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \sin \left ( \frac{x – x – h}{2} \right ) \right )}{\cos x \cos (x + h)} \right ]\)

 

\(= 2 \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ) \left ( \lim_{h \rightarrow 0} \frac{ 1 }{ \cos x \cos (x + h)} \right ) – 7 \left( \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ) \left ( \lim_{h \rightarrow 0} \frac{ \sin \left ( \frac{ 2x + h}{2} \right ) }{\cos x \cos (x + h)} \right )\)

 

\(= 2 . 1 . \frac{1}{\cos x \cos x} – 7 . 1 \left ( \frac{\sin x}{\cos x \cos x} \right )\)

 

\(= 2 \sec^{2} x – 7 \sec x \tan x\)

The limits and derivatives chapter introduces the students to the vast topic of calculus. Several important concepts are introduced to the students like what are derivatives and limits, limits of different functions like polynomials and rational functions, trigonometric functions, derivatives of polynomials and trigonometric functions, etc.

The topics taught in this chapter are extremely crucial as these concepts will be continued in class 12 also. Apart from that, the limits and derivatives concepts are extremely crucial for the different competitive exams like JEE as a lot of questions related to it are included.

In this chapter, several example questions are included to help the students get an idea of how limits and derivatives questions are to be solved. Then, students are required to solve the NCERT questions given in the book to get completely acquainted with the different question variations and develop confidence in tackling the calculus questions. Students can always refer to these NCERT solutions class 11 Maths limits and derivatives (chapter 13) here and clear their doubts.

Stay tuned with BYJU’S to get the complete NCERT Solutions For Class 11 Maths chapters. These NCERT solutions will not only help the students to clear all their doubts instantly but also help them to learn the in-depth maths concepts in a better and more effective way.