NCERT Solutions Class 12 Maths Chapter 5 Exercise 5.2 - Continuity and Differentiability

1. Differentiate the function with respect to $x$. $\sin \left( {{x^2} + 5} \right)$

Ans: Let $f(x) = \sin \left( {{x^2} + 5} \right),u(x) = {x^2} + 5$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v\left( {{x^2} + 5} \right)$

$ = \tan \left( {{x^2} + 5} \right) = f(x)$

Thus, $f$ is a composite of two functions.

Put $t = u(x) = {x^2} + 5$

Then, we obtain $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos \left( {{x^2} + 5} \right)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} + 5} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}(5) = 2x + 0 = 2x$

Therefore, by chain rule. $\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos \left( {{x^2} + 5} \right) \times 2x$

$ = 2x\cos \left( {{x^2} + 5} \right)$

2. Differentiate the functions with respect of ${\text{x}}$. $\cos (\sin x)$

Ans: Let $f(x) = \cos (\sin x),u(x) = \sin x$, and $v(t) = \cos t$

Then, $(\operatorname{vou} )(x) = v(u(x)) = v(\sin x) = \cos (\sin x) = f(x)$

Thus, $f$ is a composite function of two functions.

Put $t = u(x) = \sin x$

$\therefore \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[\cos t] =  - \sin t =  - \sin (\sin x)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sin x) = \cos x$

By chain rule, $\dfrac{{df}}{{dx}},\dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} =  - \sin (\sin x) \cdot \cos x$

$ =  - \cos x\sin (\sin x)$

3. Differentiate the functions with respect of ${\text{x}}$.

 $\sin (ax + b)$

Ans: Let $f(x) = \sin (ax + b),u(x) = ax + b$, and $v(t) = \sin t$

Then, $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = f(x)$

Thus, $f$ is a composite function of two functions $u$ and $v$.

Put $t = u(x) = ax + b$

Therefore, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Hence, by chain rule, we obtain

$\dfrac{{df}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a$

$ = a\cos (ax + b)$

4. Differentiate the functions with respect of ${\text{x}}$. 

$\sec (\tan (\sqrt x ))$

Ans: Let $f(x) = \sec (\tan (\sqrt x )),u(x) = \sqrt x ,v(t) = \tan t$, and $w(s) = \sec s$

Then, $(vou)(x) = w[v(u(x))] = w[v(\sqrt x )] = w(\tan \sqrt x ) = \sec (\tan \sqrt x ) = f(x)$

Thus, $f$ is a composite function of three functions, $u, v$ and $w$. Put $s = v(t) = \tan t$ and $t = u(x) = \sqrt x $

$[s = \tan t]$

Then, $\dfrac{{dw}}{{ds}} = \dfrac{d}{{ds}}(\sec s) = \sec s\tan s$

$ = \sec (\tan t) \cdot \tan (\tan t)$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x )\quad [t = \sqrt x ]$

$\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}(\tan t) = {\sec ^2}t = {\sec ^2}\sqrt x $

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)$

$ = \dfrac{1}{2} \cdot {x^{\frac{1}{2} - 1}} = \dfrac{1}{{2\sqrt x }}$

Hence, by chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dw}}{{ds}} \cdot \dfrac{{ds}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ = \sec (\tan \sqrt x ) \cdot \tan (\tan \sqrt x ) \times {\sec ^2}\sqrt x  \times \dfrac{1}{{2\sqrt x }}$

$ = \dfrac{1}{{2\sqrt x }}{\sec ^2}\sqrt x (\tan \sqrt x )\tan (\tan \sqrt x )$

$ = \dfrac{{{{\sec }^2}\sqrt x \sec (\tan \sqrt x )\tan (\tan \sqrt x )}}{{2\sqrt x }}$

5. Differentiate the functions with respect of ${\text{X}}$. $\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}$

Ans: The given function is $f(x) = \dfrac{{\sin (ax + b)}}{{\cos (cx + d)}} = \dfrac{{g(x)}}{{h(x)}},$where $g(x) = \sin (ax + b)$ and

$h(x) = \cos (cx + d)$

$\therefore f = \dfrac{{{g^\prime }h - g{h^\prime }}}{{{h^2}}}$

Consider $g(x) = \sin (ax + b)$

Let $u(x) = ax + b,v(t) = \sin t$

Then $(vou)(x) = v(u(x)) = v(ax + b) = \sin (ax + b) = g(x)$

$\therefore g$ is a composite function of two functions, $u$ and $v$.

Put $t = u(x) = ax + b$

$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\sin t) = \cos t = \cos (ax + b)$

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(ax + b) = \dfrac{d}{{dx}}(ax) + \dfrac{d}{{dx}}(b) = a + 0 = a$

Therefore, by chain rule, we obtain ${g^\prime } = \dfrac{{dg}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}} = \cos (ax + b) \cdot a = a\cos (ax + b)$

Consider $h(x) = \cos (cx + d)$

Let $p(x) = cx + d,q(y) = \cos y$

Then, $(qop)(x) = q(p(x)) = q(cx + d) = \cos (cx + d) = h(x)$

$\therefore h$ is a composite function of two functions, $p$ and $q$.

Put $y = p(x) = cx + d$

$\dfrac{{dq}}{{dy}} = \dfrac{d}{{dy}}(\cos y) =  - \sin y =  - \sin (cx + d)$

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(cx + d) = \dfrac{d}{{dx}}(cx) + \dfrac{d}{{dx}}(d) = c$

Therefore, by chain rule, we obtain ${h^\prime } = \dfrac{{dh}}{{dx}} = \dfrac{{dq}}{{dy}} \cdot \dfrac{{dy}}{{dx}} =  - \sin (cx + d){\text{xc}} =  - c\sin (cx + d)$

$\therefore {f^\prime } = \dfrac{{a\cos (ax + b) \cdot \cos (cx + d) - \sin (ax + b)\{  - c\sin cx + d\} }}{{{{[\cos (cx + d)]}^2}}}$

$ = \dfrac{{a\cos (ax + b)}}{{\cos (cx + d)}} + c\sin (ax + b) \cdot \dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} \times \dfrac{1}{{\cos (cx + d)}}$

$ = a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$

6. Differentiate the function with respect to ${\text{x}}$. 

$\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

Ans: $\cos {x^3} \cdot {\sin ^2}\left( {{x^5}} \right)$

$\dfrac{d}{{dx}}\left[ {\cos {x^3} \cdot {{\sin }^2}\left( {{x^5}} \right)} \right] = {\sin ^2}\left( {{x^5}} \right){\text{x}}\dfrac{d}{{dx}}\left( {\cos {x^3}} \right) + \cos {x^3}{\text{x}}\dfrac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^5}} \right)} \right]$

$ = {\sin ^2}\left( {{x^5}} \right){\text{x}}\left( { - \sin {x^3}} \right){\text{x}}\dfrac{d}{{dx}}\left( {{x^3}} \right) + \cos {x^3} + 2\sin \left( {{x^5}} \right) \cdot \dfrac{d}{{dx}}\left[ {\sin {x^5}} \right]$

The given function is

$ = \sin {x^3}{\sin ^2}\left( {{x^5}} \right) \times 3{x^2} + 2\sin {x^5}\cos {x^3} \cdot \cos {x^5} \times \dfrac{d}{{dx}}\left( {{x^5}} \right)$

$ = 3{x^2}\sin {x^3} \cdot {\sin ^3}\left( {{x^5}} \right) + 2\sin {x^5}\cos {x^5}\cos {x^3} \cdot x5{x^4}$

$ = 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}\sin {x^3}{\sin ^2}\left( {{x^5}} \right)$

7. Differentiate the functions with respect to ${\text{x}}$. $2\sqrt {\cot \left( {{x^2}} \right)} $

Ans: $\dfrac{d}{{dx}}\left[ {2\sqrt {\cot \left( {{x^2}} \right)} } \right]$

$ = 2 \cdot \dfrac{1}{{2\sqrt {\cot \left( {{x^2}} \right)} }} \times \dfrac{{\text{d}}}{{{\text{dx}}}}\left[ {\cot \left( {{x^2}} \right)} \right]$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}} {\text{x}} - {\operatorname{cosec} ^2}\left( {{x^2}} \right) \times \dfrac{d}{{dx}}\left( {{x^2}} \right)$

$ = \sqrt {\dfrac{{\sin \left( {{x^2}} \right)}}{{\cos \left( {{x^2}} \right)}}}  \times \dfrac{1}{{{{\sin }^2}\left( {{x^2}} \right)}}x(2x)$

$ = \dfrac{{ - 2x}}{{\sqrt {\cos {x^2}\sqrt {\sin {x^2}\sin {x^2}} } }}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sqrt {2\sin {x^2}\cos {x^2}} \sin {x^2}}}$

$ = \dfrac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}$

8. Differentiate the functions with respect to $x$ $\cos (\sqrt x )$

Ans: Let $f(x) = \cos (\sqrt x )$

Also, let $u(x) = \sqrt x $

And, $v(t) = \cos t$

Then, $(vou)(x) = v(u(x))$

$ = v(\sqrt x )$

$ = \cos \sqrt x $

$ = f(x)$

Clearly, $f$ is a composite function of two functions, $u$ and $v$, such that $t = u(x) = \sqrt x $

Then,

$\dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}(\sqrt x ) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)$

$\dfrac{1}{2}{x^{ - \dfrac{1}{2}}} = \dfrac{1}{{2\sqrt x }}$

And, $\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\cos t) =  - \sin t =  - \sin \sqrt x $

By using chain rule, we obtain $\dfrac{{dt}}{{dx}} = \dfrac{{dv}}{{dt}} \cdot \dfrac{{dt}}{{dx}}$

$ =  - \sin (\sqrt x ) \cdot \dfrac{1}{{2\sqrt x }}$

$ =  - \dfrac{1}{{2\sqrt x }}\sin (\sqrt x )$

$ =  - \dfrac{{\sin (\sqrt x )}}{{2\sqrt x }}$

9. Prove that the function $f$ given by $f(x) = |x - 1|,x \in {\mathbf{R}}$ is not differentiable at $x = 1$.

Ans: The given function is $f(x) = |x - 1|,x \in {\mathbf{R}}$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{k \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiability of the given function at $x = 1$, Consider the left hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f|I + h - 1||1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - h}}{h}\quad (h < 0 \Rightarrow |h| =  - h)$

$ =  - 1$

Consider the right hand limit of $f$ at $x = 1$

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f|I + h - 1| - |1 - 1|}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{|h| - 0}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{h}{h}\quad (h > 0 \Rightarrow |h| = h)$

$ = 1$

Since the left and right hand limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

10. Prove that the greatest integer function defined by $f = (x) = [x],0 < x < 3$ is not differentiable at $x = 1$ and $x = 2$.

Ans: The given function $f$ is $f = (x) = [x],0 < x < 3$

It is known that a function $f$ is differentiable at a point $x = c$ in its domain if both $\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(c + h) - f(c)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(c + h) - f(c)}}{h}$ are finite and equal.

To check the differentiable of the given function at $x = 1$, consider the left hand limit of $f$ at

$x = 1$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[1 + h] - [1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{0 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ - }}  = \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ $\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(1 + h) - f(1)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[1 + h][1]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 1}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right limits of $f$ at $x = 1$ are not equal, $f$ is not differentiable at $x = 1$

To check the differentiable of the given function at $x = 2$, consider the left hand limit of $f$ at

$x = 2$

$\mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{ - 1}}{h} = \infty $

Consider the right hand limit of $f$ at $x = 1$ 

$\mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(2 + h) - f(2)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{[2 + h] - [2]}}{h}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{1 - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right hand limits of $f$ at $x = 2$ are not equal, $f$ is not differentiable at $x = 2$

Conclusion

Continuity and Differentiability are crucial for a solid foundation in Math. Referring to these NCERT Solutions by Vedantu students can significantly enhance your understanding of continuity and differentiability. Class 12 Ex 5.2 focuses on the continuity of functions, a crucial topic that forms the basis for understanding differentiability and further calculus concepts. Pay attention to the step-by-step solutions provided in Class 12 Maths Ex 5.2, grasp the underlying principles, and ensure clarity on the concepts before moving forward.

Class 12 Maths Chapter 5: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.