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

Question:2. Differentiate the functions with respect to x in

$\cos ( \sin x )$

Answer:

Given function is
$f(x)= \cos ( \sin x )$
Lets take $t = \sin x$ then,
$f(t) = \cos t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ ( By chain rule)
$\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)$
$\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x$
Therefore, the answer is $-\sin(\sin x).\cos x$

Question:3. Differentiate the functions with respect to x in

$\sin (ax +b )$

Answer:

Given function is
$f(x) = \sin (ax +b )$
when we differentiate it w.r.t. x.
Lets take $t = ax+b$ . then,
$f(t) = \sin t$
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)$
$\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a$
Now,
$\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a$
Therefore, the answer is $a \cos (ax+b)$

Question:4. Differentiate the functions with respect to x in

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

Answer:

Given function is
$f(x)=\sec (\tan (\sqrt x) )$
when we differentiate it w.r.t. x.
Lets take $t = \sqrt x$ . then,
$f(t) = \sec (\tan t)$
take $\tan t = k$. then,
$f(k) = \sec k$
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ (By chain rule)
$\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)$
$(\because k = \tan t \ and \ t = \sqrt x)$
$\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)$
$\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}$
Now,
$\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} =\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}$
Therefore, the answer is $\frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}$

Question:5. Differentiate the functions with respect to x in

$\frac{\sin (ax +b )}{\cos (cx + d)}$

Answer:

Given function is
$f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}$
We know that,
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}$
$g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$
Lets take $u = (ax+b) \ and \ v = (cx+d)$
Then,
$\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c$
$g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}$ (By chain rule)
$\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)$
$\frac{du}{dx} = \frac{d(ax+b)}{dx} = a$
$g^{'}(x)=a\cos (ax+b)$ -(i)
Similarly,
$h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}$
$\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))$
$\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c$
$h^{'}(x)=-c\sin(cx+d)$ -(ii)
Now, put (i) and (ii) in
$f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}$
$= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}$
$= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$
Therefore, the answer is $a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)$

Question:6. Differentiate the functions with respect to x in

$\cos x^3 . \sin ^ 2 ( x ^5 )$

Answer:

Given function is
$f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )$
Differentitation w.r.t. x is
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)$
$g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)$
Lets take $u = x^3 \ and \ v = x^5$
Our functions become,
$\cos x^3 = \cos u$ and $\sin^2(x^5) = \sin^2v$
Now,
$g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}$ ( By chain rule)
$\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)$
$\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2$
$g^{'}(x) = -\sin x^3.3x^2$ -(i)
Similarly,
$h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}$
$\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5 \ \ \ (\because v = x^5)$

$\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4$
$h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5$ -(ii)
Put (i) and (ii) in
$f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) = -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5\cos x^5$
Therefore, the answer is $10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$

Question:7. Differentiate the functions with respect to x in

$2 \sqrt { \cot ( x^2 )}$

Answer:

Give function is
$f(x)=2 \sqrt { \cot ( x^2 )}$
Let's take $t = x^2$
$f(t) = 2\sqrt{\cot t}$
Now, take $\cot t = k^2$
$f(k) = 2k$
Differentiation w.r.t. x
$\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx}$ -(By chain rule)
$\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2$
$\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}} \ \ \ (\because t = x^2)$
$\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x$
So,
$\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} }$ ( Multiply and divide by $\sqrt 2$ and multiply and divide $\sqrt {\cot x^2}$ by $\sqrt{\sin x^2}$

$(\because \cot x = \frac{\cos x}{\sin x} \ and \ cosec x = \frac{1}{\sin x } )$
$=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}} \ \ \ \ (\because 2\sin x\cos x=\sin2x)$
There, the answer is $\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}$

Question:8 Differentiate the functions with respect to x in

$\cos ( \sqrt x )$

Answer:

Let us assume : $y\ =\ \cos ( \sqrt x )$

Differentiating y with respect to x, we get :

$\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}$

or $=\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}$

or $=\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}$

Question:9. Prove that the function f given by$f (x) = |x-1 | , x \epsilon R$ is not differentiable at x = 1.

Answer:

Given function is

$f(x) = |x - 1|, \quad x \in \mathbb{R}$

We know that any function is differentiable when both

$\lim\limits_{h \to 0^-} \frac{f(c + h) - f(c)}{h}$ and $\lim\limits_{h \to 0^+} \frac{f(c + h) - f(c)}{h}$ are finite and equal.

The required condition for the function to be differentiable at $x = 1$ is

$\lim\limits_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}$.

Now, the Left-hand limit of the function at $x = 1$ is

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

$= \lim\limits_{h \to 0^-} \frac{-h}{h} = -1 \quad (\because h < 0)$.

The Right-hand limit of the function at $x = 1$ is

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

$= \lim\limits_{h \to 0^+} \frac{h}{h} = 1$.

Now, it is clear that

$\text{R.H.L. at } x = 1 \neq \text{L.H.L. at } x = 1$.
Therefore, function $f (x) = |x-1 |$ is not differentiable at x = 1

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

x = 1 and x = 2.

Answer:

Given function is

$f(x) = \lfloor x \rfloor, \quad 0 < x < 3$

We know that any function is differentiable when both

$\lim\limits_{h \to 0^-} \frac{f(c + h) - f(c)}{h}$ and $\lim\limits_{h \to 0^+} \frac{f(c + h) - f(c)}{h}$ are finite and equal.

The required condition for the function to be differentiable at $x = 1$ is

$\lim\limits_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}$.

Now, the Left-hand limit of the function at $x = 1$ is

$\lim\limits_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim\limits_{h \to 0^-} \frac{\lfloor 1 + h \rfloor - \lfloor 1 \rfloor}{h} = \lim\limits_{h \to 0^-} \frac{0 - 1}{h}$

$= \lim\limits_{h \to 0^-} \frac{-1}{h} = -\infty \quad (\because h < 0 \Rightarrow 1 + h < 1, \therefore \lfloor 1 + h \rfloor = 0)$.

The Right-hand limit of the function at $x = 1$ is

$\lim\limits_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim\limits_{h \to 0^+} \frac{\lfloor 1 + h \rfloor - \lfloor 1 \rfloor}{h} = \lim\limits_{h \to 0^+} \frac{1 - 1}{h}$

$= \lim\limits_{h \to 0^+} \frac{0}{h} = 0 \quad (\because h > 0 \Rightarrow 1 + h > 1, \therefore \lfloor 1 + h \rfloor = 1)$.

Now, it is clear that

$\text{R.H.L. at } x = 1 \neq \text{L.H.L. at } x = 1$, and $\text{L.H.L.}$ is not finite as well.

Therefore, the function $f(x) = \lfloor x \rfloor$ is not differentiable at $x = 1$.

Similarly, for $x = 2$,

The required condition for the function to be differentiable at $x = 2$ is

$\lim\limits_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} = \lim\limits_{h \to 0^+} \frac{f(2 + h) - f(2)}{h}$.

Now, the Left-hand limit of the function at $x = 2$ is

$\lim\limits_{h \to 0^-} \frac{f(2 + h) - f(2)}{h} = \lim\limits_{h \to 0^-} \frac{\lfloor 2 + h \rfloor - \lfloor 2 \rfloor}{h} = \lim\limits_{h \to 0^-} \frac{1 - 2}{h}$

$= \lim\limits_{h \to 0^-} \frac{-1}{h} = -\infty \quad (\because h < 0 \Rightarrow 2 + h < 2, \therefore \lfloor 2 + h \rfloor = 1)$.

The Right-hand limit of the function at $x = 2$ is

$\lim\limits_{h \to 0^+} \frac{f(2 + h) - f(2)}{h} = \lim\limits_{h \to 0^+} \frac{\lfloor 2 + h \rfloor - \lfloor 2 \rfloor}{h} = \lim\limits_{h \to 0^+} \frac{2 - 2}{h}$

$= \lim\limits_{h \to 0^+} \frac{0}{h} = 0 \quad (\because h > 0 \Rightarrow 2 + h > 2, \therefore \lfloor 2 + h \rfloor = 2)$.
Now, it is clear that
R.H.L. at x= 2 $\neq$ L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function $f(x) = [x]$ is not differentiable at x = 2