Ncert Solutions For Class 12 Maths Ex 5.2

Ncert Solutions For Class 12 Maths Chapter 5 Ex 5.2

Q1: Differentiate the function with respect to \(x\)

\(f(x)=\ sin (x^{2}+5)\)

Sol:The given function is \(f(x)=\ sin (x^{2}+5)=y\)

Let \(t=(x^{2}+5)\)

\(So f(t)=\sin t\)

\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) ……..(i)

\(\frac{\mathrm{d} y}{\mathrm{d} t}=\cos t\) ……..(ii)

And \(\frac{\mathrm{d} t}{\mathrm{d} x}=2x\) …….(iii)

Substituting equation (ii) and (iii) in (i) we have,

\(\frac{\mathrm{d} y}{\mathrm{d} x}=\cos t \times (2x)\)

And we know \(t=x^{2}+5\)

Thus \(\frac{\mathrm{d} y}{\mathrm{d} x}=(2x)\times \cos x\)

Q2: Differentiate the function with respect to \(x\)

\(f(x)=\cos (\sin x)\)

Sol:

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

Here \(f\) is a composite function which can be written in the form of two composite function \(u and v\).

\(u(x)=\sin x\; \; and\; \; v(t)=\cos t\) \((vou)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)\)

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

\(therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\cos t)}{\mathrm{d} x}=-\sin t = -\sin(\sin x)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x\)

By chain rule ,

\(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) \(\frac{\mathrm{d} f}{\mathrm{d} x}=-\sin (\sin x)\times \ cos x=-\cos x\sin (\sin x)\)

Q3: Differentiate the function with respect to \(x\)

\(f(x)=\sin (ax+b)\)

Sol:

Let \(f(x)=\sin (ax+b)\)

Here \(f\) is a composite function which can be written in the form of two composite function \(u and v\).

\(u(x)=ax+b\; \; and\; \; v(t)=\sin t\) \((vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)=f(x)\)

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

\(therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} x}=\cos t = \cos(ax+b)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a\)

By chain rule ,

\(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\) \(\frac{\mathrm{d} f}{\mathrm{d} x}=a\times \ cos (ax+b)\)

Q4: Differentiate the function with respect to \(x\)

\(\sec (\tan (\sqrt{x}))\).

Sol:

Let \(\sec (\tan (\sqrt{x}))\)

Here \(f\) is a composite function which can be written in the form of three composite function \(u,v and w\).

\(u(x)=\sqrt{x}\; \; v(t)=\tan t\; \; and w(s)=\sec s\) \((wovou)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w[\tan( \sqrt{x})]=\sec (\tan (\sqrt{x}))=f(x)\)

Put \(s=v(t)=\tan t and t=u(x)=\sqrt{x}\)

then ,\(\frac{\mathrm{d} w}{\mathrm{d} s}=\frac{\mathrm{d} (\sec s)}{\mathrm{d} s}=\sec s\tan s =\sec(\tan t). \tan (\tan t)\) ( as \(s=\tan t\)

\(=\sec(\tan \sqrt{x}). \tan (\tan \sqrt{x})\) \(\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{\mathrm{d} (\tan t)}{\mathrm{d} t}=\sec ^{2}t=\sec ^{2}\sqrt{t}\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{\mathrm{d} (x^{\frac{1}{2}})}{\mathrm{d} x}=\frac{1}{2}.x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}\)

By chain rule ,

\(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} w}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}\)

=\(\sec (\tan\sqrt{x}).\tan (\tan\sqrt{x})\times \sec ^{2} \sqrt{x}\times \frac{1}{2\sqrt{x}}\)

=\(\frac{\sec ^{2}\sqrt{x}.\sec (\tan \sqrt{x}.\tan (\tan \sqrt{x}))}{2\sqrt{x}}\)

Q5: Differentiate the function with respect to \(x\)

\(f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}\)

Sol:

The given function is \(f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}= \frac{g(x)}{h(x)}\),

where \(g(x)= \sin (ax+b) \) and \(h(x)=\cos (cx+d)\)

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

Here \(g\) is a composite function which can be written in the form of two composite function \(u and v\).

\(u(x)=ax+b\; \; v(t)=\sin t\) \((vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)]=g(x)\)

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

\(\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t=\cos (ax+b)\) \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a\)

By chain rule ,

\(g^{‘}=\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\cos (ax+b).a=a \cos (ax+b)\)

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

Here \(h\) is a composite function which can be written in the form of two composite function \(p and q\).

\(p(x)=cx+d\; \; q(y)=\cos y\) \((qop)(x)=q(p(x))=q(cx+d)=\cos (cx+d)]=h(x)\)

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

\(\frac{\mathrm{d} q}{\mathrm{d} y}=\frac{\mathrm{d} (\cos y)}{\mathrm{d} y}=-\sin y=-\sin (cx+d)\) \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (cx+d)}{\mathrm{d} x}=c\)

By chain rule ,

\(h^{‘}=\frac{\mathrm{d} h}{\mathrm{d} x}=\frac{\mathrm{d} q}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (cx+d).c=-c \sin (cx+d)\)

Therefore by chain rule , we obtain

\(f^{‘}=\frac{a\cos (ax+b).\ cos (cx+d)-\sin (ax+b)(-\sin(cx+d))}{[\cos(cx+d)]^{2}}\)

=\(=\frac{a\cos (ax+b)}{\ cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}\)

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

Q6: Differentiate the function with respect to \(x\)

\(f(x)=\cos x^{3}.\sin ^{2}(x^{5})\)

Sol:

The given function is \(f(x)=\cos x^{3}.\sin ^{2}(x^{5})\)

\(\frac{\mathrm{d} }{\mathrm{d} x}\left [ \cos x^{3}.\sin ^{2}(x^{5}) \right ]=\sin ^{2}(x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\cos x^{3}+\cos x^{3}\times \frac{\mathrm{d} }{\mathrm{d} x}\sin ^{2}(x^{5})\)

=\(\sin ^{2}(x^{5})\times (-\sin x^{3})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{3})+\cos x^{3}\times 2\sin (x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \sin (x^{5}) \right ]\)

\(=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\frac{\mathrm{d} }{\mathrm{d} x} (x^{5})\) \(=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\times 5x^{4}\) \(=10x^{4}\sin x^{5}.\cos x^{3}.\cos x^{5}\times -3x^{2}\sin x^{3}.\sin ^{2}(x^{5} )\)

Q7: Differentiate the functions with respect to x.

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

Sol:

The given function is \(2\sqrt{\cot (x^{2})}\)

\(\frac{\mathrm{d} }{\mathrm{d} x}2\sqrt{\cot (x^{2})}\) \(=2.\frac{1}{2\sqrt{\cot (x^{2})}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \cot (x^{2}) \right ]\) \(=\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times -cosec^{2} (x^{2})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})\) \(=-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times \frac{1}{\sin ^{2(x^{2})}}\times(2x)\) \(=\frac{-2x}{\sqrt{\cos (x^{2})}\sqrt{\sin (x^{2})}\sin (x^{2})}\) \(=\frac{-2\sqrt{2}x}{\sqrt{2\cos (x^{2})\sin (x^{2})}\sin (x^{2})}\) \(=\frac{-2\sqrt{2}x}{\sin (x^{2})\sqrt{\sin 2(x^{2})}}\)

Q8: Differentiate the functions with respect to x.

\(\cos\sqrt{x}\)

Sol:

The given function \(f(x)\) is \(\cos\sqrt{x}\).

Let \(u(x)=\sqrt{x}\)

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

\((vou)(x)=v(u(x))\) \(=v(\sqrt{x})\) \(=\cos (\sqrt{x})\)

=\(f(x)\)

Clearly, \(f\) is a composite function of two functions, \(u\; and \; v\), such that

\(t=u(x)=\sqrt{x}\)

Then, \(\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}\)

\(=\frac{1}{2\sqrt{x}}\)

And, \(\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(\cos t)=-\sin t\)

\(=-\sin (\sqrt{x})\)

By chain rule we have,

\(\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}.\frac{\mathrm{d} t}{\mathrm{d} x}\).

\(=-\sin (\sqrt{x}).\frac{1}{2\sqrt{x}}\) \(=-\frac{1}{2\sqrt{x}}\sin (\sqrt{x})\) \(=-\frac{\sin (\sqrt{x})}{2\sqrt{x}}\)

Q9: Prove that the function f given by \(f(x)=\left | x-1 \right |,x\in \mathbb{R}\), is not differentiable at \(x=1\).

Sol:

The given function is \(f(x)=\left | x-1 \right |,x\in \mathbb{R}\).

It is known that a function \(f\) is differentiable at a point \(x=c\) in its domain if the right hand limit and the left hand limit are finite and equal.

To check the differentiability of the given function at x=1,

The right hand and the left hand limits where x=c are

\(\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}\) and \(\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}\)

Considering the right hand limit of the given function at x=1

\(\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{+}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}\) \(=\lim_{h \to 0^{+}}\frac{\left | h \right |-\left | 0 \right |}{h}\) \(=\lim_{h \to 0^{+}}\frac{h}{h}\) \(=1\)

Considering the left hand limit of the given function at x=1

\(\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{-}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}\) \(=\lim_{h \to 0^{-}}\frac{\left | h \right |-\left | 0 \right |}{h}\) \(=\lim_{h \to 0^{-}}\frac{-h}{h}\) \(=-1\)

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

Q10: Prove that the greatest integer function defined by \(f(x)=\left [ x \right ], 0<x<3\) is not differentiable at x = 1 and x = 2.

Sol:

The function f is \(f(x)=\left [ x \right ], 0<x<3\)

It is known that a function f is differentiable at a point \(x=c\) in its domain if both the left hand and the left hand limit are equal

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

To check the differentiability of the given function at \(x=1\), consider the right hand limit of f at \(x=1\)

\(\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{+}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}\) \(=\lim_{h \to 0^{+}}\frac{1-1}{h}\) \(=\lim_{h \to 0^{+}}\frac{1-1}{h}=\lim_{h \to 0^{+}}(0)=0\)

Now consider the left hand limit of f at \(x=1\)

\(\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}\) \(\lim_{h \to 0^{-}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}\) \(=\lim_{h \to 0^{-}}\frac{0-1}{h}\) \(=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty \)

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

Now to check the differentiability of the given function at x=2,

consider the left hand limit at x=2.

\(\lim_{h \to 0^{-}}\frac{f(2+h)-f(2)}{h}\) \(\lim_{h \to 0^{-}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}\) \(=\lim_{h \to 0^{-}}\frac{1-2}{h}\) \(=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty \)

Now consider the right hand limit of f at \(x=2\)

\(\lim_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}\) \(\lim_{h \to 0^{+}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}\) \(=\lim_{h \to 0^{+}}\frac{2-2}{h}\) \(=\lim_{h \to 0^{+}}\frac{0}{h}=\lim_{h \to 0^{+}}(0)=0\)

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