Maths M.C.Q (1 Marks)

$ \Rightarrow \,\mathop {\lim }\limits_{x \to \pi /2} \,\left( {\frac{{k\cos x}}{{\pi - 2x}}} \right) = f\left( {\frac{\pi }{2}} \right)\,\, $

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

$f(0 + ) = \mathop {\lim }\limits_{x \to 0 + } \,\cos x = 1$

$\therefore \,\,\,f(0) = \cos x = 1$

Hence no value of $k$ can make $f(0 - ) = 1.$

$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \,\frac{h}{{{e^{1/h}} + 1}} = 0.$

$ = \mathop {\lim }\limits_{h \to 0} - \frac{h}{h} + a = a - 1.$

$ = \mathop {\lim }\limits_{x \to 4 + } f(x) = \mathop {\lim }\limits_{h \to 0} \,\,f(4 + h) = \mathop {\lim }\limits_{h \to 0} \,\frac{{4 + h - 4}}{{|4 + h - 4|}} + b = b + 1$

and $f(4) = a + b$

Since $f(x)$ is continuous at $x = 4$

Therefore $\mathop {\lim }\limits_{x \to 4 - } f(x) = f(4) = \mathop {\lim }\limits_{x \to 4 + } f(x)$

$ \Rightarrow \,\,a - 1 = a + b = b + 1\,\, \Rightarrow \,\,b = - 1$ and $a = 1.$

$f(0) = \mathop {\lim }\limits_{x \to 0} \,f(x) = \mathop {\lim }\limits_{x \to 0} \,\frac{{{{(27 - 2x)}^{1/3}} - 3}}{{9 - 3\,{{(243 + 5x)}^{1/5}}}}\,$, $\left( {{\rm{Form}} \,\, \frac{{\rm{0}}}{{\rm{0}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\frac{1}{3}\,{{(27 - 2x)}^{ - 2/3}}( - 2)}}{{ - \frac{3}{5}\,{{(243 + 5x)}^{ - 4/5}}(5)}} = 2.$

$ \Rightarrow \,\,\mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{x}\,\,\mathop {\lim }\limits_{x \to 0} \,\,\log \,\cos x = \log k$

$ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{x} \times 0 = {\log _e}k\,\, \Rightarrow \,k = 1$ .

If function $f(x)$ is continuous at $x = 1,$ then

$\mathop {\lim }\limits_{x \to {1^ - }} \,f(x) = \mathop {\lim }\limits_{x \to {1^ + }} \,f(x)$

$ \Rightarrow \,\,\,1 + \sin \frac{\pi }{2} = a + b$

$\therefore \,\,\,a + b = 2$.....$(i)$

If at $x = 3,$ function is continuous, then

$\mathop {\lim }\limits_{x \to {3^ - }} \,f(3) = \mathop {\lim }\limits_{x \to {3^ + }} \,f(x)$ $ \Rightarrow \,\,3a + b = 6\tan \frac{{3\pi }}{{12}}$

$\therefore \,\,\,3a + b = 6$.....$(ii)$

From $(i)$ and $(ii),$ $a = 2,\,\,b = 0$ .

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

$ \Rightarrow \,f(0) = \,\mathop {\lim }\limits_{x \to {0^ - }} f(x)$

$k = \mathop {\lim }\limits_{h \to 0} f(0 - h) = \mathop {\lim }\limits_{h \to 0} \,\frac{{\cos \frac{\pi }{2}\,[0 - h]}}{{[0 - h]}}$

$k = \mathop {\lim }\limits_{h \to 0} \,\frac{{\cos \frac{\pi }{2}\,[ - h]}}{{[ - h]}} = \mathop {\lim }\limits_{h \to 0} \,\frac{{\cos \frac{\pi }{2}\,[ - h - 1]}}{{[ - h - 1]}}$

$k = \mathop {\lim }\limits_{h \to 0} \,\frac{{\cos \,\left( { - \frac{\pi }{2}} \right)}}{{ - 1}}$;    $k = 0$.

==>$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = f(1)$

==> $5 \times 1 - 4 = 4 \times 1 + 3 \times b \times 1$

==> $1 = 4 + 3b$ ==> $3b = - 3$ ==> $b = - 1$.

$f\left( { - \frac{\pi }{2}} \right) = \mathop {{\rm{lim}}}\limits_{x \to {{( - \pi /2)}^ - }} ( - 2\sin x)$

$ = \mathop {{\rm{lim}}}\limits_{x \to {{( - \pi /2)}^ + }} (A\sin x + B)$

==> $2 = - A + B$ .....$(i)$

and $f\left( {\frac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to {{(\pi /2)}^ - }} (A\sin x + B)$ 

$ = \mathop {{\rm{lim}}}\limits_{x \to {{(\pi /2)}^ + }} (\cos x)$

==> $0 = A + B$.....$(ii)$

From $(i)$ and $(ii),$ $A = - 1$ and $B = 1$.

$\mathop {\lim }\limits_{x \to \alpha } \,\frac{{1 - \cos \,(a{x^2} + bx + c)}}{{{{(x - \alpha )}^2}}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{x\cot x}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} {\left\{ {{{(1 + x)}^{\frac{1}{x}}}} \right\}^{\mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{x}{{\tan x}}} \right)}}$ $ = {e^1} = e$.

If $f(x)$ is continuous from right at $x = 2$ then $\mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2) = k$

==> $\mathop {\lim }\limits_{x \to {2^ + }} {\left[ {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right]^{ - 1}} = k$

==> $k = \mathop {\lim }\limits_{h \to 0} f(2 + h)$

==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {{{(2 + h)}^2} + {e^{\frac{1}{{2 - (2 + h)}}}}} \right]^{\, - 1}}$

==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {\,4 + {h^2} + 4h + {e^{ - 1/h}}\,} \right]^{\, - 1}}$

==> $k = {[4 + 0 + 0 + {e^{ - \infty }}]^{\, - 1}}$

==> $k = \frac{1}{4}$.

$ = \tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)$ at $x = \pi $, $f(\pi ) = - \tan \frac{\pi }{4} = - 1$.

==> $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{\frac{1}{x}}} - 1}}{{{e^{\frac{1}{x}}} + 1}} = \frac{{{e^\infty } - 1}}{{{e^\infty } + 1}} = - 1$

==> $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{1}{x}}} - 1}}{{{e^{\frac{1}{x}}} + 1}} = \frac{{1 - {e^{ - \frac{1}{x}}}}}{{1 + {e^{\frac{1}{x}}}}} = \frac{{1 - {e^{ - \infty }}}}{{1 + {e^\infty }}} = 1$

So, $\mathop {\lim }\limits_{x \to 0} f(x)$ exists at $x = 0$, but at $x = 0$ it is not continuous.

If $f$ is continuous at $x=3,$ then

$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3)$              ............. $(1)$

Also, $\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} f(ax + 1) = 3a + 1$

$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(bx + 1) = 3b + 3$

$f(3)=3 a+1$

Therefore, from $(1),$ we obtain

$3 a+1=3 b+3=3 a+1$

$\Rightarrow 3 a+1=3 b+3$

$\Rightarrow 3 a=3 b+2$

$\Rightarrow a=b+\frac{2}{3}$

Therefore, the required relationship is given by, $a=b+\frac{2}{3}$

The given function $f$ is continuous at $x=\frac{\pi}{2},$ it is defined at $x=\frac{\pi}{2}$ and if the value of the $f$ at $x=\frac{\pi}{2}$ equals the limit of $f$ at $x=\frac{\pi}{2}$

It is evident that $f$ is defined at $x=\frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=3$

$\mathop {\lim }\limits_{x \to 2} \frac{\pi }{2}f(x) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi  - 2x}}$

Put $x=\frac{\pi}{2}+h$

Then, $x \rightarrow \frac{\pi}{2} \Rightarrow h \rightarrow 0$

$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f(x) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi  - 2x}} = \mathop {\lim }\limits_{h \to 0} \frac{{k\cos \left( {\frac{\pi }{2} + h} \right)}}{{\pi  - 2\left( {\frac{\pi }{2} + h} \right)}}$

$ = k\mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \,h}}{{ - 2h}} = \frac{k}{2}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h} = \frac{k}{2} \cdot 1 = \frac{k}{2}$

$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f(x) = f\left( {\frac{\pi }{2}} \right)$

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

$\Rightarrow k=6$

Therefore, the required value of $k$ is $6.$

$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 - h)$

$ = \mathop {\lim }\limits_{h \to 0} \,\,(1 - h) = 1$

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \,f(1 + h) = 1$

Hence function is continuous in $(0, 2).$

Now $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \,(0 + h) = 0 = f(0)$

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \,(2 - h) = 1 = f(2)$

Hence function is continuous in $[0, 2]$

Clearly, from graph it is not differentiable at $x = 1.$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {{{(1 + h)}^3} - 1} \right\} - 0}}{h} = 3$

$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {(1 - h) - 1} \right\} - 0}}{{ - h}} = 1$

$\therefore \,\,\,Rf'(1) \ne Lf'(1)$ $ \Rightarrow \,\,f(x)$ is not differentiable at $x = 1.$

Now, $f(1 + 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 + h) = 0$

and $f(1 - 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 - h) = 0$

$\therefore \,\,\,f(1 + 0) = f(1 - 0) = f(0)$

$ \Rightarrow \,\,f(x)$ is continuous at $x = 1.$

Hence at $x = 1,\,\,\,f(x)$ is continuous and not differentiable.

Hence, $f(x)$ will be continuous at $x = 0$.

$\mathop {{\rm{lim}}}\limits_{x \to {0^ - }} ({e^x} + ax) = \mathop {{\rm{lim}}}\limits_{x \to {0^ + }} b{(x - 1)^2}$

==> ${e^0} + a \times 0 = b{(0 - 1)^2}$ ==> $b = 1$….. $(i)$

But $f(x)$ is differentiable at $x = 0$, then

$Lf'(x) = Rf'(x)$ ==> $\frac{d}{{dx}}({e^x} + ax) = \frac{d}{{dx}}b{(x - 1)^2}$

==> ${e^x} + a = 2b(x - 1)$

At $x = 0,$ ${e^0} + a = - 2b$ ==> $a + 1 = - 2b$ ==> $a = - 3$

==> $(a,\,\,b) = ( - 3,\,\,1)$.

Clearly, $f(x)$ is continuous and differentiable for all non zero $x.$

Now $\mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0} {e^x} = 1$,

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x){e^{ - x}} = 1$

Also, $f(0) = {e^0} = 1$. So, $f(x)$ is continuous for all $x$.

($LHD $ at $x = 0)$ $ = {\left( {\frac{d}{{dx}}({e^x})} \right)_{x = 0}} = 1$

( $RHD $  at $x = 0)$  $ = {\left( {\frac{d}{{dx}}({e^{-x}})} \right)_{x = 0}} = 1$

So, $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is not differentiable at $L\,f'\,(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$.

Hence $f(x) = {e^{ - \,|\,x\,|}}$ is everywhere continuous but not differentiable at $x = 0$.

Also $Lf'(0) = Rf'(0)$

==> $\mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h}$

==> $\mathop {\lim }\limits_{h \to 0} \left( {\frac{{{e^{ - 2h}} - 1 + 1}}{{ - h}}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ah + \frac{{b{h^2}}}{2} - 1 + 1}}{h}} \right)$

==> $\mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - 2{e^{ - 2h}}}}{{ - 1}}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {a + \frac{{bh}}{2}} \right)$

==> $2 = a + 0$ ==> $a = 2,\,\,b$ any number.

but $ - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1$ and $x \to 0$

$\therefore $ $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 0 = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0)$

Therefore $f(x)$ is continuous at $x = 0$. 

Also, the function $f(x) = {x^2}\sin \frac{1}{x}$ is differentiable because 

$Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin \frac{1}{h} - 0}}{h} = 0$, 

$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin (1/ - h)}}{{ - h}} = 0$.

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{m\,{{(1 - h)}^2} - m}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{m\,[1 + {h^2} - 2h - 1]}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,m\,(2 - h) = 2m$ and $R\,f'\,(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f\,(1 + h) - f(1)}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{2\,(1 + h) - m}}{h}$.

For differentiability, $L\,f'(1) = R\,f'\,(1)$.

But for any value of $m,\,\,R\,\,f'\,(1) = L\,f'(1)$ not possible.

$\left[ \because f(x+y)=f(x)+f(y) \right]$

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

$[ \because g $ is continuous therefore $\mathop {\lim }\limits_{h \to 0} g(h) = g(0)]$.

$ = \left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}&{}\\{x - x + 1,}&{0 \le x < 1}& = \\{x + x - 1,}&{x \ge 1}&{}\end{array}} \right.\left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}\\1&{0 \le x < 1}\\{2x - 1,}&{x \ge 1}\end{array}} \right.$

Clearly, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1$. 

So, $f(x)$ is continuous at $x = 0,\,\,1.$

Now $f'(x) = \left\{ {\begin{array}{*{20}{l}}{ - 2,\,\,\,\,x < 0}\\{\,\,0,\,\,\,\,\,0 \le x < 1}\\{\,\,2,\,\,\,\,\,x \ge 1}\end{array}} \right.$

Here $x = 0$, $f'({0^ + }) = 0$ while $f'({0^ - }) = - 2$

and at $x = 1$, $f'({1^ + }) = 2$ while $f'({1^ - }) = 0$

Thus, $f(x)$ is not differentiable at $x = 0$ and $1.$

$f(x) = \left\{ {\begin{array}{*{20}{c}}{ - (x - 1) - (x - 3),}&{x < 1}\\{(x - 1) - (x - 3),}&{x > 1}\\{(x - 1) - (x - 3),}&{x < 3}\\{(x - 1) + (x - 3),}&{x > 3}\end{array}} \right.$

$ = \left\{ {\begin{array}{*{20}{c}}{4 - 2x,}&{x < 1}\\{2\,\,\,\,\,\,\,\,\,,}&{1 < x < 3}\\{2x - 4,}&{x > 3}\end{array}} \right.$

At $x = 2$, $f(x) = $ $2$ .

Hence $\,f'(x) = 0$.

$\mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 - h} \right) = f\left( 0 \right)$

$ \Rightarrow \quad \mathop {\lim }\limits_{h \to 0} {e^{ - h}} + a =  - 3\quad  \Rightarrow a =  - 4$

For the value of $a, f$ is diff at $x=0$

Hence $\frac{d}{{dx}}\left\{ {\log |x|} \right\} = \frac{1}{x}$, if $x > 0$

$ = \left( {\frac{1}{{ - x}}} \right)( - 1) = \frac{1}{x}$, if $x < 0$

Thus $\frac{d}{{dx}}\left\{ {\log |x|} \right\} = \frac{1}{x}$, if $x \ne 0$.

Differentiating with respect to  $ x$, we get

$\frac{{dy}}{{dx}} = a\cos x - b\sin x$

Now ${\left( {\frac{{dy}}{{dx}}} \right)^2} = {(a\cos x - b\sin x)^2}$

$ = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x - 2ab\sin x\cos x$

and ${y^2} = {(a\sin x + b\cos x)^2}$

$ = {a^2}{\sin ^2}x + {b^2}{\cos ^2}x + 2ab\sin x\cos x$

So, ${\left( {\frac{{dy}}{{dx}}} \right)^2} + {y^2} = {a^2}({\sin ^2}x + {\cos ^2}x) + {b^2}({\sin ^2}x + {\cos ^2}x)$

Hence ${\left( {\frac{{dy}}{{dx}}} \right)^2} + {y^2} = ({a^2} + {b^2}) =$ constant.

But $f(x) = f'(x) \Rightarrow {x^2} - 3x = 2x - 3$

==> ${x^2} - 5x + 3 = 0$

$\therefore x = \frac{{5 \pm \sqrt {25 - 12} }}{2} = \frac{{5 \pm \sqrt {13} }}{2}$.

==>$y = x(\cos x + \sin x) + \frac{1}{{2\sqrt x }}$

Differentiating w.r.t.  $x$, we have

$\frac{{dy}}{{dx}} = x\frac{d}{{dx}}(\cos x + \sin x) + (\cos x + \sin x) - \frac{1}{4}{x^{ - 3/2}}$

==>$\frac{dy}{dx}=$  $(1+x)\cos x+(1-x)\sin x-\frac{1}{4x\sqrt{x}}$

$\frac{{dy}}{{dx}} = \frac{{ - x}}{y} = \frac{{ - xy}}{{(1 - {x^2})}}$ or 

$(1 - {x^2})\frac{{dy}}{{dx}} + xy = 0$.

$ = \frac{d}{{dx}}[\cos 2x] = - 2\sin 2x$.

$ = {\tan ^{ - 1}}\frac{{5x - x}}{{1 + 5x.x}} + {\tan ^{ - 1}}\frac{{\frac{2}{3} + x}}{{1 - \frac{2}{3}.x}}$

=> $\frac{{dy}}{{dx}} = \frac{5}{{1 + 25{x^2}}}$.

==> $\frac{{dy}}{{dx}} = \frac{5}{{1 + 25{x^2}}}$.

$ = \frac{{\sqrt {1 - {{\cos }^2}x} }}{{2\sqrt {\cos x} \sqrt {1 - \cos x} }} = \frac{1}{2}\sqrt {\frac{{1 + \cos x}}{{\cos x}}} $.

$\Rightarrow y = \frac{{{{(\sqrt {a + x} - \sqrt {a - x} )}^2}}}{{(a + x) - (a - x)}}$

$ \Rightarrow y = \frac{{(a + x) + (a - x) - 2(\sqrt {{a^2} - {x^2}} )}}{{2x}}$

$ = \frac{{2a - 2\sqrt {{a^2} - {x^2}} }}{{2x}}$ or $y = \frac{{a - \sqrt {{a^2} - {x^2}} }}{x}…..(i)$

Differentiating w.r.t. $ x $ of $ y,$  we get

$\frac{{dy}}{{dx}} = \frac{{x\left[ { - \frac{1}{{2\sqrt {{a^2} - {x^2}} }}( - 2x)} \right] - (a - \sqrt {{a^2} - {x^2}} )}}{{{x^2}}}$

$ = \frac{{{x^2} - a\sqrt {{a^2} - {x^2}} + {a^2} - {x^2}}}{{{x^2}\sqrt {{a^2} - {x^2}} }} = \frac{{a(a - \sqrt {{a^2} - {x^2}} )}}{{{x^2}\sqrt {{a^2} - {x^2}} }}$

$ = \frac{a}{{x\sqrt {{a^2} - {x^2}} }}\left[ {\frac{{a - \sqrt {{a^2} - {x^2}} }}{x}} \right] = \frac{{ay}}{{x\sqrt {{a^2} - {x^2}} }}$      [By $(i)$]

$\therefore \frac{{dy}}{{dx}} = \frac{3}{2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x + 3x{\cot ^2}x( - {\rm{cose}}{{\rm{c}}^2}x)]$

$ = \frac{3}{2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x - 3x{\cot ^2}x\,{\rm{cose}}{{\rm{c}}^2}x]$.

$\frac{{dy}}{{dx}} = \frac{1}{2}\frac{{\cos (\sqrt {\sin x + \cos x} )}}{{\sqrt {\sin x + \cos x} }}(\cos x - \sin x)$.

$ = \frac{{4x}}{{{{(1 - {x^2})}^2}}}\cos \left( {\frac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$.

$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {\tan \left( {\frac{\pi }{4} + x} \right)} }}{\sec ^2}\left( {\frac{\pi }{4} + x} \right)$

$ = \frac{1}{2}\sqrt {\left[ {\frac{{1 - \tan x}}{{1 + \tan x}}} \right]} {\sec ^2}\left( {\frac{\pi }{4} + x} \right)$.

$ = \frac{d}{{dx}}[2\,{\rm{cosec}}2x] = - 4\,{\rm{cosec}}2x\cot 2x$.

$\therefore y = 2{\cos ^{ - 1}}\sqrt x $ or 

$\frac{{dy}}{{dx}} = 2.\frac{{ - 1}}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt x }} \,\,etc.$

Aliter: $y = {\sin ^{ - 1}}\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x $

==> $\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - 1 + x} }}.\frac{{ - 1}}{{2\sqrt {1 - x} }} - \frac{1}{{\sqrt {1 - x} }}.\frac{1}{{2\sqrt x }}$

$ = \frac{{ - 2}}{{2\sqrt x \sqrt {1 - x} }} = \frac{{ - 1}}{{\sqrt x \sqrt {1 - x} }}$.

$ = {\cot ^{ - 1}}\left[ {\frac{{2 + 2\cos x}}{{2\sin x}}} \right] = {\cot ^{ - 1}}\left[ {\frac{{1 + \cos x}}{{\sin x}}} \right]$

$ = {\cot ^{ - 1}}\left[ {\cot \frac{x}{2}} \right] = \frac{x}{2}$

$\therefore \frac{{dy}}{{dx}} = \frac{1}{2}$ .

$f(x) = \frac{1}{{\sqrt {{x^2} + {a^2}} + \sqrt {{x^2} + {b^2}} }}.\frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}$

$f(x) = \frac{1}{{{a^2} - {b^2}}}\left[ {\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} } \right]$

$f'(x) = \frac{1}{{{a^2} - {b^2}}}\left[ {\frac{{2x}}{{2\sqrt {{x^2} + {a^2}} }} - \frac{{2x}}{{2\sqrt {{x^2} + {b^2}} }}} \right]$

$f'(x) = \frac{x}{{{a^2} - {b^2}}}\left[ {\frac{1}{{\sqrt {{x^2} + {a^2}} }} - \frac{1}{{\sqrt {{x^2} + {b^2}} }}} \right]$.

$f(x) = \left\{ \begin{array}{l} - (x - 1) - (x - 5),\,\,\,x < 1\\\,\,\,\,(x - 1) - (x - 5),\,\,\,1 < x < 5\\\,\,\,\,\,\,\,x - 1 + x - 5,\,\,\,\,x > 5\end{array} \right.$

$f(x) = \left\{ {\begin{array}{*{20}{c}}{6 - 2x,}&{x < 1}\\{4\,\,\,\,\,\,,}&{1 < x < 5}\\{2x - 6,}&{x > 5}\end{array}} \right.$

$\because$ $x = 3 \in (1,\,5)$,  

For $x = 3, $   $ f(x) = 4,\;\;f'(x) = 0$.