{-1}

\(\left\{ \frac { a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 }\left( a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } \right) } } \right\} \)

\(\left\{ \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right\} \)

If function is continuous at \(x=\frac{\pi}{2}, \text { then }\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\mathrm{RHL}=f\left(\frac{\pi}{2}\right)\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \)
\(=\lim _{h \rightarrow 0} \frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}=\lim _{h-0} \frac{1-\cos ^{3} h}{3 \sin ^{2} h} \)
\(=\lim _{h \rightarrow 0} \frac{(1-\cos h)\left(1+\cos ^{2} h+\cos h\right)}{3(1-\cos h)(1+\cos h)} \)
\(=\lim _{h \rightarrow 0} \frac{1+\cos ^{2} h+\cos h}{3(1+\cos h)}=\frac{1+1+1}{3(1+1)}=\frac{1}{2} \ldots .(i i) \)
\(\operatorname{RHL} =\lim _{h=\frac{\pi}{2}} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0} \frac{b\left\{1-\sin \left(\frac{\pi}{2}+h\right)\right\}}{\left\{\pi-2\left(\frac{\pi}{2}+h\right)\right\}^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{(\pi-\pi-2 h)^{2}}=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{4 h^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b \cdot 2 \sin ^{2} \frac{h}{2}}{4 h^{2}}=\lim _{h \rightarrow 0} \frac{b}{8}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \)
\(=\frac{b}{8} \times 1=\frac{b}{8} \)
Substituting from (ii) and (iii) in (i), we get
\(\frac{1}{2}=\frac{b}{8}=a \Rightarrow a=\frac{1}{2}, b=4\)
Hence, for \(a=\frac { 1 }{ 2 } ,\quad b=4 \text { function is continuous at} x=\frac { \pi }{ 2 } .\)

\(\frac { -cos\left( { cos }^{ 2 }\sqrt { x } \right) sin\sqrt { x } cos\sqrt { x } }{ \sqrt { x } } \)

{f(x)=|x|}

\(y={ tan }^{ -1 }\frac { 3x+2x }{ 1-3x2x } \)
= \({ tan }^{ -1 }3x+{ tan }^{ -1 }2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3 }{ 1+9{ x }^{ 3 } } +\frac { 2 }{ 1+4{ x }^{ 2 } } \)

On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
\(\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

We have, y = a^x +x^a+x^x+a^a
Let v = x^x
log v = x log x
\(\frac { 1 }{ v } \frac { dv }{ dx } =x+\frac { 1 }{ x } +logx\)
\(\frac { dv }{ dx } =v\left| 1+logx \right| \)
= x^x(1+lodx)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

We have, y = tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)
Put x = cos2\(\theta\)
\(\Rightarrow\)2\(\theta\) = cos^-1x
\(\Rightarrow\)\(\theta\) = 1/2cos^-1x
y = tan^-1\(\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
y = \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \)
(\(\because\)cos2\(\theta\) = 2cos²\(\theta\)-1 = 1-2sin²\(\theta\)
y = tan^-1(tan \(\theta\))
y = \(\theta\) = 1/2cos^-1x
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)​​​​​​​

Not derivable at x = 0.

For x = -3 function is not defined. Hence, not continuous for x ∈ R.

For the function \(f\left( x \right) ={ x }^{ 3 }-{ 6x }^{ 2 }+ax+b\), it is given that f(1) = f(3) = 0.
Find the values of a and b and hence verify Rolle's Theorem on [1, 3].

We have: \(y={ X }^{ x }\)
Taking logs \(logy\quad y=x\quad log\quad x\)
\(loy\quad y=x\quad log\quad x\)
\(\frac { 1 }{ y } \frac { dy }{ dx } =x.\frac { 1 }{ x } +log\quad x.1\)
\(\frac { dy }{ dx } =y(1+log\quad x)...(1)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\frac { d }{ dx } (1+log\quad x)+\frac { dy }{ dx } (1+log\quad x)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =y.\left( 0+\frac { 1 }{ x } \right) +\frac { dy }{ dx } .\frac { \frac { dy }{ dx } }{ y } \)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { y }{ x } +\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { 1 }{ y } { \left( \frac { dy }{ dx } \right) }^{ 2 }-\frac { y }{ x } =0\)

Let y = \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) \)
Put \(x=tan\theta \)
\(y={ tan }^{ -1 }\frac { \sqrt { 1+{ tan }^{ 2 }\theta } -1 }{ tan\theta } \)
\(={ tan }^{ -1 }\left( \frac { sec\theta -1 }{ tan\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 1-cos\theta }{ sin\theta } \right) \)
\(={ tan }^{ -1 }\left( \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ sin }\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { { sin }\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } \right) ={ tan }^{ -1 }\left( { tan }\frac { \theta }{ 2 } \right) \)
\(=\frac { \theta }{ 2 } =\frac { 1 }{ 2 } { tan }^{ -1 }x\)
Hence \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \frac { 1 }{ 1+{ x }^{ 2 } } =\frac { 1 }{ 2(1+{ x }^{ 2 }) } \)

Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x.
Then (h o g) (x) = h(g (x)) = h (1– x + | x |)
= |1– x + | x || = f(x)
we have seen that h is a continuous function.
Hence g being a sum of a polynomial function and the modulus function is continuous.
But then f being a composite of two continuous functions is continuous.

\( f(x)=\begin{cases} -x,if\quad x<0 \\ x,\quad if\quad x\ge 0. \end{cases}\)
Clearly the function is defined at 0 and f(0) = 0. Left hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (-x) } =0\)
Similarly, the right hand limit of f at 0 is
\(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ (x) } =0\)
Thus, the left hand limit, right hand limit and the value of the function coincide at x = 0.
Hence, f is continuous at x = 0.