\(\left\{ \left( x+\frac { 1 }{ x } \right) ^{ x }\left[ \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } +log\left( x+\frac { 1 }{ x } \right) \right] +{ x }^{ 1+\frac { 1 }{ x } }\left( \frac { x+1-logx }{ { x }^{ 2 } } \right) \right\} \)

{-1}

Putting x = \(cos2\theta \left\{ 2tan^{ -1 }\sqrt { \frac { 1-x }{ 1+x } } \right\} \) we get 
\(2tan^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
i,e.,\(2tan^{ -1 }\sqrt { \frac { 2sin^{ 2 }\theta }{ 2cos^{ 2 }\theta } } \)
\(=2\quad tan^{ -1 }\left( tan\theta \right) \)
\(=2\theta =cos^{ -1 }x\)
Hence \(y=e^{ sin^{ 2 } }xcos^{ -1 }x\)
\(\Rightarrow logy=sin^{ 2 }x+log\left( cos^{ 1 }x \right) \)
\(\Rightarrow \frac { 1 }{ y } \times \frac { dy }{ dx } =2sinxcosx+\frac { 1 }{ cos^{ -1 }x } \times \frac { -1 }{ \sqrt { 1-x^{ 2 } } } \)
= sin 2x \(-\frac { 1 }{ cos^{ 1 }x\sqrt { 1-x^{ 2 } } } \)
\(\Rightarrow \frac { dy }{ dx } =e^{ sin^{ 2 } }xcos^{ -1 }x\left[ sin2x-\frac { 1 }{ cos^{ -1 }x\sqrt { 1-x^{ 2 } } } \right] \)

If continuous, function should be continuous at x = 2, as for x > 2 and x < 2 function is a polynomial function, which is always continuous.
\(\therefore \ \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \)
\(\therefore \ \lim _{h \rightarrow 0}\{2 a(2-h)-b\}=\lim _{h \rightarrow 0}\left\{a(2+h)^{2}+b\right\}=2 \)
\(\Rightarrow 4 a-b=4 a+b=2 \Rightarrow b=0, a=\frac{1}{2} \)
b = 0, a = \(\frac {1}{2}\)

For function to be continuous at x = 2, we have
\(\mathrm{LHL}_{x=2}=\mathrm{RHL}_{x=2}=f(2)\)
\(\Rightarrow \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \)
\(\Rightarrow \lim _{h \rightarrow 0}\{2(2-h)+1\}=\lim _{h \rightarrow 0}\{3(2+h)-1\}=k \)
\(\Rightarrow 4+1=6-1=k \Rightarrow k=5 . \)
k = 5

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].

\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ - } }{ \left( 3ax+b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [3a(1-h)+b] } \)
\(=3a(1-0)+b\)
\(=3a+b\)
\(\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ \left( 5ax-2b \right) } \)
\(=\lim _{ h\rightarrow 0 }{ [5a(1+0)-2b] } \)
\(=5a-2b\)
\(f(1)=11\)
Also
Since'f' is continuous at x = 1
\(\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =f(1)\)
From first and third 3a + b = 11 ....(1)
From last two 5a - 2b = 11 .....(2)
Multiplying (1) by 2, 6a + 2b = 22 .....(3)
Adding (2) and (3), 11a = 33 = a = 3
Putting in (1), 3(3) + b = 11
b = 11 - 9 = 2
Hence a = 3 and b = 2.

Solution Given that \(y^x+x^y+x^x=a^b\).
Putting \(u=y^x, v=x^y\)and \(w=x^x\), we get \(u+v+w=a^b\)
Therefore \(\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0\)
Now, $u=y^x$. Taking logarithm on both sides, we have \(\log u=x \log y\)
Differentiating both sides w.r.t. x, we have
\(\frac{1}{u} \cdot \frac{d u}{d x} =x \frac{d}{d x}(\log y)+\log y \frac{d}{d x}(x) \)
\(=x \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot 1 \)
\(\frac{d u}{d x} =u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=y^x\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]\)
So
Also v = x^y
Taking logarithm on both sides, we have
\(\log v=y \log x\)
Differentiating both sides w.r.t. x, we have
\(\frac{1}{v} \cdot \frac{d v}{d x} =y \frac{d}{d x}(\log x)+\log x \frac{d y}{d x} \)
\(=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}\)
\(\frac{d v}{d x} =v\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \)
\(=x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \)
\(w =x^x\)
So
Again
Taking logarithm on both sides, we have
log w=x log x .
Differentiating both sides w.r.t. x, we have
\(\frac{1}{w} \cdot \frac{d w}{d x}  =x \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x) =x \cdot \frac{1}{x}+\log x \cdot 1\)
i.e.
\(\frac{d w}{d x}  =w(1+\log x) =x^x(1+\log x)\)
From (1), (2), (3), (4), we have
\(y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)+x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+x^x(1+\log x)=0 \)
\( \text { or } \quad\left(x \cdot y^{x-1}+x^y \cdot \log x\right) \frac{d y}{d x}=-x^x(1+\log x)-y \cdot x^{2-1}-y^x \log y \)
\(\text { Therefore } \ \frac{d y}{d x}=\frac{-\left[y^x \log y+y \cdot x^{y-1}+x^x(1+\log x)\right]}{x \cdot y^{x-1}+x^y \log x} \)

Let y=tan(2x+3)=tan t, where t=2x+3
\(\frac { dy }{ dt } ={ sec }^{ 2 }\ t\ and\ \)
\(\frac { dt }{ dx } =2(1)+0=2.\)
\(By\quad chain\quad rule,\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(={ sec }^{ 2 }\ t.2\)
\(=2 { sec }^{ 2 }(2x+3)\)

Let cϵR be arbitrary point.
\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ (k) } =k.\)
\(Also\quad f(c)=k.\)
\(Thus\quad \lim _{ x\rightarrow c }{ f(x) } =f(c).\)
Thus ′f′ is continuous at x = c.
But c is arbitrary.
Hence, f(x) is continuous at each real number.

Since f is differentiable at c, we have
\(\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}=f^{\prime}(c)\)
But for x ≠ c, we have
\(f(x)-f(c)=\frac{f(x)-f(c)}{x-c} \cdot(x-c)\)
Therefore \(\lim _{x \rightarrow c}[f(x)-f(c)]=\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c} \cdot(x-c)\right]\)
\( \lim _{x \rightarrow c}[f(x)]-\lim _{x \rightarrow c}[f(c)] =\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c}\right] \cdot \lim _{x \rightarrow c}[(x-c)] \\ =f^{\prime}(c) \cdot 0=0 \)
\(\lim _{x \rightarrow c} f(x)=f(c)\)
Hence f is continuous at x = c.

{continous}

If x = f(t) and y = g(t), then is \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } =\frac { { d }^{ 2 }y/{ dt }^{ 2 } }{ { d }^{ 2 }x/{ dt }^{ 2 } } \) ?
{No}

{1,2}

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

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)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

\(\frac{d y}{d x}=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}\)

y'=0

f(x) = [x]  is not continuous for integers.Hence not continuous at x = ±2, ±1, 0

Absolute value function f(x) = Ix - 1I is continuous at x = 1 but not differentiable at x = 1.

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