Given that,\(y=(\cos x)^{(\cos x)^{(\cos x)} \cdots^{\infty}}\) 
\(\Rightarrow \quad y=(\cos x)^{y}\) 
Taking log on both sides, we get
log y = y log(cos x)
Now, differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}\) 
\(\Rightarrow \frac{d y}{d x}=y\left\{-y \tan x+\log \cos x \cdot \frac{d y}{d x}\right\} \)
\(\Rightarrow(1-y \log \cos x) \frac{d y}{d x}=-y^{2} \tan x \)
\(\Rightarrow \frac{d y}{d x}=\frac{y^{2} \tan x}{(y \log \cos x-1)} \) 
 

Let \(u=\cos ^{-1}\left(2 x^{2}-1\right) \text { and } v=\cos ^{-1} x\) 
\( \therefore \frac{d u}{d x} =-\frac{1}{\sqrt{1-\left(2 x^{2}-1\right)^{2}}} \cdot 4 x=\frac{-4 x}{\sqrt{1-\left(4 x^{4}+1-4 x^{2}\right.}} \)
\(=\frac{-4 x}{\sqrt{-4 x^{4}+4 x^{2}}}=\frac{-4 x}{\sqrt{4 x^{2}\left(1-x^{2}\right)}}=\frac{-2}{\sqrt{1-x^{2}}} \) 
and \(\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}}\) 
\(\therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{-2 / \sqrt{1-x^{2}}}{-1 / \sqrt{1-x^{2}}}=2\)

Given y = x log x
\(\frac{d y}{d x}=\frac{x}{x}+\log x\) 
\(\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\log e+\log x \\ \Rightarrow & \frac{d y}{d x}=\log (e x) \end{array}\)

We differentiate the relationship directly with respect to x,we get 
\(\frac{d y}{d x}+\frac{d}{d x}(\sin y)=\frac{d}{d x}(\cos x)\) 
[by chain rule of derivative]
\(\frac{d y}{d x}+\cos y \cdot \frac{d y}{d x}=-\sin x\) 
This gives \(\frac{d y}{d x}=-\frac{\sin x}{1+\cos y}\) 
where \(y \neq(2 n+1) \pi\)

\( \lim _{x \rightarrow 0^{-}} f(x)=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=1\)

As y = sin x, t
= log \(x\frac { dy }{ dx } =\frac { dy }{ dx } \div \frac { dt }{ dx } \)
\(=\frac { dy }{ dx } (sin \ x)\div \frac { dt }{ dx } (log \ x)\)
= cos x \(\div \) \(\frac1x\) = x cos x

As f(x) = |x – a| is continuous at x = a but not differentiable thereat.

As differentiable functions is continuous also

Here, P(A) = 0.6 and P(B) = 0.8
\(\text { (i) } \ (c): P(A \cap B)=P(A) \cdot P(B)=(0.6)(0.8)=0.48\)
\(\text { (ii) }(\text { a) }: P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
= 0.6 + 0.8 - 0.48 = 0.92
(iii) (d): P(B I A) = P(B) (\(\because\)A and B are independent)
= 0.8
(iv) (a): p (A I B) = p(A) (\(\because\)A and B are independent)
= 0.6
(v) (c) : P(not A and not B) = \(P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}\)
\(=1-P(A \cup B)=1-0.92=0.08\)

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)

L.H.L (at x = 0) = \(\begin{equation} \lim _{x \rightarrow 0} \frac{\sin (a+1) x+\sin x}{x}\left(\frac{0}{0} \text { form }\right) \end{equation}\) 
Using L' Hospital rule, we get
L.H.L.(at x = 0)
\(\begin{equation} =\lim _{x \rightarrow 0}(a+1) \cos (a+1) x+\cos x=a+2 \end{equation}\) 
R.H.L \(\begin{equation} \text { (at } x=0)=\lim _{x \rightarrow 0} \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}}=\lim _{x \rightarrow 0} \frac{\sqrt{1+b x}-1}{b x} \end{equation}\) 
\(\begin{equation} =\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+b x}+1}=\frac{1}{2} \end{equation}\) 
Since,f(x) is continuous at x = 0.
\(\therefore\) From (i) and (ii), we get
\(\begin{equation} a+2=c=\frac{1}{2} \Rightarrow a=-\frac{3}{2}, c=\frac{1}{2} \end{equation}\) 
Also, value of b does not affect the continuity of f(x), so b can be any real number.
(i) (a)
(ii) (d) 
(iii) (b)
(iv) (c) : \(\begin{equation} a+c=-\frac{3}{2}+\frac{1}{2}=-1 \end{equation}\) 
(v) (d) :  \(\begin{equation} c-a=\frac{1}{2}+\frac{3}{2}=2 \end{equation}\)

we have,\(\begin{equation} f(x)=\left\{\begin{array}{ll} x-3 & , x \geq 3 \\ 3-x & , 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & , x<1 \end{array}\right. \end{equation}\) 
(i) (b) : \(\begin{equation} \mathrm{R} f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{3-(1+h)-2}{h}=\lim _{h \rightarrow 0}-\frac{h}{h}=-1 \end{equation}\) 
(ii) (b) : \(\begin{equation} \mathrm{L}_{\mathrm{s}}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0} \frac{-1}{h}\left[\frac{(1-h)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}-2\right] \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{1+h^{2}-2 h-6+6 h+13-8}{-4 h}\right) \end{equation}\) 
\(\begin{equation} =\lim _{h \rightarrow 0}\left(\frac{h^{2}+4 h}{-4 h}\right)=-1 \end{equation}\) 
(iii) (c) : Since, R.H.D. at x = 3 is 1 and L.H.D. at x = 3 is-1
\(\therefore\)  f(x) is non-differentiable at x = 3.
(iv) (d)
(v) (c) : From above, we have
\(\begin{equation} f^{\prime}(x)=\frac{x}{2}-\frac{3}{2}, x<1 \end{equation}\) 
\(\begin{equation} \therefore f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}=-2 \end{equation}\)