The given function is \(f(x)=\left\{\begin{array}{ll} 2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1 \end{array}\right.\)
The given function is defined at all points of the real line.
Let c be a point on the real line.
Case I: \(\text { If } c<0, \text { then } f(c)=2 c \)
\(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c \)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x < 0
Case II:\(\text { If } c=0, \text { then } f(c)=f(0)=0\)
The left hand limit of f at x = 0 is,
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2 \times 0=0\)
The right hand limit of f at x = 0 is,
Therefore, f is continuous at x = 0
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{*}}(0)=0 \)
\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)
Case III: \(\text { If } 0
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points of the interval (0, 1).
Case IV: \(\text { If } c=1, \text { then } f(c)=f(1)=0\)
The left hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(0)=0\)
The right hand limit of f at x = 1 is,
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4 \times 1=4\)
It is observed that the left and right hand limits of f at x = 1 do not coincide.
Therefore, f is not continuous at x = 1
Case V: \(\text { If } c<1, \text { then } f(c)=4 c \text { and } \lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(4 x)=4 c\)
\(\therefore \lim _{x \rightarrow c} f(x)=f(c) \)
Therefore, f is continuous at all points x, such that x > 1
Hence, f is not continuous only at x = 1

We have
f(x) = sin x + cos x,
or f'(x) = cos x – sin x

Now f'(x) = 0 gives sin x = cos x which gives that \(x=\frac{\pi}{4}, \frac{5 \pi}{4} \text { as } 0 \leq x \leq 2 \pi\)
The points \(x=\frac{\pi}{4} \text { and } x=\frac{5 \pi}{4}\) divide the interval [0, 2p] into three disjoint intervals namely,
\(\left[0, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Note that } f^{\prime}(x)>0 \text {if } x \in\left[0, \frac{\pi}{4}\right) \cup\left(\frac{5 \pi}{4}, 2 \pi\right]\)
or f is increasing in the intervals \(\left[0, \frac{\pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right]\)
\(\text {Also }f^{\prime}(x)<0 \text { if } x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)
or is plecreasing in \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

Here g(x) = x - [x]
Let a be an integer and h is very small h > 0, then
\( [a-h]=a-1,[a+h]=a \)
\(\text {and } [a]=a \) 
\(\text {At } x=a, \mathrm{LHL}=\lim _{x \rightarrow a^{-}} g(x)=\lim _{x \rightarrow a^{-}}(x-[x])\)
\(Put x=a-h ; when x \rightarrow a^{-}, then\ h \rightarrow 0\)
\( \mathrm{LHL} =\lim _{h \rightarrow 0}(a-h-[a-h]) \)
\(=\lim _{h \rightarrow 0}(a-h-(a-1)) \)

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

Here, \(f(x)=\left\{\begin{array}{cl} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.\) 
At \(x=0, \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)\) 
\(\therefore \mathrm{LHL}=\lim _{h \rightarrow 0} \lambda\left[(0-h)^{2}-2(0-h)\right]=\lim _{h \rightarrow 0}\left[\lambda\left(h^{2}+2 h\right)\right]=0\) 
\(\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)\) 
\(\therefore \mathrm{RHL}=\lim _{h \rightarrow 0}[4(0+h)+1]=\lim _{h \rightarrow 0}[4 h+1]=0+1=1\)
\(\text { [put } x=0+h \text { ; when } x \rightarrow 0^{+} \text {, then } \left.h \rightarrow 0\right] \) 
\(\therefore \mathrm{LHL} \neq \mathrm{RHL}\) 
Thus, f(x) is not continuous at x = 0 for any value of λ.
At x = 1,
 \( \mathrm{LHL} =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(4 x+1) \)
\(\therefore \mathrm{LHL} =\lim _{h \rightarrow 0}[4(1-h)+1]=\lim _{h \rightarrow 0}[5-4 h]=5-0=5 \)
[put x=1−h; when x→1−,then h→0]   
\( \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x+1) \)
\(\therefore \lim _{h \rightarrow 0}[4(1+h)+1]=\lim _{h \rightarrow 0}(5+4 h)=5+0=5 \)
[put x=1+h; when x→1, then h→0]   
Also, f(1)=4×1+1=5
\([\because f(x)=4 x+1]\)
Thus f(x) is continuous at x=1 for all values of λ.

The area of a circle (A) with radius (r) is given by.
Therefore, the rate of change of area (A) with respect to time (t) is given by,
[By chain rule]
It is given that.
Thus, when r = 8 cm,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80 π cm²/s.

Note that
f '(x) = 3x² – 6x + 4
= 3(x² – 2x + 1) + 1
= 3(x – 1)² + 1 > 0, in every interval of R
Therefore, the function f is increasing on R.

We have,
\(f(x)=x^{2}+2 x-8, x \in[-4,2]\) 
Rolle'stheoremis satisfied
\( \therefore f^{\prime}(c)=0 \)
\(\Rightarrow f^{\prime}(c)=2 c+2=0 \)
\(c+1=0 \Rightarrow c=-1 \) 

Let g (x) = IxI and h(x) = sin x
Now,\(\operatorname{hog}(x)=h[g(x)]=h(|x|)=\sin |x|=f(x)\) 
Since, g (x) and h( x) are both continuous function for all x∈ R,
so composition of g (x) and h( x) is also acontinuous function for all x ∈ R.
Hence, f(x) = sin [x] is a continuous function

Here,
\(f(x) =\frac{x^{2}-25}{x+5}, x \neq-5 \)
\(\therefore  f(x) = x - 5\) is a polynomial function, so f(x) is continuous at all values of x, provided \(x \neq-5.\)

For x = 0,
f(x) is continuous and
at x = 1, LHL = 1 and RHL = 5,
hence discontinuous and
at x = 2;
f(x) = 5 which is continuous

We have 
\(f(x)=x^{2}+a x+1 \)
\(\therefore f^{\prime}(x)=2 x+a \)
Now, function f will be increasing in \( (1,2) \text { , if } f^{\prime}(x)>0 \text { in }(1,2) \text { . }\)
\(f^{\prime}(x)>0 \)
\(\Rightarrow 2 x+a>0 \)
\(\Rightarrow 2 x>-a \)
\(\Rightarrow x>\frac{-a}{2} \)
 Therefore, we have to find the least value of a such that " id="MathJax-Element-2040-Frame" role="presentation" style="position: relative;" tabindex="0"> Therefore, we have to find the least value of a such that
\(x>\frac{-a}{2}, \text { when } x \in(1,2) \)
\(\Rightarrow x>\frac{-a}{2}(\text { when } 1
 Thus, the least value of a for f to be increasing on (1,2) is given by, " id="MathJax-Element-2043-Frame" role="presentation" style="position: relative;" tabindex="0">Thus, the least value of a for f to be increasing on (1,2) is given by,
\(\frac{-a}{2}=1 \)
\(\frac{-a}{2}=1 \Rightarrow a=-2 \)
 Hence, the required value of a is −2 . " id="MathJax-Element-2046-Frame" role="presentation" style="position: relative;" tabindex="0">Hence, the required value of a is −2 . 

To see this we use the following facts
\(\lim _{x \rightarrow 0} \sin x=0\)
We have not proved it, but is intuitively clear from the graph of sin x near 0.
Now, observe that f (x) = sin x is defined for every real number. Let c be a real number. Put x = c + h. If x \(\rightarrow\) c we know that h \(\rightarrow\) 0. Therefore
\(\lim _{x \rightarrow c} f(x) =\lim _{x \rightarrow c} \sin x \)
\(=\lim _{h \rightarrow 0} \sin (c+h) \)
\(=\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h] \)
\(=\lim _{h \rightarrow 0}[\sin c \cos h]+\lim _{h \rightarrow 0}[\cos c \sin h] \)
\(=\sin c+0=\sin c=f(c) \)
Thus \(\lim _{x \rightarrow c} f(x)=f(c)\) and hence f is a continuous function.

We have : x + y = 35 ...(1)
Let \(P=x^{ 2 }y^{ 5 }\)
\(x^{ 2 }(35-x)^{ 5 }\)
\(\therefore \frac { dP }{ dx } =x^{ 2 }.5(35-x)^{ 4 }(-1)+(35-x)^{ 5 }(2x)\)
\( =(35-x)^{ 4 }[-5x^{ 2 }+70x-2x^{ 2 }]\)
\(=(35-x)^{ 4 }(70x-7x^{ 2 })\)
\(=x(35-x)^{ 4 }(70-7x)\)
\(\frac { dP }{ dx } =0\)
\(\Rightarrow x(35-x)^{ 4 }(7x-70)=0\)
\(\Rightarrow x=0,10,35\)
Only  admissible value is x = 0 \([\therefore\) x = 0 and 35 are rejected]
At x = 10
When x < 10 Slightly \(\frac { dP }{ dx } \) = (+) (+) (+) = +ve
When x>10 slightly \(\frac { dP }{ dx } \)= (+) (+) (-) = -ve
\(\Rightarrow \frac { dP }{ dx } \) changes for +ve to -ve
\(\Rightarrow \) P is maximum at x = 10
From (1), y = 35 - 10 = 25
Hence the required numbers are 10 and 25

\(Let\quad y={ sin }^{ 3 }x+{ cos }^{ 6 }x\)
\(\frac { dy }{ dx } ={ 3sin }^{ 2 }x\frac { d }{ dx } \left( sin\quad x \right) +6{ cos }^{ 5 }x\frac { d }{ dx } (cos\quad x)\)
\(={ 3sin }^{ 2 }x(cos\quad x) +6{ cos }^{ 5 }x(-sin\quad x)\)
\(=3\quad sin\quad x.cos\quad x(sin\quad x-2{ cos }^{ 4 }x)\)

\(We\quad have:\quad y=3\quad cos(log\quad x)+4\quad sin(log\quad x)....(1)\)
\(Diff.\quad w.r.t.\quad x,{ y }_{ 1 }=-3\quad sin(log\quad x).\frac { 1 }{ x } +4cos(log\quad x).\frac { 1 }{ x } \)
\({ xy }_{ 1 }=-3\quad sin(log\quad x)+4\quad cos(log\quad x)....(2)\)
\(Again\quad diff.\quad w.r.t.\quad x,\)
\({ xy }_{ 2 }+{ y }_{ 1 }=-3\quad cos(log\quad x).\frac { 1 }{ x } -4\quad sin\quad (log\quad x).\frac { 1 }{ x } \)
\({ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }=-(3\quad cos(log\quad x)+4sin(log\quad x))\)
\(Hence,{ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }+y=0\)

\(We\quad have\quad :\quad x={ 2at }^{ 2 },y={ at }^{ 4 }\)
\(\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { { 4at }^{ 3 } }{ 4at } ={ t }^{ 2 }\)

Let S(x) be the selling price of x items and let C(x) be the cost price of x items. Then, we have

\(\mathrm{S}(x)=\left(5-\frac{x}{100}\right) x=5 x-\frac{x^{2}}{100} \)
\(\mathrm{C}(x)=\frac{x}{5}+500 \)
Thus, the profit function P(x) is given by
\(\mathrm{P}(x)=\mathrm{S}(x)-\mathrm{C}(x)=5 x-\frac{x^{2}}{100}-\frac{x}{5}-500 \)
\(i.e., \mathrm{P}(x)=\frac{24}{5} x-\frac{x^{2}}{100}-500 \)
\(or \ \mathrm{P}^{\prime}(x)=\frac{24}{5}-\frac{x}{50} \)
\(\text { Now } \mathrm{P}^{\prime}(x)=0 \text { gives } x=240 \text { . Also } \mathrm{P}^{\prime \prime}(x)=\frac{-1}{50} \text { . So } \mathrm{P}^{\prime \prime}(240)=\frac{-1}{50}<0\)
Thus, x = 240 is a point of maxima. Hence, the manufacturer can earn maximum profit, if he sells 240 items.

The given equation of curve is
\(\begin{aligned} &y=x^{3}-12 x+18\\ &\therefore \quad \frac{d y}{d x}=3 x^{2}-12\\ &\text { [on differentiating w.r.t. } x] \end{aligned}\) 
So, the slope of line parallel to the X-axis.
\( therefore \quad\left(\frac{d y}{d x}\right)=0\\\Rightarrow \quad 3 x^{2}-12=0\\ \Rightarrow \quad x^{2}=\frac{12}{3}=4\\ \therefore\\ x=\pm 2\\ \text { For } x=2, y=2^{3}-12 \times 2+18=2\\ \text { and for } x=-2, y=(-2)^{3}-12(-2)+18=34\\ \text { So, the points are }(2,2) \text { and }(-2,34) \)

We have \(y^{x}=e^{y-x}\) 
Taking log both sides, we get
\( x \log y =y-x \)
\( \Rightarrow x(\log y+1) =y \)
\(\Rightarrow \frac{y}{1+\log y} =x \) 
\(\left(\frac{(1+\log y)-\left(\frac{1}{y}\right) y}{(1+\log y)^{2}}\right) \frac{d y}{d x}=1\) 
\( \frac{\log y}{(1+\log y)^{2}} \frac{d y}{d x}=1 \)
\(\Rightarrow \frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y} \)

x - [x] = 0 when x is an integer, so that f(x) is discontinuous for all x ∈ I i.e. f(x) is discontinuous at infinite number of points.

As \(\frac{dy}{dt}\) = 2x.\(\frac{dx}{dt}\)
⇒ 6.\(\frac{dx}{dt}\) = 2x.\(\frac{dx}{dt}\) ⇒ x = 3
From curve, y = 9. Point is (3, 9)

As \(\frac { dx }{ dt } =2\) cm/s, x is side of equiolatral triangle.
A = \(\frac { \sqrt { 3 } }{ 4 } { x }^{ 2 }\)
\(\Rightarrow \frac { dA }{ dx } =\frac { \sqrt { 3 } }{ 2 } { x }\frac { dx }{ dt } \)
= \(\frac { \sqrt { 3 } }{ 2 } x2=\sqrt { 3 } x\)
∴\(|\frac{dA}{dx}|\)x=10 = 10\(\sqrt3\) cm²/s

As \(\lim _{ x\rightarrow 0 }{ \frac { \sqrt { \frac { 1 }{ 2 } (1-cosx) } }{ x } } \)
 \(=\lim _{ x\rightarrow 0 }{ \frac { |sin \ x| }{ x } } \)
and LHL ≠ RHL at x = 0

Let r be the radius and h be the height of conical part.
Here, \(\tan 45^{\circ}=\frac{r}{h} \Rightarrow 1=\frac{r}{h} \Rightarrow h=r\)
The volume of water in the tank
\(V=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^3 \mathrm{~cm}^3 \quad[\because h=r]\)
(ii) \( \frac{d V}{d t}=\frac{3}{3} \pi r^2 \frac{d r}{d t} \)
\(\Rightarrow \quad-2=\pi r^2 \frac{d r}{d t} \quad\left[\because \frac{d V}{d t}=-2\right]\)
\( \Rightarrow \quad \frac{d r}{d t}=\frac{-2}{\pi r^2}=\frac{-2}{\pi(2 \sqrt{2})^2}=\frac{-1}{4 \pi}\)
\({[r=2 \sqrt{2} \text { (given) }]}\)
Hence, the rate of decrease of radius is \(\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\).
(iii) (b) \(\therefore S=\pi r l \Rightarrow S=\pi h l\) \([\because r=h]\)
\(\Rightarrow \quad S=\pi h \sqrt{h^2+r^2}\)
(a) \(\therefore S=\pi r l \)
\(=\pi r \sqrt{h^2+r^2}=\pi r \sqrt{2 r^2} \)
\({\left[\because l=\sqrt{h^2+r^2} ; \ \because h=r\right]}\)
\(=\sqrt{2} \pi r^2\)
\(\Rightarrow \frac{d S}{d t}=\sqrt{2} \pi(2 r) \frac{d r}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi r \frac{d r}{d t}\)
\(\Rightarrow\left(\frac{d S}{d t}\right)_{r=2 \sqrt{2}}=2 \sqrt{2} \pi(2 \sqrt{2})\left(\frac{-1}{4 \pi}\right)\)
\(=-2 \mathrm{~cm}^2 / \mathrm{s} \)
Or \(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\) \({[\because r=h]}\)
\( \Rightarrow S=\pi h \sqrt{h^2+r^2} \)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2 \)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\( \Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\( \text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\( \Rightarrow \quad-2=8 \pi \frac{d h}{d t}\) \({[\because \text { from Eq. (i) }]}\)
\( \Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)
\(\text { (b) } \therefore S=\pi r l \Rightarrow S=\pi h l\)
\([\because r=h]\)
\(\Rightarrow S=\pi h \sqrt{h^2+r^2}\)
\(\Rightarrow \quad S=\sqrt{2} \pi h^2\)
\(\Rightarrow \quad \frac{d S}{d t}=\sqrt{2} \pi(2 h) \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi h \frac{d h}{d t}\)
\(\Rightarrow \quad \frac{d S}{d t}=2 \sqrt{2} \pi(2 \sqrt{2}) \frac{d h}{d t}\)
\(\text { [since, } l^2=r^2+h^2 \Rightarrow l^2=2 h^2\)
\(\left.\Rightarrow 16=2 h^2 \Rightarrow h=2 \sqrt{2}\right]\)
\(\Rightarrow \quad-2=8 \pi \frac{d h}{d t}\)
\([\because \text { from Eq. (i) }]\)
\(\Rightarrow \quad \frac{d h}{d t}=-\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s}\)

Given, \(f(x)=\left\{\begin{array}{c}|x-3|, x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, x<1\end{array}\right.\)
\(f^{\prime}(x)=\left\{\begin{array}{cc} -1, & 1<x<3 \\ \frac{x}{2}-\frac{3}{2}, & x<1 \end{array}\right.\)
(i) \(R f^{\prime}(1)=-1\)
(ii) \(L f^{\prime}(1)=\frac{1}{2}-\frac{3}{2}=\frac{-2}{2}=-1\)
(iii) (a) Since, \(L f^{\prime}(1)=R f^{\prime}(1)=-1\)
\(\Rightarrow f(x)\) is differentiable at x = -1
Or
(b) \(f^{\prime}(2)=-1,1 \leq 2<3\)
and \(f^{\prime}(-1)=\frac{-1}{2}-\frac{3}{2}, x<1=\frac{-4}{2}=-2\)

(i) (a) : Let \(\begin{equation} y=\cos \sqrt{x} \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{d}{d x}(\cos \sqrt{x})=-\sin \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \end{equation}\) 
\(\begin{equation} =-\sin \sqrt{x} \times \frac{1}{2 \sqrt{x}}=\frac{-\sin \sqrt{x}}{2 \sqrt{x}} \end{equation}\) 
(ii) (a) : Let \(\begin{equation} y=7^{x+\frac{1}{x}} \quad \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(7^{x+\frac{1}{x}}\right) \end{equation}\) 
\(\begin{equation} =7^{x+\frac{1}{x}} \cdot \log 7 \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)=7^{x+\frac{1}{x}} \cdot \log 7 \cdot\left(1-\frac{1}{x^{2}}\right) \end{equation}\) 
\(\begin{equation} =\left(\frac{x^{2}-1}{x^{2}}\right) \cdot 7^{x+\frac{1}{x}} \cdot \log 7 \end{equation}\) 
(iii) (a) : Let \(\begin{equation} y=\sqrt{\frac{1-\cos x}{1+\cos x}}=\sqrt{\frac{1-1+2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}-1+1}}=\tan \left(\frac{x}{2}\right) \end{equation}\) 
\(\begin{equation} \therefore \frac{d y}{d x}=\sec ^{2} \frac{x}{2} \cdot \frac{1}{2}=\frac{1}{2} \sec ^{2} \frac{x}{2} \end{equation}\) 
(iv) (b) : Let \(\begin{equation} y=\frac{1}{b} \tan ^{-1}\left(\frac{x}{b}\right)+\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right) \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=\frac{1}{b} \times \frac{1}{1+\frac{x^{2}}{b^{2}}} \times \frac{1}{b}+\frac{1}{a} \times \frac{1}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a} \end{equation}\) 
\(\begin{equation} =\frac{1}{b^{2}+x^{2}}+\frac{1}{a^{2}+x^{2}} \end{equation}\) 
(v) (d) : Let \(\begin{equation} y=\sec ^{-1} x+\operatorname{cosec}^{-1} \frac{x}{\sqrt{x^{2}-1}} \end{equation}\) 
Put  \(\begin{equation} x=\sec \theta \Rightarrow \theta=\sec ^{-1} x \end{equation}\) 
\(\begin{equation} \therefore \quad y=\sec ^{-1}(\sec \theta)+\operatorname{cosec}^{-1}\left(\frac{\sec \theta}{\sqrt{\sec ^{2} \theta-1}}\right) \end{equation}\) 
\(\begin{equation} =\theta+\sin ^{-1}\left[\sqrt{1-\cos ^{2} \theta}\right] \end{equation}\) 
\(\begin{equation} =\theta+\sin ^{-1}(\sin \theta)=\theta+\theta=2 \theta=2 \sec ^{-1} x \end{equation}\) 
\(\begin{equation} \therefore \quad \frac{d y}{d x}=2 \frac{d}{d x}\left(\sec ^{-1} x\right)=2 \times \frac{1}{|x| \sqrt{x^{2}-1}}=\frac{2}{|x| \sqrt{x^{2}-1}} \end{equation}\)

(b) We have, f(x) = x² - 4x + 6 or f'(x) = 2x - 4

Therefore, f'(x) = 0 gives x = 2. Now, the point x = 2 divides the real line into two disjoint intervals namely, (- \(\infty\), 2) and (2, \(\infty\)). In the interval ( -\(\infty\), 2), f'(x) = 2x - 4 < 0.
Therefore,f is strictly decreasing in this interval. Also, in the interval (2, \(\infty\)), f'(x) > 0 and so the function f is strictly increasing in this interval.
Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.