Let AB be the lamp post. Let the man CD be at distance x m from lamp post and y m be the length of his shadow at any time t.
Given \(\frac { dx }{ dt } \) = 5 km / hr = 5000 m/hr
ΔABE and CDE are similar

∴ \(\frac { DE }{ CD } =\frac { BE }{ AB } \)
⇒ \(\frac { y }{ 2 } =\frac { x+y }{ 6 } \)
⇒ 6y = 2x+2y
⇒ 4y = 2x
⇒ y = \(\frac { x }{ 2 } \)
⇒ \(\frac { dy }{ dt } =\frac { 1 }{ 2 } \frac { dx }{ dt } =\frac { 1 }{ 2 } \)(5000)
= 2500 m/hr = 2.5 km/hr.

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.

We have,
f(x) = (x-2)², then
\(f'(x)=2(x-2)=0 \) gives x = 2.
The intervals of monotonicity are \((-\infty,2)\) and \((2,\infty)\)
Since  \(f'(x)<0, \forall x \in (-\infty,2)\) for \((-\infty,2)\) the f(x) is strictly decreasing on \((2,\infty)\)
As \(f'(x)>0, \) for \(x \in(2, \infty)\) the function f(x) is strictly increasing on \((2,\infty)\)
Becasue f'(x) changes its sign from negative to positive when passing through x = 2 for the function f(x) it has a local minimum at x = 2
The local minimum value is f(2) = 0.

Given
\(f(x)=\frac { { x }^{ 2 }+6x-4 }{ 3x-6 } \)
\(\underset { x\rightarrow { 2 }^{ + } }{ lim } \frac { { x }^{ 2 }+6x-4 }{ 3x-6 } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 3(2+h)-6 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 6+3h-6 } \)
= -∞
ஃ x = 2 is the vertical asymptote,
Also

\(\therefore y=\frac { 1 }{ 3 } x+\frac { 8 }{ 3 } \) is the slanting asymptote.

\(f(x)=\frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } \) 
\(\underset { x\rightarrow \infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } \) 
= \(\frac { 3 }{ \sqrt { 1+0 } } =3\) 
ஃy = 3 is the horizontal asymptote.
Also \(\underset { x\rightarrow -\infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow { 0 }^{ - } }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } =-3\) 
ஃ y = -3 is the horizontal asymptote

Given f(x) = tan x, x ∈ [0, π]
Rolle's theorem is not applicable since tan x is not continuous at x = \(\frac{\pi}{2}\) [∵ tan \(\frac{\pi}{2}\) = ∞]

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.

(i) At the maximum height, the velocity v(t) of the particle is zero.
Now, we find the velocity of the particle at time t.
\(v(t)=\frac{ds}{dt}=128-32t\)
\(v(t)=0 \Rightarrow 128-32t=0 \Rightarrow t=4.\)
After 4 seconds, the particle reaches the maximum height.
The height at t = 4 is s(4) = 128(4) - 16(4)² = 256 ft.
(ii) When the particle hits the ground then s = 0 .
s = 0 ⇒ 128t −16t² = 0
⇒ t = 0, 8 seconds.
The particle hits the ground at t = 8 seconds. The velocity when it hits the ground v(8) = –128 ft /s.

We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

By the mean value theorem, there exists  \(c \in\left(x_{1}, x_{2}\right) \subset(a, b)\) such that,
\(\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}=f^{\prime}(c)\)
Since \(f^{\prime}(c)>0, \text { and } x_{2}-x_{1}>0 \text { we have } f\left(x_{2}\right)-f\left(x_{1}\right)>0\)
We conclude that, whenever \(x_{1}

f(x) is increasing on \(\left[ 0,\frac { \pi }{ 2 } \right] \) and \(\left[ \frac { 3\pi }{ 2 } ,2\pi \right] \)

This is an indeterminate of the form (0×∞). To evaluate this limit, we first simplify and bring it to the form \((\frac{\infty}{\infty})\) and apply L’Hôpital Rule.
\(\underset{x\rightarrow 0^{+}}{lim}xlog x= \underset{x\rightarrow 0^{+}}{(\frac{log x}{\frac{1}{x}})}\)
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{\frac{1}{x}}{-\frac{1}{x^{2}}})=\underset{x\rightarrow 0^{+}}{lim}(-x)=0.\)

This is an indeterminate of the form \(\infty, -\infty\). To evaluate this limit we first simplify and bring it in the form \((\frac{0}{0})\) and applying the l’Hôpital Rule, we get
\(\frac{x\rightarrow 0^{+}}{lim}(\frac{1}{x}-\frac{1}{e^{x}-1})=\underset{x\rightarrow 0^{+}}{lim}(\frac{e^{x}-x-1}{x(e^{x}-1)})\) \((\frac{0}{0})\) 
Now, \(\underset{x\rightarrow 0^{+}}{lim} (\frac{e^{x}-x-1}{x(e^{x}-1)})=\underset{x\rightarrow 0^{+}}{lim}(\frac{e^{x}-1}{xe^{x}+e^{x}-1})\) \((\frac{0}{0})\)
\(=\underset{x\rightarrow 0^{+}}{lim}(\frac{e^{x}}{xe^{x}+2e^{x}})=\frac{1}{2}\)

Let (x) = sin x
f^I(x) = cos x
f^II(x) = - sin x
f^III(x) = -cos x
⇒ \(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ 1 }\left( \frac { \pi }{ 4 } \right) =cos\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =-cos\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
Taylors' series for f (x) at x = \(\frac { \pi }{ 4 } \) is
\(f(x)=f\left( \frac { \pi }{ 4 } \right) +\frac { { f }^{ I }\left( \frac { \pi }{ 4 } \right) }{ 1! } \left( x-\frac { \pi }{ 4 } \right) +\frac { { f }^{ II }\left( \frac { \pi }{ 4 } \right) }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }+\frac { { f }^{ III }\left( \frac { \pi }{ 4 } \right) }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }\)+...
\(sinx=\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 } }{ 2! } -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 } }{ 3! } +...\)
\(=\frac { 1 }{ \sqrt { 2 } } \left[ 1+\left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }+..... \right] \)
\(sinx=\frac { \sqrt { 2 } }{ 2 } \left[ 1+\frac { 1 }{ 1! } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }+..... \right] \)

f'(x) = ≤1 for all 1 ≤ x ≤ 4
Using Lagrange's mean value theorem,
f'(x) = \(\frac { f(b)-f(a) }{ b-a } \) [∵ f(x) is continuous in [1, 4] and differentiable in (1, 4)]
f'(x) = \(\frac{f(4)-f(1)}{4-1}\)
f'(x) = \(\frac { f(4)-f(1) }{ 3} \)
⇒ \(\frac { f(4)-f(1) }{ 3} \) = f'(s)
⇒ \(\frac { f(4)-f(1) }{ 3} \) ≤ 1[∵ f'(x) ≤ 1]
⇒ f(4) - f(1)  ≤ 3
Hence proved

a) f(x) is continuous in [0, 9]
b) f(x) is differentiable in (0, 9)
c) f(0) = 0
\(f(9)=\sqrt { 9 } -\frac { 9 }{ 3 } =3-3=0\)
∴ f(0) = f(9)
∴ By Rolle's theorem, there exists C ∈ [0, 9] such that f'(c) = 0
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } =0\)
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }=\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ 2\sqrt { c } } =\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ \sqrt { c } } =\frac { 2 }{ 3 } \)
⇒ \(\sqrt { c } =\frac { 2 }{ 3 } \)
Squaring both sides, c = \(\frac94\) ∈ [0, 9]

Given equation of curve is y = \(\frac{x+1}{x-1}\) and the line is x + 2y = 6
Slope of the tangent to the curve
m₁ = \(\frac{dy}{dx}\)
= \(\frac { (x-1)(1)-(x-1)(1) }{ { (x-1) }^{ 2 } } \)
= \(\frac { x-1-x-1 }{ { (x-1) }^{ 2 } } =\frac { -2 }{ { (x-1) }^{ 2 } } \) [Quotient rule]
Slope of the line m₂ = \(\frac{-1}{2}\) \(\left[ \frac { co-efficient\ of\ x }{ co-efficient\ of\ y } \right] \)
Since the tangent to the curve and the lines are parallel, m₁ = m₂
⇒ \(\frac { -2 }{ { (x-1) }^{ 2 } } =\frac { -1 }{ 2 } \)
⇒ 4 = (x - 1)²
⇒ x - 1 = 土 2
⇒ x-1 = 2 or x- 1 = -2
⇒ x = 3 or x = -1
⇒ When x = 3, y = \(\frac{3+1}{3-1}\) = \(\frac42\) = 2
⇒ When x = -1, y = \(\frac{-1+1}{-1-1}=\frac{0}{-2}\) = 0
∴ Equation of the tangent at (3, 2) is
y - 2 = \(\frac{-1}{2}\)(x - 3)
⇒ 2y - 4 = -x + 3
x + 2y -7 = 0

Equation of the given curve is y² - 4xy = x² + 5..(1)
Differentiating with respect to 'x' we get,
\(2y\frac { dy }{ dx } -4\left[ x\frac { dy }{ dx } +y(1) \right] =2x\)
⇒ \(2y\frac { dy }{ dx } -4x\frac { dy }{ dx } -4y=2x\)
⇒ \(\frac { dy }{ dx } \) (2y -4x) = 2x + 4y
⇒ \(\frac { dy }{ dx } \) = \(\frac { x+2y }{ y-2x } \)
Since the tangent to the curve is horizontal, \(\frac { dy }{ dx } \) = 0 
ஃ \(\frac { x+2y }{ y-2x } \) = 0
⇒ x+ 2y = 0
⇒ x = -2y .....(2)
Substituting (2) in (1) we get,
y² - 4(-2y)y = (-1y)² + 5
⇒ y² + 8y² = 4y² + 5
⇒ y² = 4y² + 5
⇒ 5y² = 5
⇒ y² = 1
⇒ y = 土 1
From (2), When y = 1, x = - 2
When y = -1, x = 2
∴ The required points are (2, -1) and (-2,1)

Distance moved in time 't' is s = \(\frac{t^{3}}{3}-t^{2}+3\) 
Velocity at time 't ' is V = \(\frac{ds}{dt}=t^{2}-2t\)
Acceleration at time 't ' is a(t) = \(\frac{dV}{dt}=2t-2\)
Therefore, the velocity is zero when t² − 2t = 0, that is t = 0, 2. The acceleration is zero when 2t − 2 = 0 . That is at time at time t = 1

\(\frac { 35 }{ 38 } m/h\)

100 m / s, t = 4 sec, 200 m / s, −100 m / s

We have,
f'(x) = 4x³+32 = 0 gives x³ = -8
⇒ x = −2
and f′′(x) = 12 x² 
As f''(−2)>0, the function has local minimum at x = −2. The local minimum value is f (−2) = −48 
Therefore, the extreme point is (−2, −48) .

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

This an indeterminate of the form 1∞
Let \(g(x)={ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
Taking logarithm, we get,
\(log(g(x))=log{ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
\(=xlog{ \left( 1+\frac { 1 }{ x } \right) }\)
\(=\frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } =\left( \frac { 0 }{ 0 } form \right) \)

\(=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { 1 } }{ \left( 1+\frac { 1 }{ x } \right) } =1\)
But \(​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=log​​​​​​\underset { x\rightarrow \infty }{ lim } g(x)\)
\(\therefore log(\underset { x\rightarrow \infty }{ lim } g(x))=1\)
\({ e }^{ log }(\underset { x\rightarrow \infty }{ lim } g(x))={ e }^{ 1 }=e\Rightarrow (\underset { x\rightarrow \infty }{ lim } g(x))=e\)
\(\Rightarrow \underset { x\rightarrow \infty }{ lim } { \left( 1+\frac { 1 }{ x } \right) }^{ x }=e\)

\(​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }\)
This is an indeterminate of the form 0⁰
Let g(x) x^x
Taking logarithm, we get
\(log \ g(x)=log({ x }^{ 2 })=xlogx=\frac { log\quad x }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } log \ g(x)={ \left[ \frac { logx }{ \frac { 1 }{ x } } \right] }=\frac { \infty }{ \infty } \)
\(=\underset { x\rightarrow { 0 }^{ + } }{ lim } \left( \frac { \frac { 1 }{ x } }{ -\frac { 1 }{ { x }^{ 2 } } } \right) \) [by L' Hopital rule]
= \(\underset { x\rightarrow { 0 }^{ + } }{ lim } \frac { 1 }{ x } \times \frac { { x }^{ 2 } }{ 1 } =\underset { x\rightarrow { 0 }^{ + } }{ lim } -x\)
= 0
But \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=log(\underset { x\rightarrow { 0 }^{ + } }{ lim } (g(x))\)
ஃ \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=0\)
\(\Rightarrow { e }^{ log }(\underset { x\rightarrow { 0 }^{ + } }{ lim } log(g(x))={ e }^{ 0 }\)
\(\Rightarrow \underset { x\rightarrow { 0 }^{ + } }{ lim } g(x)=1\)
\(\therefore \underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }=1\)

Let x be the length of the rectangular pasture and y be the breadth of the rectangular pasture.
Given xy = 1,80,000
\(\Rightarrow y=\frac { 1,80,000 }{ x } \) 
Since fencing is not needed along the river side, 
Perimeter = 2x + y
Letf(x) = 2x+y

= \(2x+\frac { 1,80,000 }{ x } \) 
\(f'(x)=2-\frac { 1,80,000 }{ { x }^{ 2 } } \) 
f'(x) = 0
\(\Rightarrow 2=\frac { 1,80,000 }{ { x }^{ 2 } } \) 
\(\Rightarrow { x }^{ 2 }=90,000\) 
\(\Rightarrow x=\pm 300\) 
ஃ The critical number is 300, -300
\(f''(x)=-1,80,000\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 360000 }{ { x }^{ 3 } } \) 
\(f''\left( 300 \right) =\frac { 360000 }{ \left( 300 \right) ^{ 3 } } >0\) 
From (1), when x = 300, \(y=\frac { 1,80,000 }{ 300 } =600\) 
ஃ Length of the minimum needed fencing material = 2x + y
= 2(300) + 600 = 600 + 600 = 1200 m

f (x) is defined and differentiable for all x ∈ (0, 2π).
f'(x) = sin x (-sin x) + cos x (cos x)
= cos² X - sin² x
= cos2x
f'(x) = 0
\(\Rightarrow cosx=0cos=\frac { \pi }{ 2 } ,cos\frac { 3\pi }{ 2 } ,cos\frac { 5\pi }{ 2 } ,cos\frac { 7\pi }{ 2 } \)
\(\Rightarrow 2x=\frac { \pi }{ 2 } \frac { \pi }{ 2 } ,\frac { 3\pi }{ 2 } ,\frac { 5\pi }{ 2 } ,\frac { 7\pi }{ 2 } \) 
\(\Rightarrow x=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \)
The stationary points are at

\(x=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \) 
\(\left( 0,\frac { \pi }{ 4 } \right) ,\left( \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right) \left( \frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } \right) \left( \frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \)

\(\therefore\) f (x) is strictly increasing in \(\left( 0,\frac { \pi }{ 4 } \right) \)\(\left( \frac { 3\pi }{ 4 } ,5\frac { \pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) and strictly decreasing in \(\left( \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right) \left( \frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \) 
Since f'(x) changes its positionfrom positive to negative at \(x=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \) there is a local
maximum at \(x=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \) 
\(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } cos\frac { \pi }{ 4 } +5\) 
= \(\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } +5=\frac { 1 }{ 2 } +5=\frac { 11 }{ 2 } \) 
\(f\left( \frac { 5\pi }{ 4 } \right) =sin{ \frac { 5\pi }{ 4 } }cos\frac { \pi }{ 4 } +5\) 
= \(\left( \frac { -1 }{ \sqrt { 2 } } \right) \left( \frac { -1 }{ \sqrt { 2 } } \right) +5\) 
= \(\frac { 1 }{ 2 } +5=\frac { 11 }{ 2 } \) 
Also f'(x) changes its position from negative to positive at \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \) there is local mmimum
at \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \)
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =cos\frac { 3\pi }{ 4 } sin\frac { 3\pi }{ 4 } +5\)
= \(\left( \frac { -1 }{ \sqrt { 2 } } \right) \left( \frac { +1 }{ \sqrt { 2 } } \right) +5=5-\frac { 1 }{ 2 } =\frac { 9 }{ 2 } \) 
\(f\left( \frac { 7\pi }{ 4 } \right) =cos\frac { 7\pi }{ 4 } sin\frac { 7\pi }{ 4 } +5\) 
= \(cos\left( 2\pi -\frac { \pi }{ 4 } \right) sin\left( 2\pi -\frac { \pi }{ 4 } \right) +5\)
= \(\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { -1 }{ \sqrt { 2 } } \right) +5\)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)

Given \(f(x)=\frac { x }{ x-5 } \) 
f(x) is defined and differentiable for all x∈R-[5]
\(\therefore f'(x)=\frac { (x-5)(1)-x(1) }{ (x-5)^{ 2 } } \) 
= \(\frac { x-5-x }{ (x-5)^{ 2 } } =\frac { -5 }{ (x-5)^{ 2 } } \) 
f'(x) = 0
\(\Rightarrow \frac { -5 }{ (x-5)^{ 2 } } \neq 0\) 
There is no stationary point. The possible intervals are (-∞, 5) and (5, ∞).

ஃ f (x) is strictly decreasing on (-∞, 5) and (5, ∞).
Since there is no stationary point, the curve does not changes its position.
Hence there is no local extremum.

Given f(x) = 2.0 + 3x2 - 12x
The given function is defined and differentiable for all x ∈(-∞,∞)
f'(x) = 6x² + 6x - 12
The stationary points are given by
6x² + 6x - 12 = 0
\(\Rightarrow\) x² +x-2 = 0
\(\Rightarrow\) (x + 2)(x - 1) = 0
\(\Rightarrow\) x = -2, 1
Hence the intervals of monotonicity are
(-∞, - 2), (-2, 1) and (1, ∞).

ஃ f(x) is strictly increasing on (-∞, - 2)
(1, ∞) and strictly decreasing on (-2, 1).
Since f'(x) changes from positive to negative at x = -2, there is a local maximum at x = -2
\(\therefore\) f(-2) = 2 (2)³ + 3 (2)² - 12 (-2) 
= 2(-8) + 3(4) + 24 
= -16+12+24 = 20
Also f'(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
ஃ f(1) = 2 (1)³ + 3 (1)² - 12(1)
= 5 -12 = -7

Given f(x) = 3x² - 4x³ ; [-1, 2]
f'(x) = 12x³ - 12x²
f'(x) = 0
⇒12x³- 12x² = 0
⇒ 12x²(x-1) = 0
⇒ x = 0 or x = 1
Evaluatingf(x) at the end points x = -1, x = 2 and at the critical number x = 0, x = 1 we get
f(-1) = 3(-1)4-4 (-1)³
= 3 + 4 = 7
f(2) = 3(2)⁴ - 4(2³)
= 48 - 32 =16
f(0) = 0
f(1) = 3(1)⁴ - 4(1)³
= 3 - 4 = -1
From these values, the absolute maximum is 16 at x = 2 and the absolute minimum is -1 which occurs at x = 1.

Equation of the given curves are x² - y² = r²
xy = c²
x² - y² = r²
⇒ 2x - 2y\(\frac{dy}{dx}\) = 0
⇒ 2x = 2y\(\frac{dy}{dx}\)
⇒ \(\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}\)
Let (x₁, y₁) be the point of intersection of the given curves
∴ Slope of tangent to the first curve m₁ = \(\frac{x_1}{y_1}\) ..(1)
⇒ xy = c²
⇒ x.\(\frac{dy}{dx}\) = -y
⇒ \(\frac{dy}{dx}=\frac{-y}{x}\)
∴ Slope of the tangent to the first curve m₂ = \(\frac{-y_1}{x_1}\)..(2)
Consider m₁ m₂ = \(\left( \frac { { x }_{ 1 } }{ { y }_{ 1 } } \right) \left( \frac { -{ y }_{ 1 } }{ { x }_{ 1 } } \right) =-1\)
Since m₁ m₂ = -1, the given two curves cut orthogonally.

Equation of the rectangular hyperbola is xy = 2
⇒ y = \(\frac{2}{x}\)
⇒ x.\(\frac { dy }{ dx } \) + y(1) = 0
⇒ \(\frac { dy }{ dx } \) = \(\frac{-y}{x}\)
Slope of the tangent to the curve m₁ =\(\frac{-y}{x}\)
Equation of the parabola is x²+ 4y = 0 ...(2)
⇒2x + 4\(\frac { dy }{ dx } \) = 0
\(\frac { dy }{ dx } \) = \(\frac{-2x}{4}=\frac{-x}{2}\)
Slope of the tangent to the curve m₂ = \(\frac{-x}{2}\)
Substituting (1) in (2) we get,
x² + 4 \((\frac{2}{x})\) = 0 ⇒ x² + \(\frac{8}{x}\) = 0
⇒ x³ + 8 = 0
⇒ x³ = -8 = (-2)³ ⇒ x = -2
⇒ When x = -2, y = \(\frac{2}{-2}\) = -1
'∴ The point of intersection of RH and parabola is (-2, -2)
'∴ m₁ = \(\frac { -y }{ x } =\frac { -(-1) }{ -2 } =\frac { -1 }{ 2 } \)
m₂ = \(\frac { -x }{ 2 } =\frac { -(-2) }{ 2 } \) = 1
Let θ be the angle between rectangular hyperbola and parabola
tan \(\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \left| \frac { -\frac { 1 }{ 2 } -1 }{ 1+\left( \frac { 1 }{ 2 } \right) (1) } \right| \)
= \(\left| \frac { -\frac { 3 }{ 2 } }{ 1-\frac { 1 }{ 2 } } \right| =\left| \frac { -\frac { 3 }{ 2 } }{ \frac { 1 }{ 2 } } \right| =\left| -3 \right| =3\)
∴ θ = tan^-1 (3)

Given s (t) = 2t³ − 9t² + 12t ≥ 0
On differentiating we get
V(t) = 6t²-18t+ 12 ... (1)
= 6 (t² - 3t+ 2)
= 6 (t - 1) (t - 2)
Now V(t) = 0
⇒ 6 (t-1)(t- 2) = 0
⇒ t = 1, 2
The particle changes direction when V(t) changes its sign.
If 0 ≤ t < 1 then both (t - 1) and (t - 2) < 0
⇒ V(t) > 0
If 1 < t < 2 then (t -1) > 0 and (t - 2) < 0
⇒ V(t) < 0
If t > 2 then both (t - 1) and (t - 2) > 0
⇒ V(t) > 0
∴ The particle changes direction when t = 1 and t = 2 sec.
(ii) Total distance travelled by the particle in the first 4 seconds is |s(0)- s (1)| + |s (1) - s (2)| + |s (2) -s (4)|
s(0) = -4
s(1) = 2(1)3 - 9(1)2 + 12 (1) - 4
= 2 - 9 + 12 - 4 = 1
s (2) = 2 \(\times\) 2³ - 9 \(\times\) 2² + 12 \(\times\) 2 - 4
= 16 - 36 + 24 - 4
= 0
s (4) = 2(4)³ - 9(4)² + 12 (4) - 4
= 128 - 144 + 48 - 4 = 28
∴ Is (0) -s (1)|+ Is (1) -s (2)|+ Is (2) -s(4)|
= |-4 - 1| + |1 - 0| + |0 - 28|
= |-5| + |1| + |0 - 28|
= 5 + 1 + 28 = 34 m
(iii) Given s (t) = 2t³ − 9t² + 12t ≥ 0
[acceleration = \(\frac { dv }{ dt } \)]
When t = 1,
Acceleration = 12 (1) - 18 = -6 m/sec²
When t = 2
Acceleration = 12 (2) - 18 = 6 m/sec²

Given equation of the curve is y = 1 + x³ and the line is x + 12y = 12
Slope of the tangent to the curve
m₁ = \(\frac { dy }{ dx } \) = 3x² and the
Slope of the line = m₂
= \(\frac{-1}{2}\) \(\left[ \because m=\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } \right] \)
Since the slope of the tangent to the curve and the line are orthogonal, m₁ m₂ = - 1.
∴ 3x²\(\left( \frac { -1 }{ 2 } \right) \) = -1
⇒ \(\frac{x^2}{4}\) = 1
⇒ x² = 4
⇒ x = ±2
When x = 2, y = 1 + 2³ = 9
When x = -2, y = 1+ (-2)³
= 1-8 = -7 
∴ Equation of the tangent at (2, 9) is
y-9 = 12(x-2) [∵ m₁ = 3x² = 3(2)² = 12]
∴ y - 9 = 12x - 24
∴ 12x - y = 15
Equation of the tangent at (-2, -7) is
y + 7= 12(x + 2)
⇒ y + 7 = 12x + 24
⇒ 12x - y = -17