We have,
\(f'(x)=2x>0, \forall x\in(2,7)\) and
\(f'(x)=2x>0, \forall x\in(-2,0)\)
and hence the proof is completed.

f (1) = 0 and f(2) = -2. Clearly f (x) is defined and differentiable in 1< x <2.
Therefore,  by the Mean Value Theorem, there exists a c∈(1, 2) such that by
\(f'(c)=\frac{f(2)-f(1)}{2-1}=1-2c\)
That is  \(1-2c= -2 \Rightarrow c= \frac{3}{2}\)

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

(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.

This is an indeterminate of the form \(1^{\infty}\).
Let \(g(x)=(1+x)^{\frac{1}{x}}\). Taking the logarithm, we get
\(log \ g(x)=\frac{log(1+x)}{x}\)
\(\underset{x\rightarrow 0^{+}}{lim} log (g(x))=\underset{x\rightarrow 0^{+}}{lim}(\frac{log(1+x)}{x})\) \((\frac{0}{0})\) 
=\(\underset{x\rightarrow0^{+}}{lim}(\frac{\frac{1}{1+x}}{1})\)  (by 1’Hôpital Rule)
= 1.
But, \(\underset{x\rightarrow0^{+}}{lim}log g(x)=log(\underset{x\rightarrow 0^{+}}{lim} g(x))\)
Therefore, log\((\underset{x\rightarrow 0^{+}}{lim} g(x))=1\).
Hence by exponentiating, we get, \(\underset{x\rightarrow 0^{+}}{lim}g(x)=e.\)

Let the point of intersection of the two curves be (a,b) . Hence,
 \(a^{2}+4b^{2}=8\) and  \(a^{2}-2b^{2}=4\) ...(4)
It is enough if we show that the product of the slopes of the two curves evaluated at (a, b) is −1.
Differentiation of \(x^{2}+4y^{2}=8\)  with respect x, gives
\(2x+8y=\frac{dy}{dx}=0\).
Therefore \(\frac{dy}{dx}= -\frac{-x}{4y}\),
\((\frac{dy}{dx})_{(a,b)}=m_{1}= -\frac{a}{4b}\)
Differentiation of x²-2y² = 4 with respect to x, gives
\(2x-4y\frac{dy}{dx}=0\)
Therefore, \(\frac{dy}{dx}=\frac{x}{2y}\), 
at \((a,b)(\frac{dy}{dx})=m_{2}= \frac{a}{4b}\).
Therefore, \(m_{1}\times m_{2}=(-\frac{a}{4b})\times (\frac{a}{2b})= -\frac{a^{2}}{8b^{2}}\)  ...(5)
Applying the ratio of proportions in (4), we get
\(\frac{a^{2}}{-16-16}=\frac{b^{2}}{-8+4}=\frac{1}{-2-4}\)
Therefore, \(\frac{a^{2}}{b^{2}}=\frac{32}{4}=8\) Substituting in (5), we get \(m_{1}\times m_{2}=-1\) Hence, the curves cut orthogonally.

Let AB be the position of the ladder at any time t such that OA = x and OB = y
Then OA² + OB² = AB²
⇒ x² + y² = 17²
Given \(\frac { dx }{ dt } \) = 5 and x = 8
When x = 8, 8² + y² = 17²
 ⇒ y² = 289 - 64 = 225
 ⇒ y = 15
Differentiating (1) with respect to 't' we get,
\(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } =0\)
⇒ 8(5) + 15 \(\frac { dy }{ dt } \) = 0 [∵ x = 8, \(\frac { dx }{ dt } \) = 15, y = 15]
⇒ 40 + 15\(\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dt } =\frac { -40 }{ 15 } =\frac { -8 }{ 3 } \) m/sec
∴ The rate of top of the ladder moving down the wall is \(\frac{-8}{3}\) m/sec
(ii) The ladder, the wall and the floor forms a right angled triangle.
 Area = \(\frac12\)xy
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } =\frac { 1 }{ 2 } \left[ x\frac { dy }{ dx } +y\frac { dx }{ dt } \right] \)
\(=\frac { 1 }{ 2 } \left[ 8\left( -\frac { 8 }{ 3 } \right) +15(5) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64 }{ 3 } +75 \right] =\frac { 1 }{ 2 } \)
= \(\frac { 1 }{ 2 } \left[ \frac { -64+225 }{ 3 } \right] =\frac { 1 }{ 2 } \left( \frac { 161 }{ 3 } \right) \)
= \(\frac { dA }{ dt } \) = 26.83 sq.m/sec

Let h and r be the height and the base radius. Therefore h = 2r. Let V be the volume of the salt cone.

\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{12}\pi h^{3}; \frac{dV}{dt}=30\) mtr³ / min.
Hence,  \(\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}\)
Therefore,  \(\frac{dh}{dt}=4 \frac{dV}{dt}.\frac{1}{\pi h^{2}}\)
That is,  \(\frac{dh}{dt}=4\times30\times \frac{1}{100 \pi}\)
=\(\frac{6}{5\pi}\) mtr / min.