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 \(\frac{\pi}{2}\) = ∞]

(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.

\(​​​​​​\underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }\)
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 \(\underset { x\rightarrow { 0 }^{ + } }{ lim } (log(g(x))=0\)
\(\Rightarrow { e }^{ log }(\underset { x\rightarrow { 0 }^{ + } }{ lim } log(g(x))={ e }^{ 0 }\)
\(\Rightarrow \underset { x\rightarrow { 0 }^{ + } }{ lim } g(x)=1\)
\(\therefore \underset { x\rightarrow { 0 }^{ + } }{ lim } { x }^{ x }=1\)

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, \(\underset{x\rightarrow 0^{+}}{lim}g(x)=e.\)

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)

Let the two curves intersect at a point (x₀ , y₀) This leads to (a-c)x₀² + (b-d)y₀² = 0
Let us now find the slope of the curves at the point of intersection (x₀, y₀). The slopes of the curves are as follows :
For the curve ax² + by² = 1,  \(\frac{dy}{dx}= -\frac{ax}{by}\)
For the curve cx² + dy² = 1, \(\frac{dy}{dx}= -\frac{cx}{by}\)
Now, two curves cut orthogonally, if the product of their slopes intersection (x₀, y₀) is −1. Hence, for the above two curves to cut orthogonally at (x₀, y₀) if
\((-\frac{ax_{0}}{by_{0}})\times(-\frac{cx_{0}}{dy_{0}})=-1\)
That is, acx₀² + bdy₀² = 0,
together with \((a-c)x^{2}_{0}+(b-d)y_{0}^{2}=0\)
gives, \(\frac{a-c}{ac}=\frac{b-d}{bd}\)
That is,  \(\frac{1}{c}-\frac{1}{a}=\frac{1}{d}-\frac{1}{b}\).
Hence,  \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}\).

Let x represent the distance covered by the car, y represent the distance covered by the police jeep, and s represent the distance between the car and jeep.
ஃ Given = x = 0.8 km, y = 0.6 km,

\(\frac { dy }{ dt } \) = -60km/hr,
\(\frac { ds }{ dt } \) = 20 km/hr,
In ΔABC, S² = x² + y² ......(1)
⇒ S² = (0.8)² + (0.6)²
= 0.64 + 0.36
⇒ S² = 1
⇒ s = 1 ....(2)
Differentiating (1) with respect to 't' we get,
\(2s\frac { ds }{ dt } =2x\frac { dx }{ dt } +2y\frac { dy }{ dt } \)
⇒ \(s\frac { ds }{ dt } =x\frac { dx }{ dt } +y\frac { dy }{ dt } \) [Divided by 2]
⇒\(1\left( \frac { ds }{ dt } \right) =(0.8)\left( \frac { dx }{ dt } \right) +(0.6)(-60)\)
⇒ 1(20) = (0.8) \(\left( \frac { dx }{ dt } \right) \) + (0.6)(-60)
⇒ 20 = (0.8) \(\left( \frac { dx }{ dt } \right) \) - 36
⇒ 20 + 36 = (0.8) \(\frac { dx }{ dt } \)
⇒ \(\frac { dx }{ dt } =\frac { 56 }{ 0.8 } \) = 70km/hr.
⇒Speed of the car is 70 km/hr.

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. 

\(\underset { x\rightarrow \infty }{ lim } { e }^{ -x }\sqrt { x } =\underset { x\rightarrow \infty }{ lim } \frac { { \sqrt { x } } }{ { e }^{ x } } =\frac { \infty }{ \infty } \)
Which is in indeterminate form. Applying L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { e }^{ -x } }{ \sqrt { x } } \)
\(\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } e^{ -x }\sqrt { \frac { 1 }{ x } } =\frac { 1 }{ 2 } { e }^{ -\infty }(0)=0\) [When x ➝ ∞, \(\frac1x\) ➝ 0 e^-∞ = 0]

Meclaurin's expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\).................
log(1-x) = \(0-\frac { 1 }{ 1! } x-\frac { { x }^{ 2 } }{ 2! }- \frac { 2 }{ 3! } { x }^{ 3 }-\frac { { 6x }^{ 4 } }{ 4! } +.....\)
= \(-x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } \)+ .....
log(1 - x) = \(-\left( x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +.... \right) \)

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

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).