f(x) -|x+3| =  ∀ \(x\in R\).
Now, |x+3| ≥ 0  ∀ \(x\in R\).
⇒ f(x) ≥ 0  ∀ \(x\in R\).
So, the minimum value of f(x) is 0
Also, f(x) = |x + 3| does not have the maximum value.

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 \(\frac{dT}{dx}=0\)

Let OA = x m, OB = y m
Then x² + y² = 25

Differentiating, \(2x\frac { dx }{ dt } +2y\frac { dy }{ dt } \) = 0
⇒ \(\frac { dy }{ dx } =-\frac { x }{ y } \frac { dx }{ dt } \)
When \(\frac { dx }{ dt } =\frac { 1 }{ 2 } ,\frac { dy }{ dt } =\frac { -x }{ 2y } \)
When y = 4, x² = 25-y²
⇒ x =\(\sqrt { 25-16 } \) = 3
Thus \(\frac { dy }{ dt } =-\frac { 3 }{ 2\times 4 } =\frac { -3 }{ 8 } \).

Differentiating the given function, we get
f'(x) = 6x² + 6x -12
= 6(x² + x - 2)
f'(x) = 6(x + 2)(x + 1)
Thus, f'(x) = 0 \(\Rightarrow\) x = -2, 1 \(\in\) (-3, 2).
Therefore, the critical numbers are, x = -2, 1. Evaluating f(x) at the endpoints x = -3, 2 and at critical numbers x = -2,1 we get f(-3) = 9, f(2) = 4, f(-2) = 20 and f(1) = -7.
From these values, the absolute maximum is 20 which occurs at, x = -2 and the absolute minimum is −7 which occurs at x = 1

Equation of the given curve is y = x sin x
∴ slope = m = \(\left( \frac { dy }{ dx } \right) \)_{\(\left( \frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)}
= \(\frac { \pi }{ 2 } cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } \)
= \(\frac { \pi }{ 2 } \) (0) + 1 = 1
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - \(\frac { \pi }{ 2 } \) = 1 \(\left( x-\frac { \pi }{ 2 } \right) \)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
Equating of the normal is y - y₁ = \(\frac{-1}{m}\) (x - x₁)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
⇒ 2x + 2y = 2π
⇒ x + y- π = 0

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

Let r be the radius of the base, h be the height of the cone, I be the slant height and θ be the semi vertical angle.
In ΔOAB l² = h²+r²
⇒ r² =l²-h²
⇒ V =\(\frac { 1 }{ 3 } { \pi }r^{ 2 }h=\frac { \pi }{ 3 } \)(l²-h²)h
=\(\frac { \pi }{ 3 } \) (l²h - h³)
\(\frac { dv }{ dh } =\frac { \pi }{ 3 } \)(l²-3h²)

\(\frac { dv }{ dh } \)=0
⇒ \(\frac { \pi }{ 3 } \)(l² - 3h²) = 0
⇒ l² - 3h² = 0 ⇒ \(\frac { { l }^{ 2 } }{ { h }^{ 2 } } \) =3
⇒ \(\frac { l }{ h } \) = \(\sqrt { 3 } \)  ⇒ h=\(\frac { 1 }{ \sqrt { 3 } } \)
Now, \(\frac { d^{ 2 }V }{ dh^{ 2 } } =\frac { \pi }{ 3 } \)(-6h) = -2πh
at h =\(\frac { l }{ \sqrt { 3 } } ,\frac { d^{ 2 }V }{ dh^{ 2 } } =-2\pi \left( \frac { l }{ \sqrt { 3 } } \right) \) < 0
V is max at h =\(\frac { l }{ \sqrt { 3 } } \)
∴ r² =l²-\(\frac { { l }^{ 2 } }{ 3 } =\frac { 2{ l }^{ 2 } }{ 3 } \) =2h²
\(\\ \left[ \because h=\frac { l }{ \sqrt { 3 } } \Rightarrow h^{ 2 }=\frac { { l }^{ 2 } }{ 3 } \right] \)
∴ \(\frac { { r }^{ 2 } }{ { h }^{ 2 } } \) =2
⇒ \(\frac { r }{ h } =\sqrt { 2 } \)
⇒ tanθ =\(\sqrt { 2 } \)
⇒ θ = tan^-1(\(\sqrt { 2 } \))

Let us now find the slopes of the curves.
The slope m₁ for the curve y = x².
\(m_{1}=\frac{dy}{dx}=2x\).
and slope m₂ for the curve x = y² 
\(m_{2}=\frac{dy}{dx}=\frac{1}{2y}\).
Let \(\theta \)₁ and \(\theta _2\) be the acute angles at (0,0) and (1,1) respectively.
At (0,0), we come across the indeterminate form of 0×∞ in the denominator of
 tan \(\theta \)₁ = \(|\frac{2x-\frac{1}{2y}}{1+(2x)(\frac{1}{2y})}|\) and so we follow the limiting process

\(tan\theta_{1}=\underset{(x,y)\rightarrow(0,0)}{lim}|\frac{2x-\frac{1}{2y}}{1+(2x)(\frac{1}{2y})}|\)
=\(\underset{(x,y)\rightarrow(0,0)}{lim} {|\frac{4xy-1}{2(y-x)}|}\)
= \( \infty \)
which gives \(\theta_{1}=tan^{-1}(\infty)=\frac{\pi}{2}\).
At, (1,1) \(m_{1}=2, m_{2}=\frac{1}{2}\)
\(tan\theta_{2}=|\frac{2-\frac{1} {2}}{1+(2)(\frac{1}{2})}|\)
=\(\frac{3}{4}\)
which gives \(\theta_{2}=tan^{-1}(\frac{3}{4})\).

Let a(t) be the distance of car A north of P at time t, and b (t) the distance of car B east of P at time t, and let c(t) be the distance from car A to car B at time t. By the Pythagorean Theorem, c(t)² = a(t)² + b(t)²

Taking derivatives, we get 2c(t)c'(t) = 2a(t)a'(t) + 2b(t)b'(t).
So, c′ = \(\frac { { aa }^{ ' }+{ bb }^{ ' } }{ c } =\frac { { aa }^{ ' }+{ bb }^{ ' } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
Substituting known values, we get
\(c' =\frac { (10\times 80)+(15\times 100) }{ \sqrt { { 10 }^{ 2 }+{ 15 }^{ 2 } } } =\frac { 460 }{ \sqrt { 13 } } \) ≈ 127.6 km/hr at the time of intersect

Given that s(t) = t³ − 6t² + 9t + 1. On differentiating, we get v(t) = 3t² -12t + 9 and a(t) = 6t −12.
(i) The particle is at rest when v(t) = 0 . Therefore, v(t) = 3(t −1)(t − 3) = 0 gives t = 1 and t = 3.
(ii) The particle changes direction when v (t) changes its sign. Now.
if 0 ≤ t < 1 then both (t −1) and (t − 3) < 0 and hence, v(t) > 0.
If 1< t < 3 then (t −1) > 0 and (t − 3) < 0 and hence, v(t) < 0.
If t > 3 then both (t −1) and (t − 3) > 0 and hence, v(t) > 0.
Therefore, the particle changes direction when t = 1 and t = 3.
(iii) The total distance travelled by the particle from time t = 0 to t = 2 is given by,
|s(0) − s(1)| + |s(1) − s(2)| = |1− 5 | + | 5 − 3| = 6 metres.

If the motion is upward, at the maximum height the velocity is not zero.