f(x) = x³+ 2x²-1
f'(x) = 3x² + 4x = 0
⇒ x (3x + 4) = 0
⇒ x = 0 or \(\frac { 4 }{ 3 } \)
The possible intervals are \(\left( -\infty ,-\frac { 4 }{ 3 } \right) \left( -\frac { 4 }{ 3 } ,0 \right) \) and (0, ∞).

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.

(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.

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,  \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}-\frac{1}{d}\).

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

The volume of the baloon of radius r is \(V =\frac{4}{3}\pi r ^{3} \)
We are given \(\frac{dV }{dt }=1000 \) and we need to find \(\frac{dr }{dt} \) when r = 7. Now,
\(\frac{dV}{dt}=3\times\frac{4}{3}\pi r^{2}\times \frac{dr}{dt}\)
Substituting r = 7 an \(\frac{dV}{dt}\) = 1000, we get 1000 \(= 4\pi \times 49 \times \frac{dr}{dt}\)
Hence, \(\frac{dr}{dt}=\frac{1000}{4\times49\times \pi}=\frac{250}{49\pi}\)

The surface area S of the baloon is S = 4ㅠr². Therefore, \(\frac{dS}{dt}=8\pi \times r \times \frac{dr}{dt}\)
Substituting\(\frac{dr}{dt}=\frac{250}{49\pi}\) and r = 7, we get \(\frac{dS}{dt}=8\pi\times7\times \frac{250}{49\pi}=\frac{2000}{7}\)
Therefore, the rate of change of radius is \(\frac{250}{49 \pi}\) cm/sec and the change of surface area is \(\frac{2000}{7}\) cm² / sec

f(x) = x³-2x²-x+3
f'(x) = 3x² - 4x - 1
f'(c) = 3c² - 4c- 1
f(a) = f(0) = 3
f(b) = f(1) = 1³-2-1+3 = 1
Then, if atleast one C \(\in \) (0, 1) such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ 3c²-4c-1 = \(\frac { 1-3 }{ 1-0 } \)
⇒ 3c²-4c+1 = 0
⇒ c = \(\frac { 4\pm \sqrt { 16-4(3) } }{ 2(3) } \)
⇒ c = \(\frac { 4\pm 2 }{ 6 } =\frac { 6 }{ 6 } \) or \(\frac { 2 }{ 6 } \)
⇒ 1 or \(\frac { 1 }{ 3 } \).

y = sin²x
\(\frac { dy }{ dx } \) = 2 sin x cos x = sin 2x
∴ m = \(\left( \frac { dy }{ dx } \right) _{ \left( \frac { \pi }{ 3 } ,\frac { 3 }{ 4 } \right) }=sin\frac { 2\pi }{ 3 } =\frac { \sqrt { 3 } }{ 2 } \)
∴ Slope of the normal = \(-\frac { 1 }{ m } =-\frac { 2 }{ \sqrt { 3 } } \)
∴ Equation of normal is y-y₁ = \(-\frac { 1 }{ m } \)(x-x₁)
⇒ \(y-\frac { 3 }{ 4 } =-\frac { 2 }{ \sqrt { 3 } } \left( x-\frac { \pi }{ 3 } \right) \)
⇒ 12\(\sqrt { 3 } \)y-9\(\sqrt { 3 } \) = -24x + 8π [ multiply 12\(\sqrt { 3 } \)]
∴ 24x + 12\(\sqrt { 3 } \)y = 8π + 9\(\sqrt { 3 } \).

f(x) = x²e^-2x
f(x) = x² e^-2x
f'(x) = x² (-2) e^-2x + e^-2x(2x)
f''(x) = 2xe^-2x (1 -x)
f'(x) = 0
⇒ 2x e^-2x(1 - x) = 0
⇒ x = 0,1
∴ The critical numbers are x = 0, 1
f"(x) = [x e^-2x (-1) + x (-2)e^-2x(1-x)+1e^-2x(1-x)]
= 2e^-2x(-x^-2x+2x²+ 1-x)
= 2e^-2x(2x²- 4x + 1)
f"(0) = 2(1)(1) = 2 > 0
f"(1) = 2e^-2 (2 - 4 + 1)
= 2e^-2 (-1)
= \(-{ 2e }^{ 2 }=\frac { -2 }{ { e }^{ 2 } } <0\)
Since f"(0) > 0, there is a local minimum at x = 0.
ஃ(0) = 0² e⁰ = 0
Since f"(1) < 0, there is a local maximum at x = 1.
\(\therefore f(1)={ 1 }^{ 2 }e^{ -2(1) }={ e }^{ -2 }=\frac { 1 }{ { e }^{ 2 } } \)

Let P(x) = \(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\). Let  \(\alpha<\beta \) be two real zeros of P(x). Therefore, \(P(\alpha)=P(\beta)=0.\) Since P(x) is continuous in \([\alpha, \beta]\) and differentiable in \((\alpha, \beta)\) by an application of Rolle’s theorem there exists \(\gamma \in (\alpha,\beta)\) such that \(P'(\gamma)=0\). Since,
\(P'(x)=na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_{1}\) which completes the proof.

Observe that the given curve is neither a circle nor an ellipse. For your reference the curve is shown in Figure.

Now,  \(\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} } \)
= -\(\frac{6 cos2t}{6sin3t} = -\frac{cos2t}{sin3t} \).
Therefore, the tangent at any point is
\(y-3sin2t= -\frac{cos2t}{sin3t}(x-2cos3t)\)
That is, x cos 2t + y sin 3t = 3sin 2t sin 3t + 2cos 2t cos 3t.
The slope of the normal is the negative of the reciprocal of the tangent which in this case is \(\frac{sin3t}{cos2t}\). Hence, the equation of the normal is \(y-3sin2t=\frac{sin3t}{cos2t}(x-2cos3t)\).
That is, x sin 3t - y cos 2t = 2sin 3t cos3t 3sin 2t cos 2t = sin 6t - \(\frac{3}{2}\) sin 4t.

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