Let x, y be the sides of the rectangle. Hence the area of the rectangle is xy = k (given). The perimeter of the rectangle P is 2(x+ y). So the problem is to minimize 2(x+ y) suject to the condition xy = k. Let \(P(x)=2\left( x+\frac { k }{ x } \right) \)
\(P'(x)=2\left( 1-\frac { k }{ { x }^{ 2 } } \right) \)
P'(x) = 0 gives \(\left( 1-\frac { k }{ { x }^{ 2 } } \right) =0\)
As x, y are sides of the rectangle, \(x=\sqrt { k } \) is a critical number.
Now, P''(x) = \(\frac{4k}{x^3}\) and P''(\(​​\sqrt k\)) >0 \(\Rightarrow\) p(x) and has its minimum value at \(​​\sqrt k\)
Substituting \(x=\sqrt { k } \)  in xy = k we get \(y=\sqrt { k } \) . Therefore the minimum perimeter rectangle of a given area is a square.

The amount after t years \((A)=\underset { x\rightarrow { \infty } }{ lim } { A }_{ 0 }{ \left( 1+\frac { r }{ n } \right) }^{ nt }\)
This is an indeterminate of the form 1∞.
Let g(x) = \({ \left( 1+\frac { r }{ n } \right) }^{ nt }\)
Taking logarithm we get,
log (g(x) = \(log{ \left( 1+\frac { r }{ n } \right) }^{ nt }\)
= \(\frac { log{ \left( 1+\frac { r }{ n } \right) } }{ \frac { 1 }{ nt } } \)
\(\therefore \underset { x\rightarrow { \infty } }{ lim } log(g(x)=\underset { x\rightarrow { \infty } }{ lim } \frac { log{ \left( 1+\frac { r }{ n } \right) } }{ \frac { 1 }{ nt } } \left( \frac { 0 }{ 0 } from \right) \)
\(=\underset { x\rightarrow { \infty } }{ lim } \frac { \frac { 1 }{ 1+\frac { r }{ n } } \left( \frac { -r }{ { n }^{ 2 } } \right) }{ -\frac { 1 }{ { n }^{ 2 }t } } \) [By L' Hôpital rule]

\(=\underset { x\rightarrow { \infty } }{ lim } \frac { 1 }{ 1+\frac { r }{ n } } \left( \frac { -r }{ { n }^{ 2 } } \right) \left( -\frac { { n }^{ 2 }t }{ 1 } \right) \)
= \(\underset { x\rightarrow { \infty } }{ lim } \frac { 1 }{ 1+\frac { r }{ n } } (rt)\)
= \(\left( \frac { 1 }{ 1+0 } \right) (rt)={ A }_{ 0 }rt\)
But \(\underset { x\rightarrow { \infty } }{ lim } log(g(x))=log(\underset { x\rightarrow { \infty } }{ lim } g(x))\)
\(\therefore log(\underset { x\rightarrow { \infty } }{ lim } g(x))=rt\)
\(\Rightarrow { e }^{ log(\underset { x\rightarrow { \infty } }{ lim } g(x)) }=rt\)
\(\Rightarrow \underset { x\rightarrow { \infty } }{ lim } g(x)={ e }^{ rt }\)
\(\Rightarrow \underset { x\rightarrow { \infty } }{ lim } { A }_{ 0 }{ \left( 1+\frac { r }{ n } \right) }^{ nt }={ A }_{ 0 }.{ e }^{ rt }\)
\(\Rightarrow A={ A }_{ 0 }.({ e }^{ rt })\)
Hence Proved.

Given r + h = 6
⇒ h = 6 - r ...(1)
Let f(r) = V = πr²h
= πr²(6-r) = π(6r² -r³)
f'(r) = π (12r - 3r²)
∴f'(r) = 0
⇒π (12r- 3r²) = 0
⇒ 12r-3r² = 0
⇒ 3r(4-r) = 0
⇒ r = 0 or r = 4
ஃThe critical numbers are 0, 4
f"(r) = π (12 - 6r)
When r = 4, f"(r) = π(12 - 24) < 0
ஃ f(r) maximum when r = 4
When r = 4, h = 6 - 4 = 2
When r = 0, h = 6 - 0 = 6
ஃ Volume of the cylinder V = πr² h = π (4)² (2)
= 32 πCu. units
or volume of the cylinder V = π (0²) (6) = 0
Cu. Units.

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 \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \)
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =cos\frac { 3\pi }{ 4 } sin\frac { 3\pi }{ 4 } +5\)
= \(\left( \frac { -1 }{ \sqrt { 2 } } \right) \left( \frac { +1 }{ \sqrt { 2 } } \right) +5=5-\frac { 1 }{ 2 } =\frac { 9 }{ 2 } \) 
\(f\left( \frac { 7\pi }{ 4 } \right) =cos\frac { 7\pi }{ 4 } sin\frac { 7\pi }{ 4 } +5\) 
= \(cos\left( 2\pi -\frac { \pi }{ 4 } \right) sin\left( 2\pi -\frac { \pi }{ 4 } \right) +5\)
= \(\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { -1 }{ \sqrt { 2 } } \right) +5\)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)

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

We have, \(P=\frac{234+16x}{x+3}\)
Therefore, \(\frac{dP}{dt}= - \frac{186}{(x+3)^{2}}\times \frac{dx}{dt}\).
Substituting \(x=90, \frac{dx}{dt}=15\) we get\(\frac{dP}{dt}= -\frac{186}{93^{2}}\times 15= -\frac{10}{31}\approx -0.32\) repee/ week.
That is the price is changing, in fact decreasing at the rate of Rs. 0.32 per unit.

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.

The average rate of change in an interval [a, b] is \(\frac { f(b)-f(a) }{ b-a } \) whereas, the instantaneous rate of change at a point x is f′(x) for the given function. They are respectively, b + a and 2x.

\(C=\frac { 3\pi }{ 4 } \)

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 \(\frac { dy }{ dt } =-\frac { 3 }{ 2\times 4 } =\frac { -3 }{ 8 } \).

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 \(\frac { 1 }{ 3 } \).

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, \(\underset{x\rightarrow 0^{+}}{lim} (\frac{e^{x}-x-1}{x(e^{x}-1)})=\underset{x\rightarrow 0^{+}}{lim}(\frac{e^{x}-1}{xe^{x}+e^{x}-1})\) \((\frac{0}{0})\)
\(=\underset{x\rightarrow 0^{+}}{lim}(\frac{e^{x}}{xe^{x}+2e^{x}})=\frac{1}{2}\)

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

Let f (t) be the distance travelled by the trucker in 't' hours. This is a continuous function in [0, 2] and differentiable in (0, 2). Now, f (0) = 0 and f (2) =164. By an application of the Mean Value Theorem, there exists a time c such that, \(f'(c)=\frac{164-0}{2-0}=82>80\)
Therefore at some point of time, during the travel in 2 hours the trucker must have travelled with a speed more than 80 km which justifies the issuance of a speed violation ticket.

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.

Given equation of the curve is y = 1 + x³ and the line is x + 12y = 12
Slope of the tangent to the curve
m₁ = \(\frac { dy }{ dx } \) = 3 x² and the
Slope of the line = m₂
= \(\frac{-1}{2}\) \(\left[ \because m=\frac { co-efficient\ of\ x }{ co-efficient\ of\ y } \right] \)
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 = 0

The curve y = sin x intersects the positive x -axis. When y = 0 which gives, x = 
\( x=n\pi , n=1,2,3,...\)
Now, \(\frac{dy}{dx}=cos x\). The slpoe \(x=n\pi\)  are \(cos(n\pi)=(-1)^{n}\).   
Hence, the required angle of intersection is m₂ = 0
\(tan \theta = \frac{(-1)^n - 0}{1+((-1)^n(0)} = 1 ∀ n\)

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.

Absolute minimum value = 1,
Absolute maximum value = 6

(i) π
(ii) πX2^n-1
(iii) 2
(iv) 0

2x – y + 3 = 0

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, ∞).

x =\(\sqrt { t } \)
V = \(\frac { dx }{ dt } =\frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } }\) ..(1)
Acceleration = \(\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } =\frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } t^{ -\frac { 3 }{ 2 } } \right) =\frac { -t^{ -\frac { 3 }{ 2 } } }{ 4 } \)
∴ Acceleration is negative
Acceleration = \(-\frac { 1 }{ 4 } \left( { t }^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\)
= \(-2\left( \frac { 1 }{ 2 } t^{ -\frac { 1 }{ 2 } } \right) ^{ 3 }\) = 2V³    [using (1)]
Hence, acceleration is negative proportional to the cube of the velocity.

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as,
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{+}}{lim}(\frac{cos \ x}{2x})=\infty\)
\(\underset{x\rightarrow 0^{-}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{-}}{lim}(\frac{cos \ x}{2x})=\infty\)
As the left limit and the right limit are not the same we conclude that the limit does not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in \(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\) to conclude
\(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\)\(\underset{x\rightarrow 0^{+}}{lim} (\frac{-sin \ x}{2})\)=0
which is not true because it was not an indeterminate form.

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as
\(\underset{x\rightarrow 0}{lim}(\frac{sin \ mx}{x})\)=\(\underset{x\rightarrow 0}{lim}(\frac{m\times cos \ mx}{1})\)
= m
The next example tells that the limit does not exist.

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

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