(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.

(1) The domain of f (x) is R \{−1,1}.
(2) Since f (−x,−y) = f (x, y) , the curve is symmetric about the origin.
(3) Putting y = 0, we get x = 0 . Hence the x -intercept is (0, 0).
(4) Putting x = 0, we get y = 0. Hence the y -intercept is (0, 0).
(5) To determine monotonicity, we find the first derivative as \(f'(x)\frac { -3\left( { x }^{ 2 }+1 \right) }{ { ({ x }^{ 2 }-1) }^{ 2 } } \)
Hence, f'( x) does not exist at x = −1,1. Therefore, critical numbers are x = −1, 1. The intervals of monotonicity is tabulated

6) Since there is no sign change in f'( x) when passing through critical numbers. There is no local extrema.
(7) To determine the concavity, we find the second derivative as \(f"(x)=\frac { 6x({ x }^{ 2 }-3) }{ { ({ x }^{ 2 }-1) }^{ 3 } } \) f"( x) = 0\(\Rightarrow\) x = 0 and f"(x) does not exist at x = −1, 1.
The intervals of concavity is tabulated.

(8) As x = −1 and 1are not in the domain of f(x) and at x = 0, the second derivative is zero and f"(x) changes its sign from positive to negative when passing through x = 0. Therefore, the point of inflection is (0, f (0)) = (0,0) .

(9) \(\underset { x\rightarrow \pm \infty }{ lim } f(x)=\underset { x\rightarrow \pm }{ lim } \cfrac { 3x }{ { x }^{ 2 }-1 } =\underset { x\rightarrow \pm \infty }{ lim } \cfrac { 3 }{ x\frac { 1 }{ x } } =0\) Therefore y = 0 is a horizontal asymptote.
Since the denominator is zero, when x = 士1.\(\underset { x{ \rightarrow -1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow -1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty ,\underset { x{ \rightarrow 1 }^{ - } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } -\infty ,\underset { x{ \rightarrow 1 }^{ + } }{ lim } \frac { 3x }{ { x }^{ 2 }-1 } =\infty .\)
Therefore x = −1 and x = 1 are vertical asymptotes.
The rough sketch of the curve is shown on the right side.

Let the price of low-grade steel be Rs. p per tonne. Then the price of high-grade steel is Rs. 2p per tonne.
The total receipt per day is given by \(R=px+py=px+2p\left( \frac { 40-5x }{ 10-x } \right) \). Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get
\(R=p\left( \frac { 80-{ x }^{ 2 } }{ 10-x } \right) \)
\(\frac { dR }{ dx } =p\left( \frac { { x }^{ 2 }-20x+80 }{ (10-x)^{ 2 } } \right) \)
\(\frac { dR }{ dx } =-\frac { 40P }{ \left( 10-x \right) ^{ 3 } } \)
Now, \(\frac { dR }{ dx } =0\Rightarrow { x }^{ 2 }-20x+80=0\) and hence \(x=10\pm 2\sqrt { 5 } \) 
At \(x=10-2\sqrt { 5 } ,\frac { { d }^{ 2 }R }{ { dx }^{ 2 } } <0\) and hence R will be maximum. If x \(x=10-2\sqrt { 5 } \) then \(y=5-\sqrt { 5 } \)  
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are \(10-2\sqrt { 5 } \)  and \(5-5\sqrt { 5 } \)

The distance from the point (1, 1) to any point (x, y) is d =\(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 } } \). Instead of extremising d, for convenience we extremise D = d² = (x−1)²+ (y− 1)² subject to the condition,  x² + y² = 1 Now, \(\frac { dD }{ dx } \) = 2( x −1)+2(y −1) \(\times\) \(\frac { dy }{ dx } \) where the \(\frac { dy }{ dx } \) will be computed by differentiating x² y² + = 1 with respect to x . Therefore we get, 2x + 2y \(\frac { dy }{ dx } \) = 0 which gives \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \) 
Substituting this, we get \(\frac { dD }{ dx } \) = 2(x −1) +2(y −1)\(\left(- \frac { x }{ y } \right) \)
= \(\frac { 2\left[ xy-y-xy+x \right] }{ y } \)
Substituting this, we get \(\frac { dD }{ dx } \) =2\(\left[ \frac { x-y }{ y } \right] \)=0
⇒x = y
Since (x, y) lie on the circle x² + y² + =1 we get, 2x² = 1 gives x = 土\(\frac { 1 }{ \sqrt { 2 } } \).  
Hence the points at which the extremum distance occur are \(\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) \), \(\left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) \) 
To find the extrema, we apply second derivative test. So,
\(\frac { { d }^{ 2 }D }{ { dx }^{ 2 } } =2\frac { { y }^{ 2 }+{ x }^{ 2 } }{ { y }^{ 3 } } \)
The value of \({ \left( \frac { { d }^{ 2 }D }{ { dx }^{ 2 } } \right) }_{ \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) }>0;{ \left( \frac { { d }^{ 2 }D }{ { dx }^{ 2 } } \right) }_{ \left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) }<0\) 
This implies the nearest and farthest points are \(\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) \) and \(\left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) \)
Therefore, the nearest and the farthest distances are respectively \(\sqrt { 2 } \)-1 and \(\sqrt { 2 } \)+1

Factorising the given function we have
\(y=f(x)\frac { { x(x }-3x) }{ (x-1) } \)
(1) The domain and the range of f (x) are respectively R \{1} and the entire real line.
(2) Putting y = 0 we get the x = 0, 3. Therefore the x -intercept is (3,0). Putting x = 0, we get y = 0. Therefore the curve passes through the origin.
(3) \(f'(x)\frac { { x }^{ 2 }-2x+3 }{ { (x-1) }^{ 2 } } \) and hence the critical point of the curve occurs at x =1 as f′(1) does not exist. But x² -x − 2 + 3 = 0 has no real solution. Hence the only critical point occurs at x = 1.
(4) x =1 is not in the domain of the function and f'(x) ≠0 ∀x ∈R \{1}, there is no local maximum or local minimum.
(5) \(f"(x)=-\frac { 4 }{ { (x-1) }^{ 3 } } \)∀x∈R\{1}. Therefore when x <1, f"(x)>0the curve is concave upwards in (-∞,1) and x>1, f"(x)<0 the curve is concave downwards in (1,∞). Since f"(x} ≠ 0 ∈ R\{1} there is no point of infection for f(x).
(6) Since, \(\underset { x\rightarrow { 1 }^{ - } }{ lim } \frac { { x }^{ 2 }-3x }{ (x-1) } =+\infty \ and\ \underset { x\rightarrow { 1 }^{ + } }{ lim } \frac { { x }^{ 2 }-3x }{ (x-1) } =-\infty ,\ x=1\) is a vertical asymptote.
The rough sketch is shown.

Factorising the given function, we have
y = f (x) = (x − 3)(x² + 3x + 3).
(1) The domain and the range of the given function f(x) are the entire real line.
(2) Putting y = 0, we get the x = 3. The other two roots are imaginary. Therefore, the x -intercept is (3, 0) . Putting x = 0, we get y = −9. Therefore, the y-intercept is (0, −9)
(3) f'(x) = 3(x² -2) and hence the critical points of the curve occur at x = \(\pm \sqrt { 2 } \)
(4) f"(x) = 6x. Therefore at x =\(\sqrt{2}\) the curve has a local minimum because f"(\(\sqrt{2}\)) = 6\(\sqrt{2}\) > 0. Then local minimum is f(\(\sqrt{2}\)) = -4\(\sqrt{2}\)-9.
Similarly x = -\(\sqrt{2}\) the curve has a local maximum because f"(-\(\sqrt{2}\)) = -6\(\sqrt{2}\) < 0. The local maximum is f (-\(\sqrt{2}\)) = 4\(\sqrt{2}\) - 9.
(5) Since f "(x) = 6x > 0, ∀x > 0 the function is concave upward in the positive real line. As f"(x) = 6x < 0,∀x < 0 the function is concave downward in the negative real line.
(6) Since f"(x) = 0 at x = 0 and f′′(x) changes its sign when passing through x = 0. Therefore the point of inflection is (0, f (0)) = (0, 9).
(7) The curve has no asymptotes.
The rough sketch of the curve is shown on the right side.

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.

Differentiating with respect to x, we get
f'(x) = 24x⁵-24x³
= 24x³(x²-1)
= 24x³(x+1)(x-1)
f'(x) = 0 ⇒ x = −1, 0, 1.
Hence the critical numbers are x = −1, 0, 1
Now, f''(x) = 120x⁴-72x² = 24x²(5x²-3)
⇒ f''(-1) = 48, f''(0) = 0, f''(1) = 48.

As f "(−1) and f "(1) are positive by the second derivative test, the function f (x) has local minimum. But at x = 0 , f "(0) = 0.
That is the second derivative test does not give any information about local extrema at x = 0. Therefore, we need to go back to the first derivative test. The intervals of monotonicity is tabulated in the table 7.8.

Interval	(−∞,−1)	(−1,0)	(0,1)	(1,∞)
Sign of f'(x)	-	+	-	+
Monotonicity	strictly decreasing	strictly increasing	strictly decreasing	strictly increasing

By the first derivative test f(x) has local minimum at x = −1, its local minimum value is −2.
At x = 0 , the function f(x) has local maximum at x = 0 , and its local maximum value is 0. At x = 1, the function f (x) has local minimum at x = 1, and its local minimum value is −2.
Remark
When the second derivative vanishes, we have no information about extrema. We have used the first derivative test to find out the extrema of the function!

The given function is defined and differentiable at all \(x\in (-\infty, \infty) \), As
\(f(x)=\frac{x }{1+x^{2}}\)
\(f'(x)=\frac{1-x^{2} }{(1+x^{2})^{2}}\)
The stationary points are give by 1-x² = 0 that is \(x=\pm1\).
Hence the intervals of monotonicity are \((-\infty, -1) , (-1,1)\) and  \((1,\infty)\)

Therefore, f (x) strictly increasing on \((-\infty,-1)\) and \((1,\infty)\) strictly decreasing on (-1, 1).

Since f′(x) changes from negative to positive when passing through x = −1, the first derivative test tells us there is a local minimum at x = −1 and the local minimum value is \(f(-1)=-\frac{1}{2}\). As f′(x) changes from positive to negative when passing through x = 1, the first derivative test tells us there is a local maximum at x = 1 and the local maximum value is \(f(1)=\frac{1}{2}\).

Factorising the given function, we have
y = f (x) = (x − 3)(x + 2) .
(1) The domain of the given function f (x) is the entire real line.
(2) Putting y = 0 we get x = −2, 3. Therefore the x -intercepts are (−2, 0) and (3, 0) putting x = 0 we get y = −6. Therefore the y -intercept is (0, −6).
(3) f'(x) = 2x −1 and hence the critical point of the curve occurs at x = \(\frac{1}{2}\).
(4) f"(x) = 2 > 0,∀x . Therefore at xTherefore at x = \(\frac{1}{2}\) the curve has a local minimum which is \(f\left( \frac { 1 }{ 2 } \right) =-\frac { 25 }{ 4 } .\)
(5) The range of the function is \(y\ge-\frac{25}{4}\)
(6) Since f"(x) = 2 > 0,∀x the function is concave upward in the entire real line.
(7) Since f(x) = 2 ≠ 0, ∀x the curve has no points of inflection
(8) The curve has no asymptotes.
The rough sketch of the curve is shown on the right side.

The given function is a polynomial of degree 4. Now,

f′(x) = (x −1)³ + 3(x −1)² . (x − 5) 
= 4(x-1)².(x-4)
f"(x) = 4(x-1)²+2(x-1).(x-4)
= 12(x −1)  (x − 3)
Now,
f''(x) = 0 ⇒ x = 1, x = 3
The intervals of concavity are tabulated in the table 7.7.

Interval	(-∞, 1)	(1, 3)	(3, ∞)
Sign of f'(x)	+	-	+
Concavity	concave up	concave down	concave up

The curve is concave upwards on (∞, 1) and (3, ∞) .
The curve is concave downwards on (1, 3) .
As f′′(x) changes its sign when it passes through x = 1 and x = 3, (1, f(1)) = (1, 0) and (3, f(3)) = (3, −16) are points of inflection for the graph y = f(x). This may be observed from the adjoining figure of the curve f′′(x) .

 We have , f(x) = \(x^\frac{2}{3}\) then \(f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\)\(​​​​=\frac{2}{3x^{\frac{1}{3}}}, f'(x)\ne 0 \forall x \in R\) and f'(x) does not exist at x = 0.
Therefore, there are no stationary points but there is a critical point at x = 0.

Because f'(x) changes its sign from negative to positive when passing through x = 0 for the function it has a local minimum at x = 0. The local minimum value is f(0) = 0. Note that here the local minimum occurs at a critical point which is not a stationary point.

Given f(x) = 4x³ + 3x²- 6x + 1
f'(x) = 12x² + 6x - 6
f"(x) = 24x + 6
f'(x) = 0
⇒12x² + 6x - 6 = 0
⇒ 2x² + x - 1 = 0
⇒ (x + 1)(2x - 1) = 0
\(\Rightarrow x=-1,\frac { 1 }{ 2 } \)
The critical numbers are -1, \(\frac { 1 }{ 2 } \)
The possible intervals of monotonicity are
\(\left( -\infty ,-1 \right) \left( -1,\frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } ,\infty \right) \)

∴ f(x) is strictly increasing in \(\left( -\infty ,-1 \right) \left( \frac { 1 }{ 2 } ,\infty \right) \) and strictly decreasing in \(\left( -1,\frac { 1 }{ 2 } \right) \)
f"(x) = 0
\(\Rightarrow 24x+6=0\Rightarrow 24x=-6\)
\(x=\frac { -6 }{ 24 } =\frac { -1 }{ 4 } \)
The possible intervals of concavity are \(\left( -\infty ,\frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } ,\infty \right) \) 

ஃf(x) concave down in \(\left( -\infty ,\frac { -1 }{ 4 } \right) \) and concave up in \(\left( \frac { -1 }{ 4 } ,\infty \right) \)
As f"(x) changes its sign when it passes through
\(x=\frac { -1 }{ 4 } \), the point of inflection is \(\left( -\frac { 1 }{ 4 } ,f\left( -\frac { 1 }{ 4 } \right) \right) \)
Now \(f\left( \frac { -1 }{ 4 } \right) =\left( -\frac { 1 }{ 4 } \right) ^{ 3 }+3\left( \frac { -1 }{ 4 } \right) ^{ 2 }-6\left( \frac { -1 }{ 4 } \right) +1\)
= \(4\left( \frac { -1 }{ 4 } \right) +\frac { 3 }{ 16 } +\frac { 6 }{ 4 } +1\)
= \(\frac { -1 }{ 16 } +\frac { 3 }{ 16 } +\frac { 3 }{ 22 } +1=\frac { 1 }{ 8 } +\frac { 3 }{ 2 } +1\)
= \(\frac { 1+12+8 }{ 8 } =\frac { 21 }{ 8 } \)
ஃ Point of  inflection. \(\left( \frac { -1 }{ 4 } ,\frac { 21 }{ 8 } \right) \)
Since f'(x) changes its sign from positive to negative at x = -1, it has a local maximum at x = -1.
ஃf (-1) = 4 (-1)³ + 3 (-1)² - 6(-1) + 1
Since f'(x) changes its sign from negative to positive at \(x=\frac { 1 }{ 2 } \) it has a local minimum at \(x=\frac { 1 }{ 2 } \)
\(\therefore f\left( \frac { 1 }{ 2 } \right) =4\left( \frac { 1 }{ 2 } \right) ^{ 3 }+3\left( \frac { 1 }{ 2 } \right) ^{ 2 }-6\left( \frac { 1 }{ 2 } \right) +1\)
= \(\frac { 4 }{ 8 } +\frac { 3 }{ 4 } -\frac { -3 }{ 2 } +1=\frac { 1 }{ 2 } +\frac { 3 }{ 4 } -\frac { 3 }{ 2 } +1\)
= \(\frac { 2+3-6+4 }{ 4 } =\frac { 3 }{ 4 } \)

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

Let f(x) = \(y=\frac { 1 }{ 1+{ e }^{ -x } } \)
1. The domain of y is R
2. It is not symmetric
3. Putting y = 0, we get x does not exist 
∴ It has no X intercept
4. Putting x = 0, we get \(y=\frac { 1 }{ 1+{ e }^{ o } } =\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } \)
∴ Y-intercept is \(\left( 0,\frac { 1 }{ 2 } \right) \)
5. \(f'(x)=\frac { d }{ dx } \left( 1+e^{ -x } \right) ^{ -1 }=-1\left( 1+{ e }^{ -x } \right) \left( -e^{ -x } \right) \)
= \(\frac { { e }^{ -x } }{ \left( 1+e^{ -x } \right) ^{ 2 } } ={ e }^{ -x }\left( 1+{ e }^{ -x } \right) ^{ -2 }\)
\(f'(x)=0\Rightarrow { e }^{ -x }0\Rightarrow -x=log0\Rightarrow x=\infty \)
Also, f(x) > 0 ∀ X ∈ R
6. Since there is no sign change inf'(x), it has no local extreme.
7. f"(x) = e^-x(-2)(1+e^-x)^-3(^_e^-x)+(1+e^-x)^-2(^_e^-x)
= 2e^-2x (I+e^-x )^-3 -e^-x ( 1+e-x )^-2
= e^-2x (1+e-x r³ [2 - e^-x (1+e^-x)]
= e^-2X(I+e-xr³[2-e^x -1]
= \(\frac { { e }^{ -2x }\left( 1-{ e }^{ x } \right) }{ \left( 1+{ { e }^{ x } } \right) ^{ 3 } } \)
f"(x) = 0 ⇒ x = 0
The intervals of concavity is tabulated as follows.

8.\(\underset { x\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ e }^{ -x } } =1\)

⇒ y = 1 is the horizontal asymptote and
\(\underset { x\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ e }^{ -x } } =0\)
⇒ y = 0 is the horizontal asymptote
ஃ The rough sketch is

1. The domain of f(x) is R {-2,2}
2. Since I (x, y) = I (-x, y) the curve is symmetric about Y-axis. 
3. Putting y = 0, we get \({ x }^{ 2 }=-1\Rightarrow x=\pm \sqrt { -1 } \) is not possible
ஃ No X - intercept
4. Putting x = 0, we get \(y=\frac { -1 }{ 4 } \)
ஃ Y - Intercept is \(\left( 0,\frac { -1 }{ 4 } \right) \)
5. \(f'(x)=\frac { \left( { x }^{ 2 }-4 \right) (2x)-\left( { x }^{ 2 }+1 \right) \left( 2x \right) }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } \)
= \(\frac { { 2x }^{ 3 }-8x-{ 2x }^{ 3 }-2x }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } =\frac { -10x }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } \)
f'(x) = 0⇒ x = 0and the curve does not exist at x = -2, 2.
ஃ The critical numbers are 0, -2, 2.
ஃ The intervals of monotonicity are
(-∞, -2)(-2, 0)(0, 2)(2, ∞)

6. Since f'(x) changes it sign from positive to negative at x = 0, it has local maximum at x =0,
7. \(f''(x)=-10\left[ \frac { \left( { x }^{ 2 }-4 \right) ^{ 2 }(1)-x(2)\left( { x }^{ 2 }-4 \right) (2x) }{ \left( { x }^{ 2 }-4 \right) ^{ 4 } } \right] \)
= \(-10\left[ \frac { \left( { x }^{ 2 }-4 \right) ^{ 2 }-4{ x }^{ 2 }\left( { x }^{ 2 }-4 \right) }{ \left( { x }^{ 2 }-4 \right) ^{ 4 } } \right] \)
= \(-10\left( { x }^{ 2 }-4 \right) \left[ \frac { { x }^{ 2 }-4-{ 4x }^{ 2 } }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } \right] \)
= \(-10\left( \frac { -3{ x }^{ 2 }-4 }{ \left( { x }^{ 2 }-4 \right) ^{ 3 } } \right) =\frac { 10\left( { 3x }^{ 2 }+4 \right) }{ ({ x }^{ 2 }-4) } \)
\(\therefore f"(x)=0\Rightarrow { 3x }^{ 2 }+4=0\Rightarrow { x }^{ 2 }=\frac { -4 }{ 3 } \)
which is not possible and f "(x) does not exist at x = -2, 2.
The intervals of concavity is tabulated as follows

Since the curve does not exist at x = 2, x = - 2 are the vertical asymptotes and the horizontal asymptote is y = 1


1. The domain of f(x) is R {-2, 2}
2. Since f(x, y) = I (-x, y) the curve is symmetric about Y-axis.
3. Putting y = 0, we get \({ x }^{ 2 }=-1\Rightarrow x=\pm \sqrt { -1 } \) is not possible
4. Putting x = 0, we get \(y=\frac { -1 }{ 4 } \)
ஃY - Intercept is \(\left( 0,\frac { -1 }{ 4 } \right) \)
5.\(f'(x)=\frac { \left( { x }^{ 2 }-4 \right) \left( 2x \right) -\left( { x }^{ 2 }+1 \right) \left( 2x \right) }{ ({ x }^{ 2 }-4)^{ 2 } } \)
= \(\frac { { 2x }^{ 3 }-8x-{ 2x }^{ 3 }-2x }{ ({ x }^{ 4 }-4)^{ 2 } } =\frac { -10x }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } \)
f'(x) = 0⇒ x = 0 and the curve does not exist at x = -2, 2.
ஃThe critical numbers are 0,-2, 2.
ஃThe intervals of monotonicity are (-∞, -2)(2, 0)(0, 2)(2, ∞)

6. Since f'(x) changes it sign from positive to • negative at x = 0, it has local maximum at x =0,
7. \(f''(x)=-10\left[ \frac { \left( { x }^{ 2 }-4 \right) ^{ 2 }(1)-x(2)\left( { x }^{ 2 }-4 \right) (2x) }{ \left( { x }^{ 2 }-4 \right) ^{ 2 } } \right] \)
= \(-10\left[ \frac { \left( { x }^{ 2 }-4 \right) ^{ 2 }-4{ x }^{ 2 }\left( { x }^{ 2 }-4 \right) }{ \left( { x }^{ 2 }-4 \right) ^{ 4 } } \right] \)
= \(-10\left( { x }^{ 2 }-4 \right) \left[ \frac { { x }^{ 2 }-4-{ 4x }^{ 2 } }{ \left( { x }^{ 2 }-4 \right) ^{ 4 } } \right] \)
= \(10\left( \frac { -3{ x }^{ 2 }-4 }{ \left( { x }^{ 2 }-4 \right) ^{ 3 } } \right) =\frac { 10(3{ x }^{ 2 }+4) }{ \left( { x }^{ 2 }-4 \right) ^{ 3 } } \)
\(\therefore f''(x)=0\Rightarrow { 3x }^{ 2 }+4=0\Rightarrow { x }^{ 2 }=\frac { -4 }{ 3 } \)
which is not possible.
ஃ There is no concavity.
8.Since the curve does not exist at x = 2, x = - 2 are the vertical asymptotes and the horizontal asymptote is y = 1.

⇒y² = x²(4-x)
1. The domain of f(x) is {x∈R/-∞ ⇒x = 0, 4
2. Puttingy = 0, we get \(0=x\sqrt { 4-x } \)
⇒x = 0.4
ஃ The X-intercept is (4,0) and the curve passes through the origin.
3. Putting x = 0, we get y=0
∴ Y - intercept is (0,0)
4.\(f'(x)=x\left( \frac { 1 }{ 2 } \right) \left( 4-x \right) ^{ -\frac { 1 }{ 2 } }\left( -1 \right) +\sqrt { 4-x\left( 1 \right) } \) 
\(f'(x)=\frac { -4 }{ 2\sqrt { 4-x } } +\sqrt { 4-x } \)
\(\frac { -x+2(4-x) }{ 2\sqrt { 4-x } } =\frac { -3x+8 }{ 2\sqrt { 4 } -x } \)
⇒ f'(x) = 0
⇒ -3x+8 = 0
⇒ -3x = -8
\(\Rightarrow x=\frac { 8 }{ 3 } \)
ஃ The possible interval are \(\left( -\infty ,\frac { 8 }{ 3 } \right) \left( \frac { 8 }{ 3 } ,4 \right) \)

5. As f'(x) changes from positive to negative, it has local maximum at \(x=\frac { 8 }{ 3 } \)
6. There is no asymptote fer the curve 
ஃ The rough sketch of the function is

Given\(y=-\frac { 1 }{ 3 } \left( { x }^{ 3 }-3x+2 \right) \)
Factorising the given function, we have
\(y=f(x)=-\frac { 1 }{ 3 } \left( x-1 \right) \left( { x }^{ 2 }+x-2 \right) \)

1. The domain and the range of the given function fix) are the entire real time.
2. Putting y = 0, we get x = 1. The other two roots are imaginary.
ஃ X intercept is (1, 0)
Putting x = 0, we get \(y=-\frac { 2 }{ 3 } \)
∴Y intercept is \(\left( 0,-\frac { 2 }{ 3 } \right) \)
3. \(f'\left( x \right) =\frac { -1 }{ 3 } \left( { 3x }^{ 2 }-3 \right) =-{ x }^{ 2 }+1\)
f'(x) = 0
\(\Rightarrow -{ x }^{ 2 }=-1\)
\(\Rightarrow x=\pm 1\)
ஃThe critical points are at x = 1, x = -1.
4. \(f''(x)=\frac { -1 }{ 3 } (6x)=-2x\)
f''(1) = -2, f''(-1) = 2.
∴f (x) is local minimum at x = 1,
\(f(1)=\frac { -1 }{ 3 } \left( 1-3+2 \right) \)
\(\Rightarrow f(1)=\frac { -1 }{ 3 } \left( 0 \right) =0\)
f (x) is local minimum at x = -1,
\(\Rightarrow f(-1)=\frac { -1 }{ 3 } \left( -1+3+2 \right) =\frac { -4 }{ 3 } \)
∴The critical points are at x = 1, x = -1.
5. Since f"(x) = -2x<0∀x>0, the function is concave downward in the positive real line and f''(x) = -2x > 0∀<0,the function is concave upward in the negative real line.
6. f''(x) = 0 at x = 0 and f''(x) changes its sign when x = 0, the point of inflection is \(\left( 0,f(0) \right) =\left( 0,\frac { -2 }{ 3 } \right) \)
7. The curve has no asymptotes.

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.

Since the open box has square area, let the length, breadth and height of the box be I, l and b cm respectively.
ஃ Surface area = l² + 41b = 108
\(\Rightarrow l+4b=\frac { 108 }{ l } \) 
\(\Rightarrow 4b=\frac { 108 }{ l } -1\) 
\(\Rightarrow b=\frac { 108 }{ 4l } -\frac { l }{ 4 } \) 
\(\Rightarrow b=\frac { 27 }{ l } -\frac { l }{ 4 } \) 
Let f(1) = Volume of the box = 1 \(\times\) 1\(\times\) b = l² b
=  \({ l }^{ 2 }\left( \frac { 27 }{ l } -\frac { l }{ 4 } \right) \)
\(f(1)=27l-\frac { { l }^{ 3 } }{ 4 } \) 
\(f'(l)=27-\frac { { 3l }^{ 2 } }{ 4 } \) 
f'(1) = 0
\(\Rightarrow 27-\frac { 3{ l }^{ 2 } }{ 4 } =0\) 
\(\Rightarrow \frac { 3l^{ 2 } }{ 4 } =27\)
\(\Rightarrow l^{2}=\not 27 \times \frac{4}{\not 3}=36\) 
\(\Rightarrow 1=\pm 6\) 
\(\Rightarrow 1=6\) 
ஃThe critical number is 6
\(f''(l)=\frac { 6l }{ 4 } =\frac { 3l }{ 2 } \) 
\(\therefore f''(6)=-\frac { 3(6) }{ 2 } <0\)
ஃ f"(1) is maximum when l = 6
When l = 6,\(b=\frac { 27 }{ 6 } -\frac { 6 }{ 4 } \) 
= \(\frac { 9 }{ 2 } -\frac { 3 }{ 2 } =\frac { 6 }{ 2 } =3cm\)
Hence the dimensions of the required box are 6 cm, 6 cm and 3 cm respectively.

Let us take the semi circle with center (0, 0) an radius r

x = r cos θ,
y = r sin θ
ஃ Length of the rectangle = 2x = 2r cos θ
Breadth of the rectangle = y = r sin θ
ஃ Area = 2xy = 2r² cos θ sin θ
= r² sin 2θ
Letf'(θ) = r² 2 cos 2θ
f(θ) = r²cosθ sinθ
⇒ 2r² cos 2θ = 0
\(\Rightarrow cos2\theta =0=cos\frac { \pi }{ 2 } \)
\(\Rightarrow 2\theta =\frac { \pi }{ 2 } \)
\(\Rightarrow \theta =\frac { \pi }{ 4 } \) 
\(f''={ 2r }^{ 2 }(2)-(-sin2\theta )\) 
= \(-4{ r }^{ 2 }sin2\theta \) 
\(\therefore f''\left( \frac { 4\pi }{ 4 } \right) =-4{ r }^{ 2 }sin2\left( \frac { \pi }{ 4 } \right) \) 
= \(-4{ r }^{ 2 }sin\frac { \pi }{ 2 } ={ 4r }^{ 2 }<0\) 
ஃ f(θ) is maximum when \(\theta =\frac { \pi }{ 4 } \)
ஃ Length of the rectangle \((x)=2ros\frac { \pi }{ 4 } \)
= \(2r\times \frac { 1 }{ \sqrt { 2 } } =\sqrt { 2 } r\) 
ஃ Breadth of the rectangle = y = r sin θ
= \(rsin\frac { \pi }{ 4 } =\frac { r }{ \sqrt { 2 } } \)

Let x and y be the length and breadth of the rectangle.
ஃ Perimeter P = 2x + 2y
\(\Rightarrow 2y=P-2x\Rightarrow y=\frac { P-2x }{ 2 } \) 
Let \(f(x)=Area=xy=x\left( \frac { P-2x }{ 2 } \right) \)
\(f(x)=\frac { Px-{ 2x }^{ 2 } }{ 2 } \)
\(f'(x)=\frac { 1 }{ 2 } \left[ P-4x \right] \)
f'(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left[ P-4x \right] =0\)
\(\Rightarrow P=4x\) 
\(\Rightarrow x=\frac { P }{ 4 } \)
∴The critical number is \(\frac { P }{ 4 } \) 
 Now,\(f''\left( \frac { P }{ 4 } \right) =-2<0\)
ஃf(x) is maximum when \(x=\frac { P }{ 4 } \)
When \(x=\frac { P }{ 4 } \)
\(\Rightarrow y=\frac { P-2\left( \frac { P }{ 4 } \right) }{ 2 } =\frac { P-\frac { P }{ 2 } }{ 2 } =\frac { P }{ 4 } \)
\(\therefore x=y=\frac { P }{ 4 } \)
ஃThe rectangle is a square when the area is maximum for a given perimeter.

Let the point on the circle be

Let the length of the rectangle be 'x' cm
Breadth of the rectangle be 'y' cm
\( x^{2}+y^{2} =(20)^{2} \)
\(y^{2} =400-x^{2}\)
[radius of the circle is 10 cm]
\(y=\sqrt{400-x^{2}}\)
Now, Area of the rectangle A = xy
\( A =x \sqrt{400-x^{2}} \)
\(\frac{d A}{d x} =x \frac{(-2 x)}{2 \sqrt{400-x^{2}}}+\sqrt{400-x^{2}} \) .....(1)
\(=\frac{-x^{2}+400-x^{2}}{\sqrt{400-x^{2}}}=\frac{-2 x^{2}+400}{\sqrt{400-x^{2}}}\)
For maximum or minimum,
\( \frac{d A}{d x} =0 \Rightarrow \frac{-2 x^{2}+400}{\sqrt{400-x^{2}}}=0 \)
\(x^{2} =200 \)
\(x =\pm 10 \sqrt{2} \)
\(x =-10 \sqrt{2} \) (is not possible)
\(\therefore x =10 \sqrt{2} \)
Now, \( \frac{d^{2} A}{d x^{2}}= \frac{\sqrt{400-x^{2}}(-4 x)-\left(-2 x^{2}+400\right)\left(-\frac{2 x}{2 \sqrt{400-x^{2}}}\right)}{400-x^{2}} \)
\(=\frac{2 x^{3}-1200 x}{\left(400-x^{2}\right) \sqrt{400-x^{2}}}\)
At \(x=10 \sqrt{2}, \quad \frac{d^{2} A}{d x^{2}}<0\)
Area of the rectangle is maximum when x \(=10 \sqrt{2}\)
\(y=\sqrt{400-200}=\sqrt{200}=10 \sqrt{2}\)
\(\therefore x=y=10 \sqrt{2}\)
Length of the rectangls = 10\(\sqrt{2}\) cm
Breadth of the rectangls = 10\(\sqrt{2}\) cm

Let \(f(x)=\frac{1}{x}\). then the Taylor series of f (x) is
\(f(x)=\sum^{n=\infty}_{n=0}a_{n}(x-2)^{n},\) where \(a_{n}=\frac{f^{(n)}(2)}{n!}\)
Various derivatives of the function f (x) evaluated at x = 2 are given below.

Substituting these values, we get the required expansion of the function as:
\(\frac{1}{x}=\frac{1}{2}-\frac{1}{4}\frac{(x-2)}{1!}+\frac{1}{4}\frac{(x-2)^{2}}{2!}-\frac{3}{8}\frac{3(x-2)^{3}}{3!}+...\)
which is, \(\frac{1}{x}=\frac{1}{2}-\frac{(x-2)}{4}+\frac{(x-2)^{2}}{8}-\frac{(x-2)^{3}}{16}+...\)

Let f (x) = tan x, then the Mclaurin series of f (x) is 
\(f(x)=\sum^{n=\infty}_{n=0} a_{n}x^{n}\), where , \(a_{n}=\frac{f^{(n)}(0)}{n!}\).
Various derivative’s of the function f (x) evaluated at x = 0 is given below :
Now,
\(f'(x)=\frac{d}{dx}(tanx)=sec^{2}(x)\)
\(f''(x)=\frac{d}{dx}sec^{2}(x))=2sec x.sec x.tan x= 2sec^{2}x.tanx\)
\(f'''(x)=\frac{d}{dx}(2sec^{2}(x).tan x)= 2sec^{2}(x).sec^{2}x+tan x.4secx.secx.tanx\)
=\(2sec^{4}x+4sec^{2}x.tan^{2}x\)
\(f^{(iv)}(x)=8sec^{3}(x).secx+tan x+4sec^{2}x.2tanx sec^{2}x+8secx.secx.tanx.tan^{2}x\)
=\(16sec^{4}xtanx+8sec^{2}x.tan^{3}x\)
\(f^{(v)}(x)=16sec^{4}x.sec^{2}x+64sec^{3}x.sec x.tan x.tan x+8 sec^{2}x.3tan^{2}x.sec^{2}x+16secx.secx.tanx.tan^{3}x\)
=\(16sec^{6}x+88sec^{4}x.tan^{2}x+16sec^{2}x.tan^{4}x\).

Substituting the values and on simplification we get the required expansion of the function as
\(tan x=x+\frac{1}{3}x^{3}+\frac{2}{15}x^{5}+...; -\frac{\pi}{2}

Let x be the length of the rectangular pasture and y be the breadth of the rectangular pasture.
Given xy = 1,80,000
\(\Rightarrow y=\frac { 1,80,000 }{ x } \) 
Since fencing is not needed along the river side, 
Perimeter = 2x + y
Letf(x) = 2x+y

= \(2x+\frac { 1,80,000 }{ x } \) 
\(f'(x)=2-\frac { 1,80,000 }{ { x }^{ 2 } } \) 
f'(x) = 0
\(\Rightarrow 2=\frac { 1,80,000 }{ { x }^{ 2 } } \) 
\(\Rightarrow { x }^{ 2 }=90,000\) 
\(\Rightarrow x=\pm 300\) 
ஃ The critical number is 300, -300
\(f''(x)=-1,80,000\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 360000 }{ { x }^{ 3 } } \) 
\(f''\left( 300 \right) =\frac { 360000 }{ \left( 300 \right) ^{ 3 } } >0\) 
From (1), when x = 300, \(y=\frac { 1,80,000 }{ 300 } =600\) 
ஃ Length of the minimum needed fencing material = 2x + y
= 2(300) + 600 = 600 + 600 = 1200 m

Let f(x) = log(1+x) then the Maclaurin series of f (x) is f (x) \(\sum _{ n=0 }^{ n=\infty }{ { a }_{ n }{ x }^{ n } } \) where, \(a_{n}=\frac{f^{n}(0)}{n!}\) f(x) various derivatives of the function f(x) evaluated at x = 0 are given below:

Substituting the values and on simplification we get the required expansion of the function given by,
\((log(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+... -1\)

Let f(x) = e^-x, x ∈ [a, b]
a) e-x is continuous in [a, b]
b) e-x is differentiable in (a, b)
c) f(b) = e^-b, f(a) = e^-a
By Lagrange's mean value I theorem, there exerise c ∈ [a, b] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ -e^-c =\(\frac { { e }^{ -b }-{ e }^{ -a } }{ b-a } \)
⇒ -e^-c = \(\frac { { e }^{ -a }-{ e }^{ -b} }{ b-a } \)
⇒ |e^-c| = \(\left| \frac { { e }^{ -b }-{ e }^{ -a } }{ b-a } \right| \)
⇒ \(\frac { \left| { e }^{ -a }-{ e }^{ -b } \right| }{ \left| a-b \right| } \) < 1
[∵ |-a^-c| < 1 for c ∊ [a, b] a > 0, b > 0]
⇒ |e^-a - e^-b| < |a -b|
Hence proved.

Let x and y be the length, breadth of the printed rectangular page

Given xy = 24
\(\Rightarrow y=\frac { 24 }{ x } \)    ..(1)
Length of the page with margin
= x+1+1 = x+2
breadth of the page with margin
= y + 1.5 + 1.5 = y + 3
Area of the page = (x + 2) (y + 3)
Let f(x) = (x + 2) (y + 3)
= \((x+2)\left( \frac { 24 }{ x } +3 \right) \) 
= \(24+3x+\frac { 48 }{ x } +30\) 
= \(3x+\frac { 48 }{ x } +30\) 
\(f'\left( x \right) =3-\frac { 48 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=16\Rightarrow x=\pm 4\)
ஃ The critical number are 4,-4
\(f''\left( x \right) =-48\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 96 }{ { x }^{ 3 } } \) 
\(f''\left( 4 \right) =\frac { 96 }{ 64 } >0\) 
ஃ f(x) is minimum when x = 4
When \(x=4,y=\frac { 24 }{ 4 } =6\)  [From (1)]
ஃ Length of the page = x + 2 = 4 + 2 = 9 cm
Breadth of the page = y + 3 = 6 + 3 = 6 cm

Given f \(x(x+3){ e }^{ \frac { \pi }{ 2 } }\) -3 ≤ x ≤ 0
a) f(x) is continuous in [-3, 0] .
b) f(x) is differentiable in (-3,0)
c) f(0) = 0(0 + 3) \({ e }^{ \frac { \pi }{ 2 } }\) = 0
f(-3) = -3 (-3 + 3)\({ e }^{ \frac { \pi }{ 2 } }\) = 0
∴ By Rolle's theorem, there exists c ∈ [-3, 0] such that f'(c) = 0
(2c + 3)\({ e }^{ \frac { \pi }{ 2 } }\) = 0 [∵ f(x) = (x2 + 3x)\({ e }^{ \frac { \pi }{ 2 } }\) f'(x) = (2c + 3) \({ e }^{ \frac { \pi }{ 2 } }\)]
⇒ 2c + 3 = 0 f(x) = x (x + 3) _e^{\(\frac{\pi}{2}\)}, 
⇒ 2c = -3
⇒ c = \(\frac{-3}{2}\)
Hence, there exist a point on the curve f(x) = x (x + 3) _e^{\(\frac{\pi}{2}\)}, -3 ≤ x ≤ 0 where the tangent is parallel to the X - axis.

Given f(0) = -1, f(2) = 4
∴ f(x) is a continuous function in [0,2]
By Lagrange's mean value theorem,
f'(x) = \(\frac { f(b)-f(a) }{ b-a } \) = \(\frac{f(2)-f(0)}{2-0}\)
= \(\frac{4-(-1)}{2}\) = \(\frac{5}{2}\) = 2.5
= 2.5 ∉ [0, 2]
Since f'(x) cannot be 2.5 at any point in [0, 2], there does not exist a differentiable function f(x).

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

f(x) is defined and differentiable for all x ∈ (0, ∞).
\(\therefore f'(x)=\frac { { 3x }^{ 2 } }{ 3 } -\frac { 1 }{ x } ={ x }^{ 2 }-\frac { 1 }{ x } \) 
\(\Rightarrow { x }^{ 2 }-\frac { 1 }{ x } =0\Rightarrow \frac { { x }^{ 3 }-1 }{ x } =0\)
\(\Rightarrow { x }^{ 3 }-1=0\Rightarrow { x }^{ 3 }1\) 
\(\Rightarrow\) x = 1
\(\therefore\) The Stationary point is at x = 1
The possible intervals are (0, 1) and (1, ∞).

Since f '(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
\(\therefore f'\left( 1 \right) =\frac { { 1 }^{ 3 } }{ 3 } -log1\) 
= \(\frac { 1 }{ 3 } -o=\frac { 1 }{ 3 } \)

f(x) is defined and differentiable in (-∞, ∞)
\(\therefore f'(x)=\frac { (1-{ e }^{ x }){ e }^{ x }-{ e }^{ x }\left( -{ e }^{ x } \right) }{ (1-{ e }^{ x })^{ 2 } } \) 
= \(\frac { 1-{ e }^{ 2 }+{ e }^{ 2 }x }{ (1-{ e }^{ x })^{ 2 } } =\frac { 1 }{ \left( 1-{ e }^{ x } \right) ^{ 2 } } \) 
\(f'\left( x \right) =0\Rightarrow \frac { 1 }{ (1-{ e }^{ x })^{ 2 } } \neq 0\)
Thus there is no stationary point 
Since \(f'\left( x \right) =\frac { 1 }{ \left( 1-{ e }^{ x } \right) ^{ 2 } } >0\)  for-all x ∈ (-∞, ∞)
Since there is no stationary print, the curve does not change its position.
Hence there is no local extremum.

Given \(f(x)=\frac { x }{ x-5 } \) 
f(x) is defined and differentiable for all x∈R-[5]
\(\therefore f'(x)=\frac { (x-5)(1)-x(1) }{ (x-5)^{ 2 } } \) 
= \(\frac { x-5-x }{ (x-5)^{ 2 } } =\frac { -5 }{ (x-5)^{ 2 } } \) 
f'(x) = 0
\(\Rightarrow \frac { -5 }{ (x-5)^{ 2 } } \neq 0\) 
There is no stationary point. The possible intervals are (-∞, 5) and (5, ∞).

ஃ f (x) is strictly decreasing on (-∞, 5) and (5, ∞).
Since there is no stationary point, the curve does not changes its position.
Hence there is no local extremum.

Given f(x) = 2.0 + 3x2 - 12x
The given function is defined and differentiable for all x ∈(-∞,∞)
f'(x) = 6x² + 6x - 12
The stationary points are given by
6x² + 6x - 12 = 0
\(\Rightarrow\) x² +x-2 = 0
\(\Rightarrow\) (x + 2)(x - 1) = 0
\(\Rightarrow\) x = -2, 1
Hence the intervals of monotonicity are
(-∞, - 2), (-2, 1) and (1, ∞).

ஃ f(x) is strictly increasing on (-∞, - 2)
(1, ∞) and strictly decreasing on (-2, 1).
Since f'(x) changes from positive to negative at x = -2, there is a local maximum at x = -2
\(\therefore\) f(-2) = 2 (2)³ + 3 (2)² - 12 (-2) 
= 2(-8) + 3(4) + 24 
= -16+12+24 = 20
Also f'(x) changes from negative to positive at x = 1, there is a local minimum at x = 1.
ஃ f(1) = 2 (1)³ + 3 (1)² - 12(1)
= 5 -12 = -7

Equation of the given curves are x² - y² = r²
xy = c²
x² - y² = r²
⇒ 2x - 2y\(\frac{dy}{dx}\) = 0
⇒ 2x = 2y\(\frac{dy}{dx}\)
⇒ \(\frac{dy}{dx}=\frac{2x}{2y}=\frac{x}{y}\)
Let (x₁, y₁) be the point of intersection of the given curves
∴ Slope of tangent to the first curve m₁ = \(\frac{x_1}{y_1}\) ..(1)
⇒ xy = c²
⇒ x.\(\frac{dy}{dx}\) = -y
⇒ \(\frac{dy}{dx}=\frac{-y}{x}\)
∴ Slope of the tangent to the first curve m₂ = \(\frac{-y_1}{x_1}\)..(2)
Consider m₁ m₂ = \(\left( \frac { { x }_{ 1 } }{ { y }_{ 1 } } \right) \left( \frac { -{ y }_{ 1 } }{ { x }_{ 1 } } \right) =-1\)
Since m₁ m₂ = -1, the given two curves cut orthogonally.

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 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 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 us now find the point of intersection of the two given curves. Equating x² = (x - 3)² we get, x = \(\frac32\). Therefore, the point of intersection is \({(\frac{3}{2},\frac{9}{4})}\). Let θ be the acute angle between the curves. The slopes of the curves are as follows:

For the curve y = x² 
\(\frac{dy}{dx}=2x\) 
\(m_{1}=(\frac{dy}{dx})\ at\ {(\frac{3}{2},\frac{9}{4})}=3\).
For the curve y = (x-3)² 
\(\frac{dy}{dx}=2(x-3)\) 
\(m_{2}=(\frac{dy}{dx}) \ at \ {(\frac{3}{2},\frac{9}{4})}= -3\)
Using (3), we get
\(tan \theta=|\frac{3-(-3)}{1-9}|=\frac{3}{4}\) 
Hence,  \(\theta\) =  \(tan^{-1}(\frac{3}{4})\).

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 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

Since the beacon makes one revolution (360°) in 10 sec.

At the point of observation, let 0 be the angle of deviation of the beacon of light from OA.
Let AB x km,when \(\angle\)AOB = 0
\(\frac { d\theta }{ dt } =\frac { 2\pi }{ 10 } \)
\(=\frac { \pi }{ 5 } \) rad /sec.
Let AB = x
Then, tan θ \( =\frac { x }{ 5 } \)
x = 5 tan θ
We know that velocity = \(\frac { dx }{ dt } \)
Differentiating (1) with respect to 't' we get,
\(\frac { dx }{ dt } =5{ sec }^{ 2 }\theta ,\frac { d\theta }{ dt } \)
\(=5({ sec }^{ 2 }{ 45 }^{ o })\left( \frac { \pi }{ 5 } \right) \)
\(=5({ \sqrt { 2 } })^{ 2 }\left( \frac { \pi }{ 5 } \right) \)
\(=5(2)\left( \frac { \pi }{ 5 } \right) \)
= 2π km/sec.

Given s (t) = 2t³ − 9t² + 12t ≥ 0

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

Given s (t) = 16t², height = 400 ft.
⇒ t² = \(\frac { 400 }{ 16 } =\frac { 100 }{ 4 } \)
t² = 25
t = 5 sec
(ii) Average velocity = \(\frac { ds }{ dt } \) = 32 t
When t = 2 sec
Average in the last
2 sec = \(\frac { V \ at \ t=3+V \ at \ t=5 }{ 2 } \)
= \(\frac { 32(3)+32(5) }{ 2 } \)
= \(\frac { 96+160 }{ 2 } =\frac { 256 }{ 2 } \)
= 128 f/sec
(iii) Instantaneous Velocity
=\(\frac { ds }{ dt } \) = 32t
When t = 5 sec
Velocity = \(\frac { ds }{ dt } \) = 32(5)
= 160 ft/sec

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

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. 

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

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.

Let (x) = sin x
f^I(x) = cos x
f^II(x) = - sin x
f^III(x) = -cos x
⇒ \(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ 1 }\left( \frac { \pi }{ 4 } \right) =cos\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
\({ f }^{ II }\left( \frac { \pi }{ 4 } \right) =-cos\frac { \pi }{ 4 } =-\frac { 1 }{ \sqrt { 2 } } \)
Taylors' series for f (x) at x = \(\frac { \pi }{ 4 } \) is
\(f(x)=f\left( \frac { \pi }{ 4 } \right) +\frac { { f }^{ I }\left( \frac { \pi }{ 4 } \right) }{ 1! } \left( x-\frac { \pi }{ 4 } \right) +\frac { { f }^{ II }\left( \frac { \pi }{ 4 } \right) }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }+\frac { { f }^{ III }\left( \frac { \pi }{ 4 } \right) }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }\)+...
\(sinx=\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 } }{ 2! } -\frac { 1 }{ \sqrt { 2 } } \frac { { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 } }{ 3! } +...\)
\(=\frac { 1 }{ \sqrt { 2 } } \left[ 1+\left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 2 }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }^{ 3 }+..... \right] \)
\(sinx=\frac { \sqrt { 2 } }{ 2 } \left[ 1+\frac { 1 }{ 1! } \left( x-\frac { \pi }{ 4 } \right) -\frac { 1 }{ 2! } { \left( x-\frac { \pi }{ 4 } \right) }-\frac { 1 }{ 3! } { \left( x-\frac { \pi }{ 4 } \right) }+..... \right] \)

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

Let f (t) represents the distance covered at 't' hour.
Given f(0) = 20 and f(2) = ?
Also speed = f' (t) ≥ 150
The distance function is continuous as well as differentiable.
By Lagrange's mean value theorem, there exists c such that
⇒ f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ f'(c) = \(\frac{f(2)-20}{2-0}\) ≥ 150 [Man speed is 150 km/hr and f' (c) represents speed]
⇒ \(\frac{f(2)-20}{2-0}\) ≥ 150
⇒ f(2) - 20 ≥ 300
⇒ f(2) ≥ 320
Hence, in the next two hours he can cover 320km.

Given f(x) = Ax² + Bx + C, x ∈ [a, b]
a) f (x) is continuous in [a, b]
b) f(x) is differentiate in (a, b)
c) f(b) = Ab²+ Bb + c,
f(a) = Aa²+ Ba + c
Using mean value theorem, there exists c ∈ [a, b] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
2Ac + B = \(\frac { (A{ b }^{ 2 }+Bb+C)-(A{ a }^{ 2 }+Ba+C) }{ b-a } \)
\(=\frac{A b^{2}+B b+\not C-A a^{2}-B a-\not C}{b-a}\)
= \(\frac { a(b+a)(b-a)+B(b-a) }{ b-a } \)
\(=\frac{(\not b-a)[\mathrm{A}( b+a)+\mathrm{B}]}{\not b-a}\)
\(2 \mathrm{~A} c+\not \mathbf{B}=\mathrm{A}(a+b)+\not \mathrm{B}\)
⇒ 2Ac = A(a + b)
= \(\frac{a+b}{2}\) ∈ [a, b]

Given \(f(x)=\frac{1}{x}\), x ∈ [a, b]
a) f(x) is continuous in [a, b]
b) f(x) is differentiable in (a, b)
c) f(b) = \(\frac1b\), f(b) = \(\frac1a\)
Using mean value theorem, there exists c ∈ [a, b] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
\(\frac { -1 }{ { c }^{ 2 } } =\frac { \frac { 1 }{ b } -\frac { 1 }{ a } }{ b-a } \) 
⇒ \(\frac { 1 }{ { c }^{ 2 } } =\frac { a-b }{ ab(b-a) } =-\frac { (b-a) }{ ab(b-a) } \)
= - \(\frac{1}{ab}\)
⇒ c² = cb
⇒ c = 土 \(\sqrt { ab} \)
ஃ c = \(\sqrt { ab} \) ∈ [a, b]
[∵ c = - \(\sqrt { ab} \) ∈ [a, b]]

Given f (x) = (x − 2)(x − 7), x ∈ [3,11]
a) f(x) is continuous in [3, 11]
b) f(x) is differentiable in (3, 11)
c) f(11) = (11-2)(11-7)
= (9) (4) = 36
f(3) = (3 - 2)(3 - 7)
= (1)(-4) = - 4
∴ By Lagrange's mean value theorem, there exists c ∈ [3,11] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
\(\left[ \begin{matrix} f(x)\begin{matrix} = & (x \end{matrix}- & 2) & \begin{matrix} (x & - \end{matrix}7) \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 7x & -2x\begin{matrix} + & 14 \end{matrix} \\ \begin{matrix} = & { x }^{ 2 } \end{matrix}- & 9x & +\begin{matrix} 14 & \end{matrix} \end{matrix} \right] \)
⇒ 2c - 9 = \(\frac{36+4}{11-3}\)
⇒ 3c - 9 = \(\frac{40}{8}\) = 5
⇒ 2c = 14
⇒ c = 7 ∈ [3 , 11]

f(x) = x³ − 3x + 2, x ∈ [-2, 2]
a) f(x) is continuous in [-2, 2]
b) f(x) is differentiable in (-2, 2)
f(-2) = (-2)³ - 3 (-2) + 2
= -8 + 6 + 2 = 0
f(2) = 2³ - 3(2) +2
= 8 - 6 + 2 = 4
By Lagrange's mean value theorem, there exists c ∈ [-2,2] such that f'(c) = \(\frac { f(b)-f(a) }{ b-a } \)
⇒ 3c³ - 3 = \(\frac { 4-0 }{ 2-(-2) } =\frac { 4 }{ 4 } =1\)
⇒ 3²-3 = 1
⇒ 3² =4 ⇒ c² = \(\frac43\)
⇒ c = 土 \(\frac { 2 }{ \sqrt { 3 } } \) ∈ [-2, 2] 

Given equation of curve is y = \(\frac{x+1}{x-1}\) and the line is x + 2y = 6
Slope of the tangent to the curve
m₁ = \(\frac{dy}{dx}\)
= \(\frac { (x-1)(1)-(x-1)(1) }{ { (x-1) }^{ 2 } } \)
= \(\frac { x-1-x-1 }{ { (x-1) }^{ 2 } } =\frac { -2 }{ { (x-1) }^{ 2 } } \) [Quotient rule]
Slope of the line m₂ = \(\frac{-1}{2}\) \(\left[ \frac { co-efficient\ of\ x }{ co-efficient\ of\ y } \right] \)
Since the tangent to the curve and the lines are parallel, m₁ = m₂
⇒ \(\frac { -2 }{ { (x-1) }^{ 2 } } =\frac { -1 }{ 2 } \)
⇒ 4 = (x - 1)²
⇒ x - 1 = 土 2
⇒ x-1 = 2 or x- 1 = -2
⇒ x = 3 or x = -1
⇒ When x = 3, y = \(\frac{3+1}{3-1}\) = \(\frac42\) = 2
⇒ When x = -1, y = \(\frac{-1+1}{-1-1}=\frac{0}{-2}\) = 0
∴ Equation of the tangent at (3, 2) is
y - 2 = \(\frac{-1}{2}\)(x - 3)
⇒ 2y - 4 = -x + 3
x + 2y -7 = 0

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

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)

Given curve is y = x³ − x² + x + 3 and the line is x + y = 1729
Slope of the tangent to the curve is
\(\frac { dy }{ dt } \) = 3x² - 12 x + 1
Slope of the normal to the curve is
m₁ = \(\frac { -1 }{ { 3x }^{ 2 }-12x+1 } \)
Slope of the line is
m₂ = \(\frac{-1}{1}\) = -1
Since the slope of the normal to the curve and the lines are parallel, m₁ = m₂
⇒ \(\frac { -1 }{ { 3x }^{ 2 }-12x+1 } \) = -1
⇒ 3x² - 12x + 1 = 1
⇒ 3x² -12x = 0
⇒ 3x (x-4) = 0
⇒ x = 0 or x = 4,....(1)
When x = 0, y = 0³ - 6(0)² + 0 + 3
⇒ y = 3 [From (1)]
⇒ When x = 4, y = 4³ - 6(4)² + 4 + 3
= 64 - 96 + 7 = - 25
∴ (0, 3) and (4, -25) are the required points.

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.