Given x + y = 10
⇒ y = 10 -x ...(1)
Let f(x) = x²+y
= x²+ (10 -x)²
= x + 100 +x²- 20x
f(x) = 2x² - 20x+ 100
f'(x) = 4x - 20
f'(x) = 0
4x-20 = 0
4x = 20
⇒ x = 5
∴The critical number is 5
f"(x) = 4
ஃ f"(5) = 4 > 0
ஃf(x) is minimum when x = 5
When x = 5, y = 10- 5 = 5
[From (1)]
ஃ Smallest possible value of x²+y²
= 5² + 5² = 25 + 25 = 50

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).

Givenf(x) = sin x + cos x, 0 f'(x) = cos x - sin x
f"(x) = sin x - cos x
\(\therefore\) f"(x) = 0
\(\Rightarrow\) sin x - cos X = 0
\(\Rightarrow\) -sin x = cosx
\(\Rightarrow\) sin (-x) = cos x
\(\Rightarrow\) \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } ,2\pi \)
ஃ The possible intervals are \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 

ஃf (x) is concave upward in \(\left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \) and concave downward in \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 
Since f" (x) changes its sign from negative to positive at \(\frac { 3\pi }{ 4 } \) and positive at \(\frac { 7\pi }{ 4 } \) f(x) has point of inflection at \(\left( \frac { 3\pi }{ 4 } ,f\left( \frac { 3\pi }{ 4 } \right) \right) \) and 

\(\left( \frac { 7\pi }{ 4 } ,f\left( \frac { 7\pi }{ 4 } \right) \right) \) 
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =sin\frac { 3\pi }{ 4 } -cos\frac { \pi }{ 4 } \) 
= \(sin\left( \pi -\frac { \pi }{ 4 } \right) +cos\left( \frac { \pi }{ 4 } \right) \) 
= \(\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } =0\) 
ஃ The points of inflection are \(\left( \frac { 3\pi }{ 4 } ,0 \right) \) and \(\left( \frac { 7\pi }{ 4 } ,0 \right) \)

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.

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.

Given s (t) = 2t³ − 9t² + 12t ≥ 0

Given \(f(x)=|\frac{1}{x}|, x\in [-1,1]\)
Rolle's theorem is not applicable since \(f(x)=(\frac{1}{x})\) is not continuous at x = 0 in [-1, 1] and not differentiable in (-1, 1)

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

Let the two positive numbers be x and y.
Given xy = 20
\(\Rightarrow y=\frac { 20 }{ x } \) 
Let f(x) = x+y
\(f(x)=x+\frac { 20 }{ x } \) 
\(f'(x)=1-\frac { 20 }{ { x }^{ 2 } } \)
f'(x) = 0
\(\Rightarrow 1-\frac { 20 }{ { x }^{ 2 } } =0\) 
\(\Rightarrow 1=\frac { 20 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=20\) 
\(\Rightarrow x=\pm \sqrt { 20 } \) 
\(\Rightarrow x=\pm 2\sqrt { 5 } \)
∴ The critical number are \(2\sqrt { 5 } -2\sqrt { 5 } \) 
\(f''(x)=\frac { 40 }{ { x }^{ 3 } } \) 
When \(x=2\sqrt { 5 } \)
\(f''\left( x \right) =\frac { 40 }{ \left( 2\sqrt { 5 } \right) ^{ 3 } } >0\) 
ஃ f(x) is minimum when \(x=2\sqrt { 5 } \) 
When \(x=2\sqrt { 5 } ,y=\frac { 20 }{ 2\sqrt { 5 } } \) 
= \(\frac { 10 }{ \sqrt { 5 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\frac { 10\sqrt { 5 } }{ 5 } =2\sqrt { 5 } \) 
Hence the required positive numbers are \(2\sqrt { 5 } \)
Minimum sum = \(2\sqrt { 5 } \) + \(2\sqrt { 5 } \)  = \(4\sqrt { 5 } \)

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]]

a) f(x) is continuous in [0, 9]
b) f(x) is differentiable in (0, 9)
c) f(0) = 0
\(f(9)=\sqrt { 9 } -\frac { 9 }{ 3 } =3-3=0\)
∴ f(0) = f(9)
∴ By Rolle's theorem, there exists C ∈ [0, 9] such that f'(c) = 0
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } =0\)
⇒ \({ \frac { 1 }{ 2 } c }^{ \frac { 1 }{ 2 } -1 }=\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ 2\sqrt { c } } =\frac { 1 }{ 3 } \)
⇒ \(\frac { 1 }{ \sqrt { c } } =\frac { 2 }{ 3 } \)
⇒ \(\sqrt { c } =\frac { 2 }{ 3 } \)
Squaring both sides, c = \(\frac94\) ∈ [0, 9]

Given \(f(x)=\frac{x^{2}-2x}{x+2}, x\in [-1,6]\)
a) f(x) is continuous in [-1, 6]
b) f(x) is differentiable in (-1,6)
c) \(f(-1)=\frac { ({ 1) }^{ 2 }-2(-1) }{ -1+2 } =\frac { 1+2 }{ 1 } =3\)
\(f(6)=\frac { { 6 }^{ 2 }-2(6) }{ 6+2 } \)
\(=\frac { 36-12 }{ 8 } =\frac { 24 }{ 8 } =3\)
∴ (-1) = f(6)
By Rolle's theorem, there exists c ∈ [-1, 6] such that
f'(c) = 0
⇒ \(\frac { (c+2)(2c-2)-({ c }^{ 2 }-2c)(1) }{ (c+{ 2 })^{ 2 } } \) = 0 [By Quotient Rule]
⇒ 2c² + 4c - 2c - 4 - c² + 2c = 0
⇒ c²+4c-4 = 0
⇒ \(c=\frac { -4\pm \sqrt { { 4 }^{ 2 }-4(1)(-4) } }{ 2(1) } \) [∵ x = \( {-b \pm \sqrt{b^2-4ac} \over 2a}\)]
⇒ \(\frac { -4+\sqrt { 16+16 } }{ 2(1) } =\frac { -4\pm \sqrt { 2\times 16 } }{ 2 } \)
\(=\frac{-4 \pm 4 \sqrt{2}}{2}=+\not 2 \frac{(-2 \pm 2 \sqrt{2})}{\not 2}\)
\(c=-2\pm \sqrt { 2 } \)
\(c=-2+\sqrt { 2 } \) ∈ [-1, 6]
[∵  \(-2+\sqrt { 2 } \) ∉ [-1, 6]]

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