\(\underset { x\rightarrow \infty }{ lim } \frac { { 2x }^{ 2 }-3 }{ { x }^{ 2 }-5x+3 } =\frac { 2-\frac { 3 }{ { x }^{ 2 } } }{ 1+\frac { 5 }{ x } +\frac { 3 }{ { x }^{ 2 } } } \)
[Dividing the numerator and denominator by x²]
= \(\frac { 2-0 }{ 1-0+0 } =\frac { 2 }{ 1 } =2\)

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]

Observe that, f(0) = 0 = f (1), is continuous in the interval [0,1] and is differentiable in (0,1). Now,
 \(f'{x}=2x(1-x)(1-2x)\).
Therefore,  \(f'(c)=0 \) gives c = 0, 1 and \(\frac{1}{2}\)
which \(\Rightarrow c= \frac{1}{2}\in (0,1)\).

We have,
\(\frac{d}{dt}W(t)=-20\times(1-0.1t)\)
Therefore at t = 2, \(\frac{d}{dt}W(t)=-16\)
That is, the person forgets at the rate of 16 words after 2 days of studying.

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.

(x-2)+(x-2)^z

Rolle’s theorem is not valid.

(1, 0)

\(\left( \frac { 9 }{ 8 } ,\frac { 9 }{ 2 } \right) \)

Since f(x) = x² - 2x - 3 , \(f'(x)=2x-2>0 \forall x\in (2, \infty)\). Hence f (x) is strictly increasing in \((2, \infty)\)

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)