We have,
f(x) = (x-2)², then
\(f'(x)=2(x-2)=0 \) gives x = 2.
The intervals of monotonicity are \((-\infty,2)\) and \((2,\infty)\)
Since  \(f'(x)<0, \forall x \in (-\infty,2)\) for \((-\infty,2)\) the f(x) is strictly decreasing on \((2,\infty)\)
As \(f'(x)>0, \) for \(x \in(2, \infty)\) the function f(x) is strictly increasing on \((2,\infty)\)
Becasue f'(x) changes its sign from negative to positive when passing through x = 2 for the function f(x) it has a local minimum at x = 2
The local minimum value is f(2) = 0.

f(x) = |3x + 1|, x ∈ |-1, 3|

Since \(LHL \neq RHL, f'(-\frac{1}{3})\) does not exist.
Hence, Lagrange's mean value theorem is not applicable.

f(x) = \(\frac { x+1 }{ x } \), x ∈ [-1, 2]
f(0) = undefined
Lagrange's mean value theorem is not applicable since f (x) is not continuous at x = 0

Given \(f(x)=x-2logx, x\in [2,7]\)
(i) f(x) is continuous in [2, 7]
(ii) f(x) is differentiable in (2, 7)
f(2) = 2 - 2 log 2
= 2 - log 2² = 2 - log 4
f (7) = 7 - 2 log 7
= 7 - log 7² = 7 - log 49
Since f(2) ≠ f (7), Rolle's theorem is not applicable.

Given f(x) = tan x, x ∈ [0, π]
Rolle's theorem is not applicable since tan x is not continuous at x = \(\frac{\pi}{2}\) [∵ tan \(\frac{\pi}{2}\) = ∞]

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 x = a cos³ t; y = b sin³ t
\(\frac { dx }{ dt } =-3a \ { cos }^{ 2 } \ t \ sin \ t\)
\(\frac { dy }{ dt } =3b \ { sin }^{ 2 } \ t \ cot \ t\)
\(\therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \)
\(\frac { 3b{ sin }^{ 2 }tcost }{ 3a{ cos }^{ 2 }tsint } =\frac { -b }{ a } \) tan t
Slope of the tangent at t = \(\frac { \pi }{ 2 } \) is
\({ m=\left( \frac { dy }{ dx } \right) }_{ t=\frac { \pi }{ 2 } }\)
\(\frac { -b }{ a } tan\frac { \pi }{ 2 } =\frac { -b }{ a } \times \infty =\infty \)
\(\therefore m=\infty \)

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

We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

The series expansion of f (x) , in powers of (x − a) , be given by
\(f(x)=A_{0}+\sum_{n=1}^{\infty} A_{n}(x-a)^{n}\)
Substituting x = a gives A₀ = f(a). Differentiation of (7) given
\(f^{\prime}(x)=1 ! A \mid+\sum_{n=2}^{\infty} n A_{n}(x-a)^{n-1}\)
Substituting x = a gives A₁ = f'(a). Differentiation of (8) given
\(f^{\prime \prime}(x)=2 ! A_{2}+\sum_{n=3}^{\infty} n(n-1) A_{n}(x-a)^{n-2}\)
Substituting x = a given \(A_{2}=\frac{f^{\prime \prime}(a)}{2 !}\)  Differentiation of (9) gives
\(f^{\prime \prime \prime}(x)=3 ! A_{3}+\sum_{n=4}^{\infty} n(n-1)(n-2) A_{n}(x-a)^{n-3}\)
Differentiation of (10) (k − 3) times gives
\(f^{(k)}(x)=k ! A_{k}+\sum_{n=k+1}^{\infty} n(n-1) \ldots(n-k+1) A_{n}(x-a)^{n-k}\)
Substituting x = a gives \(A_{k}=\frac{f^{(k)}(a)}{k !}\) which completes the proof of the theorem.
In order to expand a function around a point say x = a , equivalently in powers of (x − a) we need to differentiate the given function as many times as the required powers and evaluate at x = a. This will give the value for the coefficients of the required powers of (x − a).

By the mean value theorem, there exists  \(c \in\left(x_{1}, x_{2}\right) \subset(a, b)\) such that,
\(\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}=f^{\prime}(c)\)
Since \(f^{\prime}(c)>0, \text { and } x_{2}-x_{1}>0 \text { we have } f\left(x_{2}\right)-f\left(x_{1}\right)>0\)
We conclude that, whenever \(x_{1}

As \(\underset { x\rightarrow -{ 4 }^{ + } }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =-\infty \ and\ \underset { x\rightarrow { 4 }^{ + } }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\infty \) 
Therefore x = −4 and x = 4 are vertical asymptotes 
As \(\underset { x\rightarrow \infty }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\underset { x\rightarrow \infty }{ lim } \frac { 2-\frac { 8 }{ { x }^{ 2 } } }{ 1-\frac { 16 }{ { x }^{ 2 } } } =2\) and \(\underset { x\rightarrow -\infty }{ lim } \frac { { 2x }^{ 2 }-8 }{ { x }^{ 2 }-16 } =\underset { x\rightarrow -\infty }{ lim } \frac { 2-\frac { 8 }{ { x }^{ 2 } } }{ 1-\frac { 16 }{ { x }^{ 2 } } } =2\)
Therefore, y = 2 is a horizontal asymptote.
This can also be obtained by synthetic division

Since \(f'(x)=1-cosx\ge0\) and zero at the points \(x=2n\pi, n\in Z\) and hence the function is increasing on the real line.
Since there is no sign change in f′(x) when passing through \(x=2n\pi, n\in Z\) by the first derivative test there is no local extrema.

Differentiating the given function, we get \(f'(x)= -3 sin\ x\)
Thus, \(f'(x)=0 \Rightarrow sinx\)\( =0 \Rightarrow x=\pi \in (0,2\pi)\).
Evaluating  f(x) -at the endpoints \(x=0,2\pi\)  and at critical number  \(x=\pi\) 
we get \(f(0)=3, f(2\pi)=3 \) and \(f(\pi)=-3\)
From these values, the absolute maximum is 3 which occurs at \(x=0,2\pi\) and the absolute minimum is −3 which occurs at \(x=\pi\)

Differentiating the given function, we get
f'(x) = 6x² + 6x -12
= 6(x² + x - 2)
f'(x) = 6(x + 2)(x + 1)
Thus, f'(x) = 0 \(\Rightarrow\) x = -2, 1 \(\in\) (-3, 2).
Therefore, the critical numbers are, x = -2, 1. Evaluating f(x) at the endpoints x = -3, 2 and at critical numbers x = -2,1 we get f(-3) = 9, f(2) = 4, f(-2) = 20 and f(1) = -7.
From these values, the absolute maximum is 20 which occurs at, x = -2 and the absolute minimum is −7 which occurs at x = 1

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.
\(\therefore f(1)={ 1 }^{ 2 }e^{ -2(1) }={ e }^{ -2 }=\frac { 1 }{ { e }^{ 2 } } \)

\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2 }{ { x }^{ 2 }-1 } -\frac { x }{ x-1 } \right) =\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-x(x+1) }{ { x }^{ 2 }-1 } \right) \)
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } \)
Form which is indeterminate,
Applying L' Hopital rule we get,
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \left( \frac { 2-{ x }^{ 2 }-x }{ { x }^{ 2 }-1 } \right) =\frac { 0 }{ 0 } =\frac { -2-1 }{ 2(1) } =\frac { -3 }{ 2 } \)

\(\underset { x\rightarrow 0 }{ lim } \left( \frac { 1 }{ sinx } -\frac { 1 }{ x } \right) =\underset { x\rightarrow 0 }{ lim } \left( \frac { x-sinx }{ xsinx } \right) =\frac { 0 }{ 0 } \) form
Which is indeterminate Applying L' Hopital rule we get,
\(0+\frac { sinx }{ -xsinx+cosx } +cosx=\frac { 0 }{ 2 } =0\)

\(\underset { x\rightarrow { 1 }^{ + } }{ lim } \frac { 1-cosx }{ xcosx+sinx } =\frac { 1-cos0 }{ 0+sin0 } =\frac { 1-1 }{ 0 } =\frac { 0 }{ 0 } \)form

\(\underset { x\rightarrow \infty }{ lim } { e }^{ -x }\sqrt { x } =\underset { x\rightarrow \infty }{ lim } \frac { { \sqrt { x } } }{ { e }^{ x } } =\frac { \infty }{ \infty } \)
Which is in indeterminate form. Applying L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { e }^{ x } } =\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } \frac { { e }^{ -x } }{ \sqrt { x } } \)
\(\frac { 1 }{ 2 } \underset { x\rightarrow \infty }{ lim } e^{ -x }\sqrt { \frac { 1 }{ x } } =\frac { 1 }{ 2 } { e }^{ -\infty }(0)=0\) [When x ➝ ∞, \(\frac1x\) ➝ 0 e^-∞ = 0]

\(\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { secx }{ tanx } =\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { \frac { 1 }{ \frac { cosx }{ sinx } } }{ \frac { sinx }{ cosx } } =\underset { x\rightarrow \frac { { \pi }^{ - } }{ 2 } }{ lim } \frac { 1 }{ sinx } \)
\(=\frac { 1 }{ sin\frac { \pi }{ 2 } } =\frac { 1 }{ 1 } =1\)

\(\underset { x\rightarrow \infty }{ lim } \frac { x }{ logx } =\frac { \infty }{ \infty } \)
Which is in indeterminate form
∴ By L' Hopital rule we get,
\(\underset { x\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1 } }{ x } =\underset { x\rightarrow \infty }{ lim } x=\infty \)

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

\(\underset { x\rightarrow 0 }{ lim } \frac { 1-cosx }{ { x }^{ 2 } } =\frac { 1-cos0 }{ 0 } =\frac { 1-1 }{ 0 } =\frac { 0 }{ 0 } \)
Indeterminate form, Applying L' Hopital rule we get,
\(\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ 2x } =\frac { 1 }{ 2 } \underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } \)\(\left[ \because ​​​​​​\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } =1 \right] \)
\(=\frac { 1 }{ 2 } \times 1\)
= \(\frac12\)

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

This is an indeterminate form \(\frac{\infty}{\infty}\) and hence we use the l’Hôpital’s Rule to evaluate.
\(\underset{x\rightarrow 1^{-}}{lim}(\frac{log(1-x)}{cot(\pi x)})=\underset{x\rightarrow1^{-}}{lim}(\frac{-\frac{1}{1-x}}{-\pi cosec^{2}(\pi x)})\) \((\frac{\infty}{\infty})\) 
On Simplication,
\(=\underset{x\rightarrow-1}{lim}(\frac{sin^{2}(\pi x)}{\pi (1-x)})\) \((\frac{0}{0})\)
again applying the l’Hôpital Rule 
\(= \underset{x\rightarrow 1^{-}}{lim}(\frac{2\pi sin(\pi x). cos(\pi x)}{-\pi})\)
=\(\underset{x\rightarrow -1}{lim}(-2 sin(\pi x).cos (\pi x))\)
= 0.

f(x) = x² - 3x + 2
f^I(x) = 2x - 3
f^II(x) = 2
⇒ f(1) = 0
f^I(1) = 2 - 3 = -1
f^II(1) = 2
Taylor's series forf(x) at x = 1 is
f(x) = \(f(1)+\frac { { f }^{ 1 }(1) }{ 1! } (x-1)+\frac { { f }^{ II }(1) }{ 2! } ({ x-1) }^{ 2 }\)
= \(0-\frac { 1 }{ 1! } { (x-1) }+\frac { 2 }{ 2! } { (x-1) }^{ 2 }\)
f(x) = -(x-1)+(x-1)²

Meclaurin's expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\).................
log(1-x) = \(0-\frac { 1 }{ 1! } x-\frac { { x }^{ 2 } }{ 2! }- \frac { 2 }{ 3! } { x }^{ 3 }-\frac { { 6x }^{ 4 } }{ 4! } +.....\)
= \(-x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } \)+ .....
log(1 - x) = \(-\left( x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +.... \right) \)

\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }\) + .........
cos x = 1 + 0 - \(\frac { { x }^{ 2 } }{ 2! } +0+\frac { { x }^{ 4 } }{ 4! } +0+\frac { { x }^{ 6 } }{ 6! } +...\)
cos x = 1 - \(\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 4 } }{ 4! } +\frac { { x }^{ 6 } }{ 6! } \)

Let f(x) = sin x ⇒ f(0) = sin 0 = 0
f^I(x) = cos x ⇒ f^I(0) = cos 0 = 1
f^Il(x) = - sin x ⇒ f^II(0) = -sin 0 = 0
f^IlI(x) = - cos x ⇒ f^III(0) = -cos 0 = 0
f^IV(x) = + sin x ⇒ f^IV(0) = sin 0 = 0
f^V(x) = cos x ⇒ f^v(0) = cos 0 = 0
f^VI(x) = - sin x ⇒ f^vI(0) = - sin 0 = 0
f^VII(x) = - cos x ⇒ f^vII(0) = - cos 0 = 0
∴ Maclaurins' series expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\frac { { f }^{ IV }(0) }{ 4! } { x }^{ 4 }+.....\)
\(\therefore sinx=0+\frac { 1 }{ 1! } x+0-\frac { 1 }{ 3! } { x }^{ 3 }+0-\frac { 1 }{ 5! } { x }^{ 5 }+0-\frac { 1 }{ 7! } { x }^{ 7 }+........\)
\(x-\frac { { x }^{ 3 } }{ 3! } -\frac { { x }^{ 5 } }{ 5! } -\frac { { x }^{ 7 } }{ 7! } +...\)

\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+........\)
∴ Maclaurins' series expansion of
\({ e }^{ x }=1+\frac { 1 }{ 1! } x+\frac { 1 }{ 2! } { x }^{ 2 }+...\)

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]

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 f(x) = x² − x, x ∈ [0, 1] 
(i) f(x) is continuous in [0, 1]
(ii) f(x) is differentiable in (0, 1)
(iii) f(0) = 0² - 0 = 0
f(1) = 12-1 = 1-1 = 0
∴ f(0) = f(1)
By Rolle's theorem, there exists C ∈ [0, 1] such that
f'(c) = 0
⇒ 2c - 1 = 0
⇒ 2c = 1
⇒ c = \(\frac12\) ∈ [0, 1]

Let f (t) be the temperature at time t. By the mean value theorem, we have
\(f'(c)=\frac{f(b)-f(a)}{b-a}\)
= \(\frac{100-(-10)}{22}\)
= \(\frac{110}{22}\)
= 5°C per second.
Hence the instantaneous rate of change of temperature at some time t should be 5°C per second.

Let f (x) = sin x which is a differentiable function in any open interval. Consider an interval \([\alpha, \beta]\). Applying the mean value theorem there exists  \(c \in (\alpha, \beta)\) such that, 
\(\frac{sin \beta - sin \alpha}{\beta-\alpha}=f'(c)=cos(c)\)
Therefore, \(\frac{sin \beta - sin \alpha}{\beta-\alpha}=|cos(c)|\le1\)
Hence, \(|sin\alpha-sin\beta|\le |\alpha-\beta|\)
Remark
If we take \(\beta=0\) in the above problem, we ge \(|sin \alpha|\le |\alpha|\) 

By the mean value theorem we have, there exists 'c'∈(2, 7) such that,
\(\frac { f(7)-f(2) }{ 7-2 } \) = f'(c) ≤ 29
Hence, f(7) ≤ 5× 29 +17 = 162
Therefore, the maximum value of f (7) is 162.

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.

P(x) = \(x^{4}-6x^{3}-11x^{2}+24x+28\), \(\alpha\) = 2, \(\beta\) = 7 
and observing \(\frac{P'(x)}{2}=2x^{3}-9x^{2}-11x+12=Qx\), (say).
This implies that there is a zero of the polynomial Q(x) in the interval (2, 7)
For verification,
Q(2) = 16 - 36 - 22 +12 = 28 - 58 = -30 < 0
Q(7) = 686 - 441 - 77 +12 = 698 - 518 = 180 > 0
From this we may see that there is a zero of the polynomial Q(x) in the interval (2, 7)

Let f (x) = x⁴ + 2x³-2
Then f (x) is continuous in [0, 1] and differentiable in (0, 1)  
Now, f'(x) = 4x³+6x² 
If f'(x) = 0, then
2x²(2x+3) = 0
Therefore,  \(x=0, -\frac{3}{2}\) but  \(0,-\frac{3}{2} \notin (0,1)\).
Thus, \(f'(x)>0, \forall x\in (0,1)\).
Hence by the Rolle’s theorem there do not exist \(a,b \in(0,1)\) such that, f(a) = 0 = f(b). Therefore the equation f(x ) = 0 cannot have two roots in the interval (0, 1) . But, f (0, 2) = −2 < 0 and f (1) = 1 > 0 tells us the curve y f = (x) crosses the x -axis between 0 and 1 only once by the Intermediate value theorem. Therefore the equation x⁴ + 2x³ − 2 = 0 has only one real root in the interval (0, 1) .

Observe that, f (2) = 0 = f (3) and f (x) is continuous in the interval [2, 3] and differentiable in (2, 3). Now,
\(f'(x)=\frac{x^{2}-6}{x(x^{2}+6)} \)
Therefore, \(f'{(c)}=0\) gives
\(\frac{c^{2}-6}{c(c^{2}+6)}=0\)
which implies \( c=\pm\sqrt{6}\)
Now c = \(\pm\sqrt{6} \in (2,3).\)
Observe that \(-\sqrt{6}\notin (2,3)\) and hence \(c=\pm\sqrt{6}\) satisfies the Rolle’s theorem.
Rolle’s theorem can also be used to compute the number of roots of an algebraic equation in an interval without actually solving the equation.

We have, f (x) is continuous in \([\frac{1}{2},2 ]\) and differentiable in \((\frac{1}{2},2 )\) with  \(f(\frac{1}{2})=\frac{5}{2}=f(2) \). 
By  the Rolle’s theorem there must exist a \(c \in (\frac{1}{2},2 )\) such that, \(f'(c)=1-\frac{1}{c^{2}}=0 \Rightarrow c^{2}=1 \) gives \(\Rightarrow c=\pm1, \) As \(1\in(\frac{1}{2},2)\) we choose c = 1.

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 \(\Rightarrow c= \frac{1}{2}\in (0,1)\).

Equation of the given curve is x = cos t; y = 2 sin² t
\(\frac { dx }{ dt } \) = - sin t;
\(\frac { dy}{ dt } \) = 4 sin t cos t
\(\therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \) = \(\frac{4 \ sin \ t \ cos \ t}{-sin \ t}\)
= 4 cost
∴ Slope = m = \(\left( \frac { dy }{ dx } \right) \)t = \(\frac { \pi }{ 3 } \)
= -4 cos \(\frac { \pi }{ 3 } \) = -4 \(\left( \frac { 1 }{ 2 } \right) \)
Equation of the tangent is y - y₁ = m (x - x₁)
When t = \(\frac { \pi }{ 3 } \), x cos \(\frac { \pi }{ 3 } \) = \(\frac12\)
when t =  \(\frac { \pi }{ 3 } \), y = 2 \({ \left( sin\ \frac { \pi }{ 3 } \right) }^{ 2 }\)
= 2 \({ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }\) = 2 \(\frac34\) = \(\frac32\)
∴ Equating of the tangent is y - \(\frac32\) = -2 (x-\(\frac12\))
\(\Rightarrow \frac{2 y-3}{\not 2}=\frac{1}{2}\left(\frac{2 x-1}{\not 2}\right)\)
⇒ 2y - 3 = -4x + 2
⇒ 4x + 2y = 5
Equating of the normal is y - y₁ = \(\frac{-1}{m}\) (x-x1)
⇒ y - \(\frac32\) = \(\frac12\)(x - \(\frac12\))
\(\Rightarrow \frac{2 y-3}{\not 2}=\frac{1}{2}\left(\frac{2 x-1}{\not 2}\right)\)
⇒ 4y-6 = 2x-1
⇒ 2x - 4y +5 = 0

Equation of the given curve is y = x sin x
∴ slope = m = \(\left( \frac { dy }{ dx } \right) \)_{\(\left( \frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)}
= \(\frac { \pi }{ 2 } cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } \)
= \(\frac { \pi }{ 2 } \) (0) + 1 = 1
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - \(\frac { \pi }{ 2 } \) = 1 \(\left( x-\frac { \pi }{ 2 } \right) \)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
Equating of the normal is y - y₁ = \(\frac{-1}{m}\) (x - x₁)
\(\Rightarrow \frac{2 y-\pi}{\not 2}=\frac{2 x-\pi}{\not 2}\)
⇒ 2x + 2y = 2π
⇒ x + y- π = 0

Equating of the given curve is y = x⁴+ 2e^x
\(\frac { dy }{ dx } \) =  x⁴+ 2e^x
m = \(\frac { dy }{ dx } \)_{(0, 2)}
= 4(0) + 2e⁰ = 2(1) = 2
∴ Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - 2 = 2 (x - 0)
⇒ y-2 =2x
⇒ 2x - y = -2
Equating of the normal is y -y₁ = \(\frac{-1}{m}\) (x-x₁)
⇒ y - 2 = -\(\frac12\) (x - 0)
⇒ 2y - 4 = -x
⇒ x + 2y - 4 = 0

Given y = x² - x⁴
\(\frac { dy }{ dx } \) = 2x-4x³
m = \(\frac { dy }{ dx } \)_{(1, 0)} = 2(1) - 4 (1)³
= 2 - 4 = - 2
Equating of the tangent is y - y₁ = m (x - x₁)
⇒ y - 0 = - 2 (x - 1)
⇒ y = -2x+2
⇒ 2x+y = 2
Equating of the normal is y - y1 = \(\frac{1}{m}\)(x - x1)
⇒ y- 0 = \(\frac12\) (x - 1)
⇒ 2y = x - 1
⇒ x- 2y = 1

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

We have, \(\frac{dy}{dx}=2x+3\). Hence at (1, 2), \((\frac{dy}{dx})=5\)

Therefore, the required equation of tangent is.
\((y-2)=5(x-1)\Rightarrow 5x-y-3=0\) 
The slope of the normal at the point (1, 2) is -\(\frac{1}{5}\).
therefore, the required equation of normal is
\((y-2)=-\frac{1}{5}(x-1)\Rightarrow x+5y-11=0\) 

Distance moved in time 't' is s = \(\frac{t^{3}}{3}-t^{2}+3\) 
Velocity at time 't ' is V = \(\frac{ds}{dt}=t^{2}-2t\)
Acceleration at time 't ' is a(t) = \(\frac{dV}{dt}=2t-2\)
Therefore, the velocity is zero when t² − 2t = 0, that is t = 0, 2. The acceleration is zero when 2t − 2 = 0 . That is at time at time t = 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.

Let the given function be written as f (x) = x² ( 10− x)² . Now
f(x) = x²(100 - 20x + x²) = x⁴-20x³+100x²
Therefore, f'(x) = 4x³- 60x³ + 200x = 4x(x²-15x+50)
f'(x) = 4x(x²-15x + 50) = 0 ⇒ x = 0, 5, 10
and f"(x) = 12x²-120x + 200
The stationary points of f(x) are x = 0, 5, 10 at these points the values of f′′(x) are respectively 200, −100 and 200. At x = 0, it has local minimum and its value is f(0) = 0. At x = 5, it has local maximum and its value is f(5) = 625. At x = 10, it has local minimum and its value is f(10) = 0.

Let x = length of the cut on each side of the little squares.
V = the volume of the folded box.
The length of the base after two cuts along each edge of size x is 12 − 2x. The depth of the box  after folding is x, so the volume is V = x \(\times\) (12 - 2x)² Note that,
when x = 0 or 6, the volume is zero and hence there cannot be a box. Therefore the problem is to maximize, V = x \(\times\) (12 - 2x)², x ∈ (0, 6) 
 \(\frac { dV }{ dx } ={ (12-2x) }^{ 2 }-4x(12-2x)\)
= (12 − 2x)(12 − 6x).

\(\frac { dV }{ dx } \) = 0 gives the stationary points x = 2, 6. Since 6 ∉ (0, 6) the only stationary point is at x = 2∈(0, 6). Further, \(\frac { dV }{ dx } \)- changes its sign from postive to negative when passing through x = 2 .
Therefore at x = 2 the volume V is local maximum. The local maximum volume value is V = 128 units. Hence the maximum cut can only be 2 units.

The given function is defined and is differentiable at all  \(x\in (-\infty, \infty) \). As
\(f(x)=\frac{1}{1+x^{2}}\).
We have \(f'(x)=-\frac{2x}{(1+x^{2})^{2}}\)
The stationary points are given by \(-\frac{2x}{(1+x^{2})^{2}}=0\) that is x = 0
Hence the intervals of monotonicity are \((-\infty,0)\) and \((0,\infty)\)
On the interval \((-\infty,0)\) the function strictly increases because f'(x) > 0 in that interval.
The function f(x) strictly decreases in the interval \((0,\infty)\) because f'(x) < 0 in that interval.
Since f′(x) changes from positive to negative when passing through x = 0, the first derivative test tells us there is local maximum at x = 0 and the local maximum value is f (0) = 1.

The given function is defined and is differentiable at all \(x \in(0, \infty)\) 
f(x) = x log x + 3x.
Therefore f'(x) = log x+1+3 = 4 + log x.
The stationary points are given by 4 + log x = 0
That is x = e^-4
Hence the intervals of monotonicity are (0, e^-4) and \((e^{-4}, \infty)\) 
At \(x=e^{-5}\in(0,e^{-4})\), f'(e^-5) = -1<0 and hence in the interval (0, e^-4) -the function is strictly decreasing.
At \(x=e^{-3}\in(0,e^{-4})\), f'(e^-3) = 1>0 and hence strictly increasing in the interval \((e^{-4}, \infty)\). 
Since f′(x) changes from negative to positive when passing through x = e⁻⁴, the first derivative test tells us there is a local minimum at x = e^-4 and it is f(e-⁴) = -e^-4.

The given function is a periodic function with period 2π and hence there will be stationary points and points of inflections in each period interval. We have,

\(\frac { dy }{ dx } =cosx\) and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =sinx\)
Now,\( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-sinx=0\Rightarrow x=n\pi \)
We now consider an interval, (-π, π ) by splitting into two sub intervals \(\left( -\pi ,0 \right) \) and \(\left( 0,\pi \right) \)
In the interval \(\left( -\pi ,0 \right) ,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } >0\) and hence the function is concave upward
In the interval \(\left( -\pi ,0 \right) ,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } <0\) and hence the function is concave downward. Therefore (0,3) is a point of inflection. The general intervals need to be considered to discuss the concavity of the curve are \(\left( n\pi ,\left( n+1 \right) \pi \right) \), where n is any integer which can be discussed as before to conclude that \(\left( n\pi ,3 \right) \) are also points of inflection.

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

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.

f(x) = x log x
\(f'\left( x \right) =x.\frac { 1 }{ x } +logx(1)\)
= 1 + log x
f''(x) = 0
⇒ 1 + logx = 0
⇒ logx = 1
\(\Rightarrow x={ e }^{ -1 }=\frac { 1 }{ e } \)
ஃ The critical numbers is \(\frac { 1 }{ e } \)
\(f''(x)=\frac { 1 }{ x } \) 
\(f''\left( \frac { 1 }{ e } \right) =\frac { 1 }{ \frac { 1 }{ e } } =e>0\) 
Since \(f''\left( \frac { 1 }{ e } \right) >0\) there is a local minimum at \(x=\frac { 1 }{ e } \) 
\(f''\left( \frac { 1 }{ e } \right) =\frac { 1 }{ e } log\frac { 1 }{ e } \)
= \(\frac { 1 }{ e } log\frac { 1 }{ e } \)
= \(\frac { -1 }{ e } { log }_{ e }e=\frac { -1 }{ e } (1)=\frac { -1 }{ e } \)
ஃ Local minimum is \(-\frac { 1 }{ e } \) which occurs at \(x=\frac { 1 }{ e } \)

f'(x) = -3.x⁵ + 5x³
f'(x) = -3x⁵ + 5x
f'(x) = - 15x⁴ + 15x²
f''(x) = 0
⇒ - 15x⁴ + 15x² = 0
⇒ - 15x² (1 - x²) = 0
⇒ x² = 0, 1-x² = 0
⇒ x = 0, x = 1, x = -1
ஃ The critical numbers are 0, 1,-1.
f"(x) - 60x³ + 30 x
f"(0) = 0
f"(1) - 60(1)³ + 30 (1)
-60 + 30 = -30
- 60(-1)³ + 30 (-1)
60- 30 = -30
Since f"(-1) < 0, it has a local maximum at
x = 1.
ஃ f(1) = -3(-1)⁵ + 5(-1)³
= -3 + 5 = 2
Since f"(-1) > 0, it has a local maximum at
x = -1.
ஃ f(-1) -3(1)⁵ + 5(1)³
3 - 5 = -2
∴ Local maximum is 2 which occurs at,
x = 1 and local minimum is -2 which occurs at 
x = -1.

This is an indeterminate of the form 1^∞
Let g(x) = \((cos{ x })^{ \frac { 1 }{ { x }^{ 2 } } }\)
Taking logarithm, we get
log g(x) = log \((cos{ x })^{ \frac { 1 }{ { x }^{ 2 } } }\)
= \(\frac { log(cosx) }{ { x }^{ 2 } } \)
Now, \(\underset { x\rightarrow { { 0 }^{ + } } }{ lim } \frac { log(cosx) }{ { x }^{ 2 } } =\left( \frac { 0 }{ 0 } form \right) \)
= \(\frac { \underset { x\rightarrow { { 0 }^{ + } } }{ lim } \frac { 1 }{ cosx } (-sinx) }{ 2x } \)
= \(\underset { x\rightarrow { { 0 }^{ + } } }{ lim } -\frac { tanx }{ 2x } \)
= \(\frac { -1 }{ 2 } \underset { x\rightarrow { { 0 }^{ + } } }{ lim } \frac { tanx }{ x } =\frac { -1 }{ 2 } (1)\) \(\left[ \because \underset { x\rightarrow { { 0 }^{ + } } }{ lim } \frac { tanx }{ x } =1 \right] \)
= \(-\frac { 1 }{ 2 } \)
But \(\underset { x\rightarrow { { 0 }^{ + } } }{ lim } log(g(x)=log{ e }^{ (\underset { x\rightarrow { { 0 }^{ + } } }{ lim } g(x)) }\)
∴ \(log\left( \underset { x\rightarrow { { 0 }^{ + } } }{ lim } g(x) \right) =-\frac { 1 }{ 2 } \)

\({ e }^{ log }\left( \underset { x\rightarrow { { 0 }^{ + } } }{ lim } g(x) \right) ={ e }^{ -\frac { 1 }{ 2 } }\)
\({ e }^{ log }\left( \underset { x\rightarrow { { 0 }^{ + } } }{ lim } g(x) \right) ={ e }^{ -\frac { 1 }{ 2 } }=\frac {1}{\sqrt e}\)

This is an indeterminate of the from 1∞
Let g(x) = (sin x)^{tan x}
Taking logarithm, we get
log (g(x)) = log (sin x)tan x
= tan x log (sin x)
Now, \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } log(g(x))\) = \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } \frac { log(sinx) }{ cotx } =\left( \frac { 0 }{ 0 } form \right) \)
= \(\frac { \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } \frac { 1 }{ sinx } \times cosx }{ -{ cosec }^{ 2 }x } \) [By L' Hôpital Rule]
= \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } -\frac { cosx }{ sinx\frac { 1 }{ { sin }^{ 2 }x } } \)
= \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } cos \ x \ sin \ x=0\)
But \(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } log(g(x))=log(\underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x))\)
\(\therefore log\left( \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x) \right) =log \ 0=1\)

\(\Rightarrow { e }^{ log }\left( \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x) \right) ={ e }^{ 1 }\)
\(\Rightarrow \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } g(x)=e\Rightarrow \underset { x\rightarrow { \frac { \pi }{ 2 } } }{ lim } { (sinx) }^{ tanx }=e^o=1\)

This an indeterminate of the form 1∞
Let \(g(x)={ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
Taking logarithm, we get,
\(log(g(x))=log{ \left( 1+\frac { 1 }{ x } \right) }^{ x }\)
\(=xlog{ \left( 1+\frac { 1 }{ x } \right) }\)
\(=\frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } \)
\(\therefore ​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { \left( 1+\frac { 1 }{ x } \right) } }{ \frac { 1 }{ x } } =\left( \frac { 0 }{ 0 } form \right) \)

\(=​​​​​​\underset { x\rightarrow \infty }{ lim } \frac { { 1 } }{ \left( 1+\frac { 1 }{ x } \right) } =1\)
But \(​​​​​​\underset { x\rightarrow \infty }{ lim } log(g(x))=log​​​​​​\underset { x\rightarrow \infty }{ lim } g(x)\)
\(\therefore log(\underset { x\rightarrow \infty }{ lim } g(x))=1\)
\({ e }^{ log }(\underset { x\rightarrow \infty }{ lim } g(x))={ e }^{ 1 }=e\Rightarrow (\underset { x\rightarrow \infty }{ lim } g(x))=e\)
\(\Rightarrow \underset { x\rightarrow \infty }{ lim } { \left( 1+\frac { 1 }{ x } \right) }^{ x }=e\)

Let \(g(x)=x^{\frac{1}{1-x}}\). This is an indeterminate of the form \(1^{\infty}\). Taking the logarithm,
\(log \ g(x)=\frac{logx}{1-x}\).
Therefore, \(\underset{x \rightarrow1}{lim \ }log \ g(x) =\underset{x\rightarrow 1}{lim}(\frac{log \ x}{1-x})(\frac{0}{0})\)
An application of l’Hôpital rule,
\(\underset{x \rightarrow1}{lim} \ (\frac{\frac{1}{x}}{-1})=-1\)
But, \(\underset{x\rightarrow 1}{lim}\) log g(x) = log (\(\underset{x\rightarrow 1}{lim}\) g(x)).
Hence on exponentiating, we get
\(\underset{x \rightarrow1}{lim}\quad x^{\frac{1}{1-x}}=e^{-1}=\frac{1}{e}\).

This is an indeterminate of the form \(\infty^{0}\).
Let \(g(x)=(1+2x)^{\frac{1}{2log \ x}}\)
Taking the logarithm, we get
log g(x) = \(\frac{log(1+2x)}{2log \ x}\)
\(\underset{x\rightarrow \infty}{lim}log \ g(x)=\underset{x\rightarrow \infty}{lim}(\frac{log(1+2x)}{2log \ x})\) \((\frac{\infty}{\infty})\) 
\(=\underset{x\rightarrow \infty}{lim}(\frac{\frac{2}{1+2x}}{\frac{2}{x}})\) (by l’Hôpital Rule)
= \(\underset{x\rightarrow \infty}{lim}(\frac{x}{1+2x})\) \((\frac{\infty}{\infty})\)
= \(\underset{x\rightarrow \infty}{lim}(\frac{1}{2})=\frac{1}{2}\) but,
\(\underset{x\rightarrow \infty}{lim}log \ g(x)=log (\underset{x\rightarrow \infty}{lim}g(x))\).
Hence by exponentiating, we get the required limit as  \(\sqrt{e}\)

This is an indeterminate of the form \(1^{\infty}\).
Let \(g(x)=(1+x)^{\frac{1}{x}}\). Taking the logarithm, we get
\(log \ g(x)=\frac{log(1+x)}{x}\)
\(\underset{x\rightarrow 0^{+}}{lim} log (g(x))=\underset{x\rightarrow 0^{+}}{lim}(\frac{log(1+x)}{x})\) \((\frac{0}{0})\) 
=\(\underset{x\rightarrow0^{+}}{lim}(\frac{\frac{1}{1+x}}{1})\)  (by 1’Hôpital Rule)
= 1.
But, \(\underset{x\rightarrow0^{+}}{lim}log g(x)=log(\underset{x\rightarrow 0^{+}}{lim} g(x))\)
Therefore, log\((\underset{x\rightarrow 0^{+}}{lim} g(x))=1\).
Hence by exponentiating, we get, \(\underset{x\rightarrow 0^{+}}{lim}g(x)=e.\)

Let (x) = cos2 x
f^I(x) = 2cos x (- sin x)
= - sin 2x ⇒ f^l(0) = 0
f^Il(x) = - 2 cos 2x ⇒ f^Il(0) = -2
f^III(x) = + 4 sin 2 x ⇒ f^III(0) = 0
f^IV(x) = 8 cos 2 x ⇒ f^IV(0) = 8
f^V(x) = -16 sin 2x ⇒ f^V(0) = 0
f^VI(x) = - 32 cos 2x ⇒ f^VI(0) = -32
∴ Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\) .......
∴ cos x = 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { 8{ x }^{ 4 } }{ 4! } +\frac { { 32x }^{ 6 } }{ 6! } \) + ...
= 1 - \(\frac { 2{ x }^{ 2 } }{ 2! } +\frac { { 2 }^{ 3 }{ x }^{ 4 } }{ 4! } +\frac { { { 2 }^{ 5 }x }^{ 6 } }{ 6! } \)+ ...

Let f(x) = tan^-1(x)
⇒ f(0) = tan-1(0) = 0
\({ f }^{ -1 }(x)=\frac { 1 }{ 1+{ x }^{ 2 } } ={ (1+{ x }^{ 2 }) }^{ -1 }\)
⇒ f^I(0) = 1
f^II(x) = -1(1 + x)^-2 (2x)
= -2x(1 + x²)^-2
⇒ f^II(0) = 0
= -2[-4x²(1 + x²)^-3 + (1 + x²)^-2]
f^III(x) = -2[x(-2)(1+x²)^-3 (2x) + (1 + x²)^-2]
⇒ f^III(0) = 2
f^IV(x) = -2[-4x²(-3) (1 + x²)^-4 (2x) + (1 + x²)-3(-8x) + (1 + x²)^-3(-2)(2x)]
⇒ f^IIl(0) = 2
= - 2 [24x³ (1 + x²)^-4-8x (1+ x²)³ -4x (1 + x²)³]
= - 2 [24x³ (1 + x²)^-4-12x (1+ x²)³]
f^IV(x) = 24[2x³ (1 + x²)-4 - x (1 + x²)^-3]
⇒ f^Il(0) = 0
f^V(x) = - 24 [2x³ (-4)(1 +x²)^-5(2x) + (1+ x²)^-4(6x² -x (- 3)(1 + x²)-4(2x) - (1 + x²)³]
= - 24 [16x⁴ (1 + x²)^-5+ 6x² (1+ x²)-4- (1 + x²)³]
= - 24 [16x⁴ (1 + x²)^-5+12x² (1+ x²)^-4- (1 + x²)³] 
⇒ fV(0) = 24
⇒ f(0) = tan-1 (0) = 0
⇒ f^I(0) = 1
⇒ f^II(0) = 0
⇒ f^III(0) = -2
⇒ f^IV(0) = 0
⇒ f^V(0) = 24
Maclaurin's series
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\frac { { f }^{ IV }(0) }{ 4! } { x }^{ 4 }+..\)
= \(0+\frac { 1 }{ 1! } x+0\frac { 2 }{ 3! } { x }^{ 3 }+0+\frac { 24 }{ 5! } { x }^{ 5 }+\)...
= x - \(\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 5 } }{ 5 } +\).......

f (x) is defined and differentiable for all x∈(-∞, ∞)
\(f'\left( x \right) =\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \)
\(f''(x)=\frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) \)
f"(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) =0\Rightarrow { e }^{ x }-{ e }^{ -x }=0\) 
\(\Rightarrow { e }^{ x }={ e }^{ -x }\Rightarrow { e }^{ x }=\frac { 1 }{ { e }^{ x } } \) 
\(\Rightarrow { e }^{ 2x }=1\Rightarrow { e }^{ 2x }={ e }^{ 0 }\) 
\(\Rightarrow 2x=0\Rightarrow x=0\) 
The possible intervals are (-∞,0) and (0,∞)

ஃ f(x) is concave up in (0, ∞) and concave down in (∞, 0).
Since f"(x) changes its position from negative to positive, when it passes through x = 0 the points of inflection is (0,1(0))
\(f(0)=\frac { 1 }{ 2 } \left( { e }^{ o }-{ e }^{ o } \right) =\frac { 1 }{ 2 } \left( 1-1 \right) =0\) 
ஃ (0, 0) is the point of inflection.

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) = x (x - 4)³
f'(x) = x .3(x - 4)² + (x - 4)³(1)
= 3x (x - 4)² + (x - 4)³
= (x - 4)² + (3x +x- 4)³
= (x - 4)² + (4x - 4)
= 4 (x-1) (x-4)²
f"(x) = 4 [(x - 1)² (x - 4) +(x - 4)² (1)]
= 4[2(x-4)(x-1)+(x - 4)²]
= 4(x- 4)[2x - 2 +x - 4)]
= 4(x - 4)(3x - 6)
= 12 (x - 4)(x- 2)
f"(x) = 0
⇒ 12 (x - 4)(x- 2) = 0
⇒ x = 2, 4
The possible intervals are (-∞, 2) (2, 4) and (4, ∞)

ஃThe curve is concave upward on (-∞, 2) (4, ∞) and concave downward on (2, 4).
As f"(x) changes its sign when it passes through x = 2 and x = 4, the points of inflection are (2, f(2)) and (4,f(4)). 
f(2) = 2(2 - 4)³
= 2(-2)³ = 2(-8) = -16
f(4) = 4 (4 - 4)³ = 0
\(\therefore\) (2, -16) and (4, 0) are the points of inflection

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.