(Each question 1 marks)

Question 511 Mark

Find the limit: $\lim \limits_{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}\right]$

Answer

Evaluating the function at 2, it is of the form $\frac{0}{0}$

Therefore, we have,

$\lim _{x \rightarrow 2} \frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{x(x-2)^{2}}{(x+2)(x-2)}$
$=\lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)}$ as $x\neq$ 2
$=\frac{2(2-2)}{2+2}=\frac{0}{4}=0$

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Question 521 Mark

Compute derivative of $g(x)=\cot x$

Answer

By definition, $g(x)=\cot x=\frac{\cos x}{\sin x}$
We use the quotient rule on this function wherever it is defined.
$\frac{d }{d x} g(x)=\frac{d}{d x}(\cot x)=\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right)$
$=\frac{(\cos x)^{\prime}(\sin x)-(\cos x)(\sin x)^{\prime}}{(\sin x)^{2}}$
$=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{(\sin x)^{2}}$
$=-\frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x}=-cosec ^{2} x$

This is the required derivative.

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Question 531 Mark

Compute derivative of $f(x)=\sin 2 x$

Answer

Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus
$\frac{d }{d x}f(x)=\frac{d}{d x}(2 \sin x \cos x)=2 \frac{d}{d x}(\sin x \cos x)$
$=2\left[(\sin x)^{\prime} \cos x+\sin x(\cos x)^{\prime}\right]$
$=2[(\cos x) \cos x+\sin x(-\sin x)]$
$=2\left(\cos ^{2} x-\sin ^{2} x\right)$
= 2cos 2x

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Question 541 Mark

Find the limit: $\lim \limits_{x \rightarrow 1}\left[\frac{x^{2}+1}{x+100}\right]$

Answer

$\lim \limits_{x \rightarrow 1} \frac{x^{2}+1}{x+100}=\frac{1^{2}+1}{1+100}=\frac{2}{101}$

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Question 551 Mark

Find the derivative of f(x) from the first principle, where f(x) is x sin x.

Answer

We have,$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$=\lim _{h \rightarrow 0} \frac{(x+h) \sin (x+h)-x \sin x}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h)(\sin x \cos h+\sin h \cos x)-x \sin x}{h}$
$=\lim _{h \rightarrow 0} \frac{x \sin x(\cos h-1)+x \cos x \sin h+h(\sin x \cos h+\sin h \cos x)}{h}$
$=\lim _{h \rightarrow 0} \frac{x \sin x(\cos h-1)}{h}+\lim _{h \rightarrow 0} x \cos x \frac{\sin h}{h}$$+\lim _{h \rightarrow 0}(\sin x \cos h+\sin h \cos x)$
$=x \cos x+\sin x$

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Question 561 Mark

Find the derivative of f(x) from the first principle, where f(x) is sin x + cos x.

Answer

$f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (x+h)+\cos (x+h)-\sin x-\cos x}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin x \cos h+\cos x \sin h+\cos x \cos h-\sin x \sin h-\sin x-\cos x}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h}(\cos x-\sin x)+\lim _{h \rightarrow 0} \sin x \frac{(\cos h-1)}{h}$ $+\lim _{h \rightarrow 0} \cos x \frac{(\cos h-1)}{h}$
$=\cos x-\sin x$

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Question 571 Mark

Find the derivative of f from the first principle, where f is given by $f(x)=x+\frac{1}{x}$

Answer

The function is not defined at x = 0. But, we have
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $=\lim _{h \rightarrow 0} \frac{\left(x+h+\frac{1}{x+h}\right)-\left(x+\frac{1}{x}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{1}{h}\left[h+\frac{1}{x+h}-\frac{1}{x}\right]$
$=\lim _{h \rightarrow 0} \frac{1}{h}\left[h+\frac{x-x-h}{x(x+h)}\right]=\lim _{h \rightarrow 0} \frac{1}{h}\left[h\left(1-\frac{1}{x(x+h)}\right)\right]$
$=\lim _{h \rightarrow 0}\left[1-\frac{1}{x(x+h)}\right]=1-\frac{1}{x^{2}}$

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Question 581 Mark

Find the derivative of f from the first principle, where f is given by $f(x)=\frac{2 x+3}{x-2}$

Answer

Note that function is not defined at x = 2. But, we have
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $=\lim _{h \rightarrow 0} \frac{\frac{2(x+h)+3}{x+h-2}-\frac{2 x+3}{x-2}}{h}$
$=\lim _{h \rightarrow 0} \frac{(2 x+2 h+3)(x-2)-(2 x+3)(x+h-2)}{h(x-2)(x+h-2)}$
$=\lim _{h \rightarrow 0} \frac{(2 x+3)(x-2)+2 h(x-2)-(2 x+3)(x-2)-h(2 x+3)}{h(x-2)(x+h-2)}$
$=\lim _{h \rightarrow 0} \frac{-7}{(x-2)(x+h-2)}=-\frac{7}{(x-2)^{2}}$

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Question 591 Mark

Compute the derivative of $f(x) = sin^2 x$.

Answer

We have,
$f(x) = sin^2x$
$\therefore \quad \frac { d f ( x ) } { d x } = \frac { d } { d x } ( \sin x \sin x )$
= (sin x)$ \frac { d } { d x } $sinx + sin x$ \frac { d } { d x } $(sinx) [using Leibnitz rule]
= (cos x)sinx + sinx(cos x)
= 2sin x cosx = sin 2x

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Question 601 Mark

Compute the derivative of tan x.

Answer

Let f(x) = tan x.
Therefore,we have,
$\frac{d f(x)}{d x}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$=\lim _{h \rightarrow 0} \frac{\tan (x+h)-\tan (x)}{h}$
$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right]$
$=\lim _{h \rightarrow 0}\left[\frac{\sin (x+h) \cos x-\cos (x+h) \sin x}{h \cos (x+h) \cos x}\right]$
$\left.=\lim _{h \rightarrow 0} \frac{\sin (x+h-x)}{h \cos (x+h) \cos x} \text { (using formula for } \sin (\mathrm{A}+\mathrm{B})\right)$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \frac{1}{\cos (x+h) \cos x}$
$=1 \cdot \frac{1}{\cos ^{2} x}=\sec ^{2} x$

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Question 611 Mark

Compute the derivative of sin x.

Answer

Let f(x) = sin x
By using the first principle of derivative, we get
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin (x + h) - \sin x}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {\frac{{x + h + x}}{2}} \right) \cdot \sin \left( {\frac{{x + h - x}}{2}} \right)}}{h}$
$\left[ {\because \sin C - \sin D = 2\cos \left( {\frac{{C + D}}{2}} \right) \times \sin \left( {\frac{{C - D}}{2}} \right)} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {x + \frac{h}{2}} \right) \cdot \sin \frac{h}{2}}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \cos \left( {x + \frac{h}{2}} \right) \cdot \mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \frac{h}{2}}}{{\frac{h}{2}}}\left[ {\begin{array}{*{20}{c}} {{\text{ as }}h \to 0,{\text{ then }}} \\ {\frac{h}{2} \to 0} \end{array}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \cos \left( {x + \frac{h}{2}} \right) \times 1\quad \left[ {\because \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1} \right]$
= cos(x + 0) = cos x [putting h = 0]
$\therefore \frac { d } { d x } ( \sin x ) $= cos x

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Question 621 Mark

Find the derivative of f(x) = $\frac{x+1}{x}$

Answer

The given function is defined everywhere except at x = 0. We use the quotient rule with u = x + 1 and v = x. Hence u′ = 1 and v′ = 1.
Therefore, we have,
$\frac{d f(x)}{d x}=\frac{d}{d x}\left(\frac{x+1}{x}\right)=\frac{d}{d x}\left(\frac{u}{v}\right)$ $=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}=\frac{1(x)-(x+1) 1}{x^{2}}=-\frac{1}{x^{2}}$

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Question 631 Mark

Find the derivative of $f(x) = 1 + x + x^2 + x^3 + ... + x^{50}$ at $x = 1$.

Answer

Given, $f(a)=1+x+x^2+x^3+\ldots+x^{50}$
On differentiating both sides w.r.t. $x$, we get
$\mathrm{f}^{\prime}(\mathrm{x})=0+1+2 \mathrm{x}+3 \mathrm{x}^2+\ldots+50 \mathrm{x}^{49}$
$\text { At } x=1$,
$\mathrm{f}^{\prime}(1)=1+2(1)+3(1)^2+\ldots+50(1)^{49}$
$=1+2+3+\ldots+50$
$= \frac { ( 50 ) ( 5 1) } { 2 } = 1275$$\left[ \because \sum _ { n } = \frac { n ( n + 1 ) } { 2 } \right]$

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Question 641 Mark

Compute the derivative of $6x^{100} – x^{55} + x$.

Answer

Using direct application of the theorem, the derivative of the function is $600 x^{99}-55 x^{54}+1$.

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Question 651 Mark

Find the limit: $\lim \limits_{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]$

Answer

$\lim \limits_{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]$ $=1+(-1)+(-1)^{2}+\ldots+(-1)^{10}$

= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 = 1

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Question 661 Mark

Find the derivative of $f ( x ) = \frac { 1 } { x }$

Answer

We have, $f ( x ) = \frac { 1 } { x }$
By using the first principle,
$f ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { f ( x + h) - f ( x ) } { h }$
$\therefore \quad {f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{x + h}} - \frac{1}{x}}}{h}$ $\left[ {\begin{array}{*{20}{c}} {\because f(x) = \frac{1}{x}} \\ {\therefore f(x + h) = \frac{1}{{x + h}}} \end{array}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{x - (x + h)}}{{x(x + h)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - h}}{{x(x + h)}}} \right]$
$ = \mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - 1}}{{x(x + h)}}} \right] = \frac{{ - 1}}{{{x^2}}}$

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Question 671 Mark

Find the limit: $\lim \limits_{x \rightarrow 3}[x(x+1)]$

Answer

We have, $\lim \limits_{x \rightarrow 3}[x(x+1)]$ = 3 (3+1) = 3 (4) = 12

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Question 681 Mark

Find the derivative of the constant function f (x) = a for a fixed real number a.

Answer

f ′(x) = $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{a-a}{h}=\lim _{h \rightarrow 0} \frac{0}{h}=0 \text { as } h \neq 0$

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Question 691 Mark

Find the limit: $\lim \limits_{x \rightarrow 1}\left[x^{3}-x^{2}+1\right]$

Answer

We have, $\lim \limits _{x \rightarrow 1}\left[x^{3}-x^{2}+1\right] = 1^3 – 1^2 + 1 = 1$

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Question 701 Mark

Find the derivative of $f(x) = x^2$.

Answer

f ′(x) = $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{2}-(x)^{2}}{h}=\lim _{h \rightarrow 0}(h+2 x)=2 x$

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MATHS (Each question 1 marks)