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Limits and Derivatives Test - 29

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Limits and Derivatives Test - 29
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  • Question 1
    1 / -0
    If $$x = a\sin 2\theta (1 + \cos 2\theta), y = b\cos 2\theta (1 - \cos 2\theta)$$, then $$\dfrac {dy}{dx} =$$
  • Question 2
    1 / -0
    Evaluate: $$\underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } } $$ 
    Solution
    $$\underset{x\rightarrow 0}{lt}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$$

    $$=\underset{x\rightarrow 0}{lt}\dfrac{x\dfrac{2\tan x}{1-\tan^2x}-2x\tan x}{(1-1+2\sin^2x)^2}$$      (since $$\tan 2\theta =\dfrac{2\tan \theta}{1-\tan^2\theta}$$)

    $$=\underset{x\rightarrow 0}{lt}\dfrac{2x\tan x(1-1+\tan^2x)}{(1-\tan^2x)y\sin^4x}=\underset{x\rightarrow 0}{lt}\dfrac{x\tan^3x}{2\sin^4x}\underset{x\rightarrow 0}{lt}\dfrac{1}{1-\tan^2x}$$

    $$=\underset{x\rightarrow 0}{lt}\dfrac{\sin^3x/\cos^3x}{2\sin^3x(\sin x/x)}\times 1=\underset{x\rightarrow 0}{lt}\dfrac{1}{2\cos^3x}\underset{x\rightarrow 0}{lt}\dfrac{1}{\sin x/x}=\dfrac{1}{2}\times 1$$

    $$=\dfrac{1}{2}$$.
  • Question 3
    1 / -0
    $$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$$ equals:

    Solution

  • Question 4
    1 / -0
    The solution of the differential equation  $$\left( \dfrac { d y } { d x } \right) ^ { 2 } - 3 x \left( \dfrac { d y } { d x } \right) - 2 y = 8$$  is
    Solution
    By option Verification
    Put $$\dfrac{dy}{dx}=4x$$
    substitute this in the equation 
    $$\implies y=2x^2-4$$
  • Question 5
    1 / -0
    Evaluate: $$\underset { x\rightarrow { 0 } }{ lim } \dfrac { x\tan 2x-2x \tan\ x\quad  }{ (1-\cos2x)^{ 2 } } $$
    Solution
    $$\rightarrow \underset{x \rightarrow 0}{lt} \dfrac{x (\tan 2x - 2 \tan  x)}{(1 - \cos 2x)^2}$$

    $$= \underset{x \rightarrow 0}{lt} \dfrac{x\left[\dfrac{2 \tan x}{1 - \tan^2 x} - 2 \tan x\right]}{(2 \sin^2 x)^2}$$

    $$= \underset{x \rightarrow 0}{lt} \dfrac{2x \tan x \left[\dfrac{1}{1 - \tan^2 x} -1 \right]}{4 \sin^4 x}$$

    $$= \underset{x \rightarrow 0}{lt} \dfrac{\dfrac{x \tan x}{1 - \tan^2 x} [x - 1 + \tan^2 x]}{2 \sin^4x}$$

    $$= \underset{x \rightarrow 0}{lt} \dfrac{x \tan^3 x \left(\dfrac{\sin^3 x}{\cos^3 x} \right) / x^4}{(1 - \tan^2  x). 2 \sin^4 x/ xy}$$

    $$= \underset{x \rightarrow 0}{lt} \dfrac{\left(\dfrac{\sin x}{x} \right)^3 . \dfrac{1}{\cos^3x}}{2 \left(\dfrac{\sin x}{x} \right)^4 (1 - \tan^2 x)} = \dfrac{1.1}{2.1 (1 - 0)} = \dfrac{1}{2}$$

    Given $$\underset{x \rightarrow 0}{lt} \dfrac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} = \dfrac{1}{2}$$
  • Question 6
    1 / -0
    Let $$f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad  when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}$$. If the function $$f(x)$$ be continous at $$x=0$$, then $$k=$$
    Solution

    We have,

    $$ {{x}^{2}}+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when,\,x\ge 0\, $$

    $$ -{{x}^{2}}-k\,\,\,\,\,\,\,\,\,\,\,\,\,when,x\le 0 $$


    Given that,

    $$L.H.L=R.H.L=f\left( x \right)$$


    At point $$x=0$$

    $$ {{x}^{2}}+k=-{{x}^{2}}-k=0 $$

    $$ \Rightarrow {{x}^{2}}+k=0 $$

    $$ \Rightarrow k=-{{x}^{2}} $$

    $$ \Rightarrow k=0 $$


    Hence, this is the answer.

  • Question 7
    1 / -0
    $$\dfrac{d}{dx}\left[\dfrac{tan x - cot x}{tan x + cot x}\right]=$$
    Solution
    $$\dfrac{d}{dx}\left(\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}}{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}}\right)=\dfrac{d}{dx}\left(\dfrac{\sin^2x-\cos^2x}{\sin^2x+\cos^2x}\right)$$
    $$=\dfrac{d}{dx}(\sin^2x-\cos^2x)=2\sin x\cos x+2\cos x\sin x=2\sin 2x\Rightarrow (A)$$
  • Question 8
    1 / -0
    Evaluate: $$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$$
    Solution
    $$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$$

    $$\displaystyle\lim_{x\to 10}\dfrac{(x-10)(x+10)}{x-10}$$

    $$\displaystyle\lim_{x\to 10}{x+10}$$

    $$=10+10=20$$
  • Question 9
    1 / -0
    Solve:
    $$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} =$$
    Solution
    $$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} \Rightarrow $$
    L' Hospital - 
    $$\underset{x \rightarrow 2}{Lt}\frac{\sqrt{2}.x^{\sqrt{2}-1}}{1}$$
    $$ = \sqrt{2}\times (2)^{\sqrt{2}-1}$$
    $$ = \frac{\sqrt{2}\times 2^{\sqrt{2}}}{2}$$
    $$ = (2)^{\sqrt{2}-\frac{1}{2}}$$

  • Question 10
    1 / -0
    The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } $$ where $$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$$, is
    Solution
    $$\mathop {f\left( x \right)}\limits_{\lim \,\,x \to 0}  = \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}$$
    $$ = \frac{{ - \sin \left( {\sin x} \right)\cos x - \sin x}}{{4{x^3}}}$$
    $$ = \frac{{ - \left( {\sin x} \right){{\cos }^2}x + \sin \left( {\sin x} \right)\sin x - \cos x}}{{12{x^2}}}$$
    $$ = \frac{{\sin \left( {\sin x} \right){{\cos }^3}x + \cos \left( {\sin x} \right)\cos 2x + \cos \left( {\sin x} \right)\sin x + \sin \left( {\sin x} \right)\cos x + \sin x}}{{24x}}$$
    $$ = \frac{{1 + 1 + Q + 1 + 1}}{{24}} = \frac{1}{6}$$
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