MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 7

Differentiate the following w.r.t.

$x$ .

$\tan^{3}x$ .

0%
$(3tan^2x)$
0%
$(3tan^2x)(sec^2\,x)$
0%
$(tan^2x)(sec^2\,x)$
0%
$(tan^2x)(sin^2\,x)$

Explanation

$\cfrac { d }{ dx } \tan ^{ 3 }{ x }$

$=3\tan ^{ 2 }{ x } .\cfrac { d }{ dx } \tan { x } .\cfrac { d }{ dx } \left( x \right)$

$=3\tan ^{ 2 }{ x }\sec ^{ 2 }{ x }$

Differentiate the following w.r.t.

$x$ .

$\tan x^{2}$ .

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$sec^2(x^2)$
0%
$sec^2(x^2)2x$
0%
$sec^2(x^3)2x$
0%
$tan^2(x^2)2x$

Explanation

$\cfrac { d }{ dx } \tan { \left( { x }^{ 2 } \right) }$

$=\sec ^{ 2 }{ \left( { x }^{ 2 } \right) } .\cfrac { d }{ dx } \left( { x }^{ 2 } \right)$

$=\sec ^{ 2 }{ \left( { x }^{ 2 } \right) } .2x$

Differentiate the following w.r.t.

$x$ .

$\sin^{2} \sqrt {x}$ .

0%
$\dfrac{1}{2\sqrt{x}}\sin(3\sqrt{x})$ .
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$\dfrac{1}{\sqrt{x}}\sin(2\sqrt{x})$ .
0%
$\dfrac{1}{2\sqrt{x}}\sin(2\sqrt{x})$ .
0%
$\dfrac{1}{2\sqrt{x}}\sin(4\sqrt{x})$ .

Explanation

$\cfrac { d }{ dx } \sin ^{ 2 }{ \sqrt { x } }$

$=2\sin { \left( \sqrt { x } \right) } .\cfrac { d }{ dx } \left( \sin { \sqrt { x } } \right) .\cfrac { d }{ dx } \sqrt { x }$

$=2\sin { \sqrt { x } } .\cos { \sqrt { x } } .\cfrac { 1 }{ 2\sqrt { x } }$

$=\sin { \left( 2\sqrt { x } \right) } .\cfrac { 1 }{ 2\sqrt { x } } \quad \left[ \because \sin { 2\theta } =2\sin { \theta } \cos { \theta } \right]$

$\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x } \right) }^{ x }+{ \left( \cos { x } \right) }^{ \frac { 1 }{ \ln { x } } }+{ \left( x\sin { x } \right) }^{ x } \right) }$ is equal to

0%
$2$
0%
$2+e$
0%
$2+\dfrac { 1 }{ e }$
0%
$3$

Explanation

$\lim_{x\rightarrow 0^+}(x\cos x)^x$

$\lim_{x\rightarrow 0^+} x^x.\cos ^x x$

$\lim_{x\rightarrow 0^+} e^{x\ln x}.\cos ^x x$

$1\times 1$

$1$

$\therefore \lim_{x\rightarrow 0^+}(x\cos x)^x=1$

------------------------------------------------------

$\lim_{x\rightarrow 0^+}(x\sin x)^x$

$\lim_{x\rightarrow 0^+} x^x.\sin ^x x$

$\lim_{x\rightarrow 0^+} e^{x\ln x}.\sin ^x x$

$1\times 1$

$1$

$\therefore \lim_{x\rightarrow 0^+}(x\sin x)^x=1$

---------------------------------------------------------

$\lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}$

$\lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$

$y= \lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$

$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\ln x}(-\ln (\sin x))$

$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\frac{1}{x}}\left ( -\frac{\cos x}{\sin x} \right )$

$\ln y =\lim_{x\rightarrow 0^+}(-\cos x)$

$\ln y = -1$

$y=e^{-1}$

$y=\dfrac{1}{e}$

$\therefore \lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}=\dfrac{1}{e}$

------------------------------------------------------------------------------------

$\lim_{x\rightarrow 0^+} ((x\cos x)^x+(\csc x)^{\frac{1}{\ln x}}+(x\sin x)^x)$

$=1+\dfrac{1}{e}+1$

$=2+\dfrac{1}{e}$

$8f(x) + 6f\left( {{1 \over x}} \right) = x + 5$ and

$y = {x^2}f(x)$ ,then

${{dy} \over {dx}}$ at x=-1 is equal to

0%
0
0%
${1 \over {14}}$
0%
$- {1 \over {14}}$
0%
None of these

Explanation

$8{f(x)}+6{f\bigg(\dfrac{1}{x}\bigg)}=x+5\implies f(-1)=\dfrac{2}{7}$

Differentiating on both sides

$8{f'(x)}-\dfrac{6}{x^{2}}f'\bigg(\dfrac{1}{x}\bigg)=1$

Put

$x=-1\implies f'(-1)=\dfrac{1}{2}$

$y=x^{2}f(x)$

$\dfrac{d{y}}{d{x}}=2{x}f(x)+x^{2}f'(x)$

$x=-1,\dfrac{d{y}}{d{x}}=2(-1)\times \dfrac{2}{7}+(-1)^{2}\dfrac{1}{2}=-\dfrac{1}{14}$

$f:\mathrm{R}\to\mathrm{R}$ is a differentiable function such that

$f'(x)\gt 2f(x)\forall x\in \mathrm{R}$ and

$f(0)=1$ , then

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$f(x)$ is increasing in $(0,\infty)$
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$f(x)$ is decreasing in $(0,\infty)$
0%
$f(x)\gt e^{2x}$ in $(0,\infty)$
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$f(x)\lt e^{2x}$ in $(0,\infty)$

Explanation

Given that,

$f'\left( x \right) >2f\left( x \right) ,f\left( 0 \right) =1$

$\int { \cfrac { df\left( x \right) }{ f\left( x \right) } } >\int { 2dx }$

$\ln { f\left( x \right) } >2x+c$

$f\left( x \right) >k{ e }^{ 2x },\quad k>0$

$f\left( 0 \right) =1$

$1>k\quad \therefore 0<k<1$

$\therefore f\left( x \right) >k{ e }^{ 2x }>0$

$f'\left( x \right) >2f\left( x \right) >0$

$\therefore f\left( x \right)$ is increasing function (since

$f'\left( x \right) >0$ ) in

$\left( 0,\infty \right)$

$\frac{d^n}{dx^n}[log(ax+b)]$ is equal to:

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$\frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$
0%
$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^{n+1}}$
0%
$\frac{(-1)^{n+1} ({n+1)}! a^{n-1}}{(ax+b)^{n+1}}$
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$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^n}$

Explanation

Let

$y=\log { \left( ax+b \right) }$ , let

$D=\cfrac { d }{ dx }$

$\cfrac { dy }{ dx } =Dy=\cfrac { a }{ ax+b }$

${ D }^{ 2 }y=\cfrac { -{ a }^{ 2 } }{ { \left( ax+b \right) }^{ 2 } } ,{ D }^{ 3 }y=\cfrac { 2{ a }^{ 3 } }{ { \left( ax+b \right) }^{ 3 } }$

${ D }^{ 4 }y=\cfrac { -3.2.1.{ a }^{ 4 } }{ { \left( ax+b \right) }^{ 4 } }$

$\therefore { D }^{ n }y=\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } }$

$\therefore \cfrac { { d }^{ n } }{ d{ x }^{ n } } \left( \log { \left( ax+b \right) } \right) =\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } }$

$\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}}$ is equal to

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$e^1$
0%
$e^{-2}$
0%
$e^2$
0%
$e^{-1}$

Function f(x)=

$\left| {x - 2} \right| - 2\left| {x - 4} \right|\,is$ discontinous at:

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$x=2,4$
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$x=2$
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No where
0%
Except $x=2$

Explanation

$f(x)=\mid x-2 \mid -2\mid x-4 \mid$

$\therefore f\left(x \right)=\begin{Bmatrix} -(x-2)+2(x-4),\qquad x<2\\ (x-2)+2(x-4), \qquad 2\le x \le 4\\ (x-2)-2(x-4) \qquad x\gt 4 \\ \end{Bmatrix}$

$x=2$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 2^{-}}=-(2-2)+2(2-4) = -4$

R.H.LRight Hand Limit/0

$\displaystyle\lim_{x\to\ 2^{+}}=(2-2)+2(2-4)=-4$

$\therefore \displaystyle\lim_{x\to\ 2^{-}}f(x) = \displaystyle\lim_{x\to\ 2^{+}}f(x)$

$x=4$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 4^{-}}=(4-2)+2(4-4)=2$

R.H.L(Right Hand Limit)

$\displaystyle\lim_{x\to\ 4^{+}}=(4-2)-2(4-4)=2$

$\therefore \displaystyle\lim_{x\to\ 4^{-}}f(x) = \displaystyle\lim_{x\to\ 4^{+}}f(x)$

$\therefore \ f(x)$ is continuous at everywhere.

Let

${P_n} = \prod\limits_{k = 2}^n {\left( {1 - {1 \over {{}^{{}^{k + 1}}{C_2}}}} \right)} .$ If

$\mathop {\lim }\limits_{x \to \infty } {P_n}$ can be expressed as lowest rational in the form

$\dfrac { a }{ b }$ , then value of

$(a+b)$ is __________.

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4
0%
8
0%
10
0%
12

Given

$y=\dfrac {3}{x}, \dfrac {dy}{dx}=$

0%
$3$
0%
$\dfrac {3}{x^{2}}$
0%
$\dfrac {-3}{x^{2}}$
0%
$3x$

Explanation

$y=\dfrac{3}{x}$

differentiating with respect to x

$\dfrac{dy}{dx}=\dfrac{-3}{x^2}$

Statement I: The function f(x) in the figure is differentiable at x = a
Statement II: The function f(x) continuous at x = a

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Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I.
0%
Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.
0%
Statement I is true but Statement II is false
0%
Statement I is false but Statement II is true.

Explanation

$f(x)$ is continuous at as it can be observed from the figure there is no discontinuous at

$x=a$ but coming to differentiability

$f(x)$ is not differentiable at

$x=a$ since the curve is sharp[i.e the left slope and right slope are not equal]

therefore the answer is statement I is false but statement II is true

$\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x =$

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$\infty$
0%
$- \infty$
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$0$
0%
does not exist

Explanation

$L=\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x$

$\Rightarrow$

$L=\tan\dfrac{\pi}{2}$ [ Applying limit ]

We know that value of

$\tan\dfrac{\pi}{2}$ is does not exist.

$\therefore$

$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x=$ does not exist

Suppose the function

$f(x)-f(2x)$ has the derivative

$5$ at

$x=1$ and derivative

$7$ at

$x=2$ . The derivative of the function

$f(x)-f(4x)$ at

$x=1$ , has the value equal to?

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$19$
0%
$9$
0%
$17$
0%
$14$

$\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x } \right) }{ \left( \tan ^{ 2 }{ 2x } \right) = } }$

0%
$1$
0%
$2$
0%
$\dfrac {1}{2}$
0%
$Does\ not\ exist$

Explanation

$\mathop {\lim }\limits_{x \to 0} {\log _{{{\tan }^2}x}}\left( {{{\tan }^2}2x} \right)$

Let,

$\begin{array}{l} y={ \log _{ { { \tan }^{ 2 } }x } }{ \tan ^{ 2 } }2x \\ =\log { \tan ^{ 2 } } 2x={ \left( { { { \tan }^{ 2 } }x } \right) ^{ y } } \\ =2\log \tan 2x=2y\log \tan x \\ \Rightarrow y=\dfrac { { \log \tan 2x } }{ { \log \tan x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \log \tan 2x } }{ { \log \tan x } } \end{array}$

Using L-Hospital Rule

$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \dfrac { 1 }{ { \tan 2x } } \times { { \sec }^{ 2 } }2x\times 2 } }{ { \dfrac { 1 }{ { \tan x } } { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2{ { \sec }^{ 2 } }2x } }{ { \tan 2x } } \times \dfrac { { \tan x } }{ { { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\tan x } }{ { \tan 2x } } \, \, \, \, \, \, \left( { { \lim }_{ x\to 0 }\sec x=4 } \right) \end{array}$

$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\dfrac { { \tan x } }{ x } \times x } }{ { \dfrac { { \tan 2x } }{ { 2x } } \times 2x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\, \, \, 1\, \, \left( { { \lim }_{ x\to 0 }\dfrac { { \tan x } }{ x } =1 } \right) \\ \Rightarrow { \lim }_{ x\to 0 }{ \log _{ { { \tan }^{ 2 } }x } }\left( { { { \tan }^{ 2 } }2x } \right) =1 \end{array}$

$\displaystyle \lim_{n \rightarrow \infty} {^{n}C_{c}}\left(\dfrac {m}{n}\right)^{x}\left(1-\dfrac {m}{n}\right)^{n-x}$ equal to

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$\dfrac {M^{x}}{x\ !}.e^{-m}$
0%
$\dfrac {M^{x}}{x\ !}.e^{m}$
0%
$0$
0%
$\dfrac {m^{x+1}}{me^{m}x\ !}$

Explanation

Given

$\displaystyle\lim_{n\rightarrow \infty}{^{n}}C_x \bigg(\dfrac{m}{n}\bigg)^{x}\bigg(1-\dfrac{m}{n}\bigg)^{n-x}=$ coefficient of

$x$ in

$\displaystyle\lim_{n\rightarrow\infty}\bigg(\dfrac{m}{n}+1-\dfrac{m}{n}\bigg)^{n}$

$=$ coeeficient of

$x$ in

$\displaystyle\lim_{n\rightarrow\infty}\bigg(1\bigg)^{n}=$ coefficient of

$x$ in

$1=0$

If [.] denotes, GIF , then

$\underset{x \rightarrow 0}{lt} \left( \left[\dfrac{2018 sin^{-1} x}{x}\right] + \left[\dfrac{2020x}{tan^{-1} x}\right]\right)$ =

0%
4038
0%
4037
0%
4036
0%
4039

The shortest distance between line

$y-x=1$ and curve

$x={y}^{2}$ is

0%
$\dfrac {\sqrt {3}}{4}$
0%
$\dfrac {3\sqrt {2}}{8}$
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$\dfrac {8}{2\sqrt {2}}$
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$\dfrac {4}{\sqrt {3}}$

Explanation

Line is

$y-x=1$ and curve

$x=y^2$

$y^2=x$

$2y\dfrac{dy}{dx}=1$

$\dfrac{dy}{dx}=\dfrac{1}{2y}=1$

$\therefore$

$y=\dfrac{1}{2}$

$x=\dfrac{1}{4}$

Tangent at

$\left(\dfrac{1}{4},\,\dfrac{1}{2}\right)$

$\dfrac{1}{2}y=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)$

$y=x+\dfrac{1}{4}$

Distance

$=\left|\dfrac{1-\dfrac{1}{4}}{\sqrt{2}}\right|=\dfrac{3}{4\sqrt{2}}=\dfrac{3\sqrt{2}}{8}$

${ x }_{ n }={ \left( 1-\cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( 1-\cfrac { 1 }{ 6 } \right) }^{ 2 }{ \left( 1-\cfrac { 1 }{ 10 } \right) }^{ 2 }...{ \left( 1-\cfrac { 1 }{ \cfrac { n(n+1) }{ 2 } } \right) }^{ 2 },n\ge 2$ . Then the value of

$\displaystyle\lim _{ n\rightarrow \infty }{ { x }_{ n } }$

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$1/3$
0%
$1/9$
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$1/81$
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$0$ (zero)

The difference of slopes of lines represent by

${y^2} - 2xy{\sec ^2}\alpha + \left( {3 + {{\tan }^2}\alpha } \right)\left( {{{\tan }^2}\alpha - 1} \right){x^2} = 0$ is

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$3$
0%
$4$
0%
$0$
0%
$2$

Explanation

$y^2-2xy\sec ^2\alpha +(3+\tan^2\alpha)(\tan ^2\alpha -1)x^2=0$

Divide by

$x^2$ & substitute

$m=y/x$

$m^2-2m\sec ^2 \alpha+(3+\tan^2\alpha)(\tan^2\alpha -1)=0$ m_1

Let

$m_1$ &

$m_2$ be note of equation

$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2\left[\begin{matrix}m_1+m_2=-b/a\\m_1m_2=c/a \end{matrix}\right.$

$(m_1-m_2)^2(-2\sec\alpha)^2-4(3\tan ^2\alpha -3+\tan ^4 \alpha -\tan ^2\alpha)$

$=4\sec ^4\alpha -4(\tan ^4\alpha +2 \tan ^2 \alpha -3)$

$=4(\sec ^4 \alpha -\tan ^4\alpha )-8\tan ^2\alpha +12$

$=\dfrac{(\sec^2\alpha -\tan ^2\alpha)}{1}(\sec ^2\alpha +\tan ^2\alpha )-8\tan ^2\alpha +12$

$4\sec^2\alpha +4\tan ^2\alpha -4\tan ^2\alpha -4\tan ^2\alpha +12$

$4(\sec ^2\alpha -\tan ^2\alpha)+12$

$4(1)+12$

$(m_1-m_2)^2=16$

$m_1-m_2=4$

$\lim _{ x\rightarrow 1 }{ \dfrac { \sqrt { 1-\cos { 2\left( x-1 \right) } } }{ x-1 } }$

0%
Exists and it equals $\sqrt {2}$
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Exists and it equals $-\sqrt {2}$
0%
Does not exist because $x-1\rightarrow 0$
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Does not exist because left hand limit is not equal to right hand limit

$f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$ , then

$\underset{a \rightarrow 0}{\lim} \dfrac{f(1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$ is

0%
$-\dfrac{53}{3}$
0%
$\dfrac{53}{3}$
0%
$-\dfrac{55}{3}$
0%
$\dfrac{55}{3}$

Explanation

$\lim_{\alpha \rightarrow 0}\dfrac{f(1-\alpha )-f(1)}{\alpha ^3+3\alpha }$

Apply L-Hospital's rule

$=\lim_{\alpha \rightarrow 0}\dfrac{f'(1-\alpha )-0}{3\alpha ^2+3 }$

$=\dfrac{f'(1 )}{3 }$

Considering, f(x)

$f(x )=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$

$f'(x )=30x^{9}-56x^7+30x^5-63x^2+6x$

Calculating required limit value:

$=\dfrac{f'(1 )}{3 }$

$=\dfrac{30(1)^{9}-56(1)^7+30(1)^5-63(1)^2+6(1)}{3 }$

$=\dfrac{-53}{3}$

$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}$ is equal to

0%
$1$
0%
$1/2$
0%
$1/3$
0%
$0$

Explanation

Solution

$\displaystyle \lim_{x\rightarrow \infty } \dfrac{1^2 + 2^2 +--- n^2}{n^3}$

we know that

$1^2 + 2^2 + --- n^2 = \dfrac{n (n+1)(2n+1)}{6}$

$\displaystyle =\lim_{x\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^3}$

$\displaystyle= \lim_{x\rightarrow \infty }\frac{(n+1)(2n+1)}{6n^2} = \lim_{x\rightarrow \infty } \frac{2n^2 +3n +1}{6n^2}$

Apply L-Hospital rule twice.

$\displaystyle = \lim_{x\rightarrow \infty } \dfrac{4n+3}{12n}$

$= \dfrac{4}{12} = \dfrac{1}{3}$

C is correct.

1063331_1116651_ans_f9e176fd269b4ce5ac81647f9adf3277.jpg

$y=|\cos x|+|\sin x|$ , then

$\dfrac {dy}{dx}$ at

$x=\dfrac {2\pi}{3}$ is

0%
$\dfrac {1-\sqrt {3}}{2}$
0%
$0$
0%
$\dfrac {\sqrt {3}-1}{2}$
0%
$\dfrac {\sqrt {3}+1}{2}$

Explanation

$y = |cos\,x|+|sin\,x|$

$, x = \dfrac{2n}{3}$

$y = -cos\,x+sin\,x$

$\left \{ x\sum (\dfrac{\pi }{2},\pi ) \right \}$

$y^{1} = sinx+cosx$

$y^{1} = sin(\dfrac{2n}{3})+cos(\dfrac{2n}{3})$

$= \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} = \dfrac{\sqrt{3}-1}{2}$ Option C is correct

1142221_1143670_ans_1376f7fff69944e79e11d99ebfe127e2.jpg

The value of

$\underset { x\longrightarrow \infty }{ Lim } \dfrac{d}{dx}\overset { \sqrt { 3 } }{ \underset { -\sqrt { 3 } }{ \int } } \dfrac{r^3}{(r+1)(r-1)}$ dr,is

0%
$0$
0%
$1$
0%
$\dfrac{1}{2}$
0%
non existent

Explanation

Given,

$\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{(r+1)(r-1)}dr$

using long division process, we get,

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{r^2-1}dr$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}r+\dfrac{r}{r^2-1}dr$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\left [ \dfrac{r^2}{2}+\dfrac{1}{2}\ln \left|r^2-1\right| \right ]_{-\sqrt{3}}^{\sqrt{3}}$

$=\lim_{x\rightarrow \infty }\dfrac{d}{dx} (0)$

$=0$

Solve

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}$

0%
$-\dfrac{5}{3}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{7}{3}$
0%
None of these

Explanation

Let

$L=\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}$

$= \displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}\times \dfrac {3x}{\tan x}\times \dfrac {5}{3}$

we know that

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}=1$ and

$\displaystyle \lim_{x\rightarrow 0}\dfrac {3x}{\tan 3x}=1$

$L=1\times 1\times \dfrac {5}{3}$

$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}= \dfrac {5}{3}$

$a{x^2} + 2hxy + b{y^2} = 0$ then

$\frac{{dy}}{{dx}}$ is equal to

0%
$\frac{y}{x}$
0%
$\frac{x}{y}$
0%
$- \frac{x}{y}$
0%
none of these

$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} =$

0%
$\dfrac{-1}{2}$
0%
$\dfrac{1}{2}$
0%
$2$
0%
$-2$

Solve:

$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=$

0%
$\dfrac{4}{3}(\sqrt{2}+1)$
0%
$\dfrac{4}{3}(\sqrt{2}-1)$
0%
$\dfrac{3}{4}(\sqrt{2}-1)$
0%
$\dfrac{3}{4}(\sqrt{2}-2)$

Explanation

$\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}}$

$=\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}\times\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}$

$=\displaystyle\int_{0}^{1}{\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}}$

$=\displaystyle\int_{0}^{1}{\left(\sqrt{x+1}+\sqrt{x}\right)dx}$

$=\left[\dfrac{{\left(x+1\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+\dfrac{{\left(x\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$

$=\left[2\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+2\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$=2\left[\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$

$=\dfrac{2}{3}\left[\left({2}^{\frac{3}{2}}-1\right)-\left(1-0\right)\right]$

$=\dfrac{2}{3}\left[2\sqrt{2}-2\right]$

$=\dfrac{4}{3}\left(\sqrt{2}-1\right)$

$y=a\ \sin\ x+b\ \cos\ x$ , then

$y^{2}+\left ( \dfrac{dy}{dx} \right )^{2}$ is

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function of $x$
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function of $y$
0%
function of $x$ and $y$
0%
constant

Explanation

$y=a\sin{x}+b\cos{x}$

$\dfrac{dy}{dx}=a\cos{x}-b\sin{x}$

${y}^{2}+{\left(\dfrac{dy}{dx}\right)}^{2}={\left(a\sin{x}+b\cos{x}\right)}^{2}+{\left(a\cos{x}-b\sin{x}\right)}^{2}$

$={a}^{2}{\sin}^{2}{x}+{b}^{2}{\cos}^{2}x+2ab\sin{x}\cos{x}+{a}^{2}{\cos}^{2}{x}+{b}^{2}{\sin}^{2}x-2ab\sin{x}\cos{x}$

$={a}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{b}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)$

$={a}^{2}+{b}^{2}$ since

${\sin}^{2}{x}+{\cos}^{2}{x}=1$

$=constant$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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