If $$3\sin(xy)+4\cos(xy)=5$$, then $$\dfrac{dy}{dx}=$$

$$\dfrac{3\sin(xy)+4\cos(xy)}{3\sin(xy)-4\cos(xy)}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 17

The extrme value of

$(x)^{\dfrac {1}{x}}$ is

0%
$e$
0%
$\left(\dfrac {1}{e}\right)^{e}$
0%
$(e)^{\dfrac {1}{e}}$
0%
$1$

$y^2$ =

$ax^2$ +

$bx$ +

$c$ , then

$y^3\dfrac{d^2y}{dx^2}$ is

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a constant
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a function of x only
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a function of y only
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a function of x and y

Explanation

${y}^{2}=a{x}^{2}+bx+c$

$\Rightarrow 2y\dfrac{dy}{dx}=\dfrac{d}{dx}\left(a{x}^{2}\right)+\dfrac{d}{dx}\left(bx\right)+\dfrac{d}{dx}\left(c\right)$

$\Rightarrow 2y\dfrac{dy}{dx}=2ax+b+0$

$\Rightarrow 2\dfrac{d}{dx}\left(y\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(2ax+b\right)$

$\Rightarrow 2{\left(\dfrac{dy}{dx}\right)}^{2}+2y\dfrac{{d}^{2}y}{d{x}^{2}}=2a$

$\Rightarrow {\left(\dfrac{dy}{dx}\right)}^{2}+y\dfrac{{d}^{2}y}{d{x}^{2}}=a$

$\Rightarrow y\dfrac{{d}^{2}y}{d{x}^{2}}=a-{\left(\dfrac{dy}{dx}\right)}^{2}$

$\Rightarrow y\dfrac{{d}^{2}y}{d{x}^{2}}=a-{\left(2ax+b\right)}^{2}$

$\Rightarrow {y}^{3}\dfrac{{d}^{2}y}{d{x}^{2}}={y}^{2}\left(a-{\left(2ax+b\right)}^{2}\right)$

$=\left(a{x}^{2}+bx+c\right)\left(a-4{a}^{2}{x}^{2}-{b}^{2}-4axb\right)$ is a function of

$x$ only

$3\sin(xy)+4\cos(xy)=5$ , then

$\dfrac{dy}{dx}=$

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$\dfrac{3\sin(xy)+4\cos(xy)}{3\sin(xy)-4\cos(xy)}$
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$\dfrac{y}{x}$
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$-\dfrac{y}{x}$
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$none\ of\ these$

Explanation

$3\dfrac{d}{dx}\left(\sin{\left(xy\right)}\right)+4\dfrac{d}{dx}\left(\cos{\left(xy\right)}\right)=\dfrac{d}{dx}\left(5\right)$

$\Rightarrow 3\cos{\left(xy\right)}\dfrac{d}{dx}\left(xy\right)-4\sin{\left(xy\right)}\dfrac{d}{dx}\left(xy\right)=0$

$\Rightarrow \left(3\cos{\left(xy\right)}-4\sin{\left(xy\right)}\right)\dfrac{d}{dx}\left(xy\right)=0$

$\Rightarrow \left(3\cos{\left(xy\right)}-4\sin{\left(xy\right)}\right)\left(x\dfrac{dy}{dx}+y\right)=0$

$\Rightarrow 3\cos{\left(xy\right)}-4\sin{\left(xy\right)}=0$ and

$x\dfrac{dy}{dx}+y=0$

$\Rightarrow 3\cos{\left(xy\right)}=4\sin{\left(xy\right)}$ and

$x\dfrac{dy}{dx}=-y$

$\Rightarrow \tan{xy}=\dfrac{3}{4}$ and

$\dfrac{dy}{dx}=\dfrac{-y}{x}$

$\therefore \dfrac{dy}{dx}=\dfrac{-y}{x}$

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$

The value of

$g(0)$ is

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0
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-3
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2
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None of these

Explanation

$f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2)$ and

$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$

$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$

$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$

$\quad f^{\prime \prime}(x)=2$

$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$

$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$

$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$

$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$

From (1),(2) and (3) we have

$a=2(3+a+b)+a$

$\text { and } b=2(3+a+b)$

$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$

Hence,

$b=0$ and

$a=-3 .$ So,

$f(x)=x^{2}-3 x$ and

$g(x)=-3 x+2$

$g(x)=-3 x+2 \Rightarrow g(0)=2$

Equation

$x^{n}-1=0, n>1, n \in N,$ has roots

$1, a_{1}, a_{2}, \dots, a_{n-1}$
The value of

$\sum_{r=1}^{n-1} \dfrac{1}{2-a_{r}}$ is

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$\dfrac{2^{n-1}(n-2)+1}{2^{n}-1}$
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$\dfrac{2^{n}(n-2)+1}{2^{n}-1}$
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$\dfrac{2^{n-1}(n-1)-1}{2^n-1}$
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None of these

Explanation

From

$(1), \log \left(x^{n}-1\right)=\log (x-1)+\log \left(x-a_{1}\right)+\cdots +\log \left(x-a_{n-1}\right)$

Differentiating w.r.t.

$x$ , we get

$\dfrac{n x^{n-1}}{x^{n}-1}=\dfrac{1}{x-1}+\dfrac{1}{x-a_{1}}+\dfrac{1}{x-a_{2}}+\dots+\dfrac{1}{x-a_{n-1}}$

Putting

$x=2$ in

$above \ equation,$ we get

$\dfrac{n 2^{n-1}}{2^{n}-1}=1+\dfrac{1}{2-a_{1}}+\dfrac{1}{2-a_{2}}+\cdots+\dfrac{1}{2-a_{n-1}}$

$\Rightarrow \dfrac{1}{2-a_{1}}+\dfrac{1}{2-a_{2}}+\cdots+\dfrac{1}{2-a_{n-1}}=\dfrac{n 2^{n-1}}{2^{n}-1}-1$

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

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

Equation

$x^{n}-1=0, n>1, n \in N,$ has roots

$1, a_{1}, a_{2}, \dots, a_{n-1}$
The value of

$\sum_{r=1}^{n-1} \dfrac{1}{1-a_{r}}$ is

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$\dfrac{n}{4}$
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$\dfrac{n(n-1)}{2}$
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$\dfrac{n-1}{2}$
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None of these

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 17 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 17 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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