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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 17 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Differentiation
Quiz 17
The extrme value of $$(x)^{\dfrac {1}{x}}$$ is
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$$e$$
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$$\left(\dfrac {1}{e}\right)^{e}$$
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$$(e)^{\dfrac {1}{e}}$$
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$$1$$
If $$y^2$$ = $$ax^2$$ + $$bx$$ + $$c$$ , 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
If $$3\sin(xy)+4\cos(xy)=5$$, 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
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