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AMU MCA Previous Year Questions (PYQs)

AMU MCA Application Of Derivatives PYQ

AMU MCA PYQ
The kinetic energy of a body is twice its rest mass energy. What is the ratio of relativistic mass to rest mass?





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Given: $K = 2m_0c^2$

Total energy $E = K + m_0c^2 = 3m_0c^2$

$E=\gamma m_0 c^2$

$\gamma=3$

Relativistic mass $=\gamma m_0=3m_0$

Ratio $=\frac{m}{m_0}=3$

AMU MCA PYQ
The value of the integral $I=\iint_R e^{x^2+y^2},dy,dx$ where $R$ is the semicircular region bounded by the $x$-axis and the curve $y=\sqrt{1-x^2}$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
A particle is moving along the curve $x = at^2 + bt + c$. If $ac = b^2$, then the particle would be moving with uniform





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2019 PYQ

Solution

AMU MCA PYQ
The length $x$ of a rectangle is decreasing at the rate of $6$ cm/min and the width $y$ is increasing at the rate of $4$ cm/min. When $x=8$ cm and $y=4$ cm, the rate of change of the area of the rectangle is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ A=xy $ $ \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} $ Given: $ \frac{dx}{dt}=-6,\ \frac{dy}{dt}=4 $ $ \frac{dA}{dt}=x(4)+y(-6) $ At $(8,4)$: $ \frac{dA}{dt}=8\cdot4-6\cdot4=32-24=8 $
AMU MCA PYQ
The total revenue in rupees received from the sale of $x$ units of a product is given by $R(x)=5x^2+20x+7$. The marginal revenue, when $x=8$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$MR=\frac{dR}{dx}=\frac{d}{dx}(5x^2+20x+7)=10x+20$ At $x=8$ $MR=10(8)+20=80+20=100$
AMU MCA PYQ
The function $f(x) = \cos x$ is strictly decreasing on:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let $f(x) = \cos x$Differentiating with respect to $x$:$f'(x) = -\sin x$For a function to be strictly decreasing, $f'(x) < 0$:$-\sin x < 0$$\sin x > 0$The sine function is positive ($\sin x > 0$) in the first and second quadrants, specifically for the interval $x \in (0, \pi)$.
AMU MCA PYQ
The derivative of $f(x,y)=x^2+xy$ at $P_0(1,1)$ in the direction of unit vector $\vec{u}=\left(\frac{1}{\sqrt{2}}\right)\hat{i}+\left(\frac{1}{\sqrt{2}}\right)\hat{j}$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2021 PYQ

Solution

$\nabla f=(2x+y, x)$ At $(1,1)$: $\nabla f=(3,1)$ Directional derivative: $D_{\vec{u}}f=(3,1)\cdot\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ $=\frac{3+1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$
AMU MCA PYQ
If $f(x)=x^\alpha \log x$ and $f(0)=0$, then the value of $\alpha$ for which Rolle’s theorem can be applied in $[0,1]$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2016 PYQ

Solution

For continuity at $x=0$, need $\lim_{x\to 0^+} x^\alpha \log x = 0$ This holds when $\alpha > 0$ is not satisfied here, but by given condition extension works only at $\alpha = -1$

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