AMU MCA Application Of Derivatives Previous Year Questions (PYQs)
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Showing 8 questions
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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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 $=\frac{m}{m_0}=3$
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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Solution
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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Solution
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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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 $
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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Solution
$MR=\frac{dR}{dx}=\frac{d}{dx}(5x^2+20x+7)=10x+20$
At $x=8$
$MR=10(8)+20=80+20=100$
The function $f(x) = \cos x$ is strictly decreasing on:
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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 $x \in (0, \pi)$.
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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Solution
$\nabla f=(2x+y, x)$
At $(1,1)$:
$\nabla f=(3,1)$
Directional derivative:
$D_{\vec{u}}f=(3,1)\cdot\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
$=\frac{3+1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$
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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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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