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MCA NIMCET Previous Year Questions (PYQs)
MCA NIMCET Application Of Derivatives PYQ
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ
Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ
Solution
Number of Roots
Given:
\( e^x \sin x = 1 \) has two real roots → say \( x_1 \) and \( x_2 \)
Apply Rolle’s Theorem:
Since \( f(x) = e^x \sin x \) is continuous and differentiable, and \( f(x_1) = f(x_2) \), ⇒ There exists \( c \in (x_1, x_2) \) such that \( f'(c) = 0 \)
Compute:
\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]
\[ \boxed{\text{At least one root}} \]
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ
Water runs into a conical tank of radius $5$ ft and height $10$ ft at a constant rate of
$2\text{ ft}^3/\text{min}$.
How fast is the water level rising when the water is $6$ ft deep?
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ
Solution
Cone volume: $V = \dfrac{1}{3}\pi r^2 h$ Similarity:
$r = \dfrac{h}{2}$
So,
$V = \dfrac{\pi}{12}h^3$
Differentiate:
$\dfrac{dV}{dt} = \dfrac{\pi}{4}h^2 \dfrac{dh}{dt}$
Put values:
$2 = \dfrac{\pi}{4}(36)\dfrac{dh}{dt}$
$2 = 9\pi\dfrac{dh}{dt}$
$\displaystyle \dfrac{dh}{dt} = \dfrac{2}{9\pi}$
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ
A box open at the top is made by cutting squares from the four corners of a $6 \times 6$ m sheet.
The height of the box for maximum volume is:
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ
Solution
Volume: $V = x(6-2x)^2$Differentiate:
$\dfrac{dV}{dx} = 0$ gives $x = 1$
So height = $1$ m → not in options.
Closest correct is None of these.
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ
Find the point at which, the tangent to the curve $y=\sqrt{4x-3}-1$ as its slope $\frac{2}{3}$
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ
Solution
MCA NIMCET PYQ
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ
The rate of increase of length of the shadow of a man $2$ meters high, due to a lamp at $10$ meters height, when he is moving away from it at $2 \text{ m/sec}$ is
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ
Solution
Let $x =$ distance of man from lampLet $y =$ length of shadow
By similar triangles:
$\dfrac{10}{x+y} = \dfrac{2}{y}$ Cross-multiply:
$10y = 2(x+y)$
$10y = 2x + 2y$
$8y = 2x$
$y = \dfrac{x}{4}$
Differentiate w.r.t time $t$:
$\dfrac{dy}{dt} = \dfrac{1}{4}\dfrac{dx}{dt}$
Given $\dfrac{dx}{dt} = 2$ m/sec
:
$\dfrac{dy}{dt} = \dfrac{1}{4} \times 2 = \dfrac{1}{2}$
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