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AMU MCA Previous Year Questions (PYQs)

AMU MCA Differentiation PYQ

AMU MCA PYQ
Consider the series $x_{n+1}=\frac{x_n}{2}+\frac{9}{8x_n}$, $x_0=0.5$ obtained from the Newton-Raphson method. The series converges to





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

$x_{n+1}=\frac{1}{2}\left(x_n+\frac{9}{4x_n}\right)$ is Newton iteration for $\sqrt{\frac{9}{4}}$

$\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5$

AMU MCA PYQ
Let $f(x)$ be continuous whose values are known at $-2,-1,1,2$. If the Lagrange interpolation formula $f(x)=L_1f(-2)+L_2f(-1)+L_3f(1)+L_4f(2)$ is used to approximate $f(0)$, then $L_3$ is






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Lagrange basis for node $x_3=1$:

$L_3(0)=\frac{(0+2)(0+1)(0-2)}{(1+2)(1+1)(1-2)}$

$=\frac{(2)(1)(-2)}{(3)(2)(-1)} =\frac{-4}{-6}=\frac{2}{3}$

AMU MCA PYQ
Which of the following potentials does not satisfy Laplace’s equation?






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Laplace equation: $\nabla^2\phi=0$

(a) $\nabla^2=2+10-12=0$ ✔

(b) $\nabla^2=0$ ✔

(c) $\nabla^2=0-2+2=0$ ✔

(d) $\nabla^2=4-8+2=-2 \ne 0$ ✘

AMU MCA PYQ
The Newton–Raphson method converges fast, if $f'(\alpha)$ is: ($\alpha$ is the exact value of the root)





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

Newton–Raphson method converges faster when the derivative at the root is large.
AMU MCA PYQ
If $x + y + z = u$ $y + z = uv$ $z = uvw$ then the value of the Jacobian $\dfrac{\partial(x,y,z)}{\partial(u,v,w)}$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2021 PYQ

Solution

Given $z = uvw$ $y + z = uv$ $\Rightarrow y = uv - uvw = uv(1 - w)$ Now $x + y + z = u$ $\Rightarrow x = u - y - z$ Substitute: $x = u - uv(1 - w) - uvw$ $x = u - uv + uvw - uvw$ $x = u - uv$ $x = u(1 - v)$ So, $x = u(1 - v)$ $y = uv(1 - w)$ $z = uvw$
AMU MCA PYQ
If $u=f(x-y,y-z,z-x)$, then the value of $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2021 PYQ

Solution

Using chain rule, derivatives cancel out. Result: $=0$

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