AMU MCA Differentiation Previous Year Question Paper and Solution with PDF

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AMU MCA Previous Year Questions (PYQs)

AMU MCA Differentiation PYQ

Qus : 1 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 $\sqrt{\frac{9}{4}}$

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

Qus : 2 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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Solution

Lagrange basis for node $x_3=1$:

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

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

Qus : 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) $\nabla^2=0$ ✔

(d) $\nabla^2=4-8+2=-2 \ne 0$ ✘

Qus : 4 AMU MCA PYQ

A curve having the condition that the slope of tangent at some point is two times the slope of the straight line joining the same point to the origin of co-ordinates, is a/an

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2019 PYQ

Solution

Qus : 5 AMU MCA PYQ

If $ y = \dfrac{1}{\sqrt{a^2 - b^2}} \cos^{-1}\left[\dfrac{a\cos(x-a) + b}{\theta}\right] $ where $ \theta = a + b\cos(x-a) $, then $ \dfrac{dy}{dx} $ is equal to

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Solution

Qus : 6 AMU MCA PYQ

If $ f(x)=\dfrac{1}{1-x} $, then the derivative of the composite function f[f{f(x)}] is equal to

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Solution

Qus : 7 AMU MCA PYQ

If $f(x)=x e^{x(1-x)}$, then $f(x)$ is

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Solution

$ f(x)=x e^{x(1-x)} $ $ f'(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2x) $ $ =e^{x(1-x)}[1+x-2x^2] $ $ e^{x(1-x)}\neq0 $ $ \Rightarrow 1+x-2x^2=0 $ $ \Rightarrow (1-x)(2x+1)=0 $ $ \Rightarrow x=1,\ -\frac{1}{2} $ For $ x\in\left[-\frac{1}{2},1\right],\ f'(x)>0 $ $ \therefore f(x)$ is increasing on $\left[-\frac{1}{2},1\right]$

Qus : 8 AMU MCA PYQ

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x=c$, then $g'(f(c))$ equals

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ g(f(x))=x $ Differentiating, $ g'(f(x))\cdot f'(x)=1 $ $ \Rightarrow g'(f(x))=\frac{1}{f'(x)} $ $ \Rightarrow g'(f(c))=\frac{1}{f'(c)} $

Qus : 9 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.

Qus : 10 AMU MCA PYQ

If $y=\sqrt{\frac{1-x}{1+x}}$, then $(1-x^2)\frac{dy}{dx}+y$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$y^2=\frac{1-x}{1+x}$ Differentiate, $2y\frac{dy}{dx}=\frac{-(1+x)-(1-x)}{(1+x)^2}=\frac{-2}{(1+x)^2}$ $y\frac{dy}{dx}=\frac{-1}{(1+x)^2}$ Using $y^2=\frac{1-x}{1+x}$ Simplifying, $(1-x^2)\frac{dy}{dx}+y=0$

Qus : 11 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, $x = u(1 - v)$ $y = uv(1 - w)$ $z = uvw$

Qus : 12 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: $=0$

Qus : 13 AMU MCA PYQ

If $y = \tan^{-1}(\sqrt{1+x^2} - x)$, then $\frac{dy}{dx}$ equals

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2016 PYQ

Solution

Let $u=\sqrt{1+x^2}-x$ Use identity ⇒ $u = \frac{1}{\sqrt{1+x^2}+x}$ So $y=\tan^{-1}u$ Differentiate ⇒ $\frac{dy}{dx} = -\frac{1}{2(1+x^2)}$ Final Answer: $\boxed{-\frac{1}{2(1+x^2)}}$

Qus : 14 AMU MCA PYQ

If $y=\sec(\tan^{-1}x)$, then $\frac{dy}{dx}$ at $x=1$ is

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2016 PYQ

Solution

Let $\theta=\tan^{-1}x$ $y=\sec\theta=\sqrt{1+x^2}$ $\frac{dy}{dx}=\frac{x}{\sqrt{1+x^2}}$ At $x=1$: $\frac{1}{\sqrt{2}}$ Final Answer: $\boxed{\frac{1}{\sqrt{2}}}$

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