Calculus


Question 1
The smallest perimeter that a rectangle with area of 4 square units can have is ______ units. (Answer in integer)
A
4
B
6
C
8
D
10
GATE ME 2023   Engineering Mathematics
Question 1 Explanation: 


Given \quad B \times L=4
Perimeter (P)=2 B+2 L
perimeter to be smallest possible
\frac{d P}{d B}=0=\frac{d}{d B}\left(2 B+2 \times \frac{4}{B}\right)=0
2-\frac{8}{B^{2}}=0
B=\pm 2 \Rightarrow B=2 \quad L=2
Smallest perimeter =2(B+L)=2(2+2)=8
Question 2
A linear transformation maps a point (x,y) in the plane to the point (\hat{x},\hat{y}) according to the rule

\hat{x}=3y,\hat{y}=2x

Then, the disc x^2+y^2 \leq 1 gets transformed to a region with an area equal to _________ . (Rounded off to two decimals)
Take \pi =3.14.
A
18.84
B
22.25
C
62.15
D
45.25
GATE ME 2023   Engineering Mathematics
Question 2 Explanation: 
\hat{x}=3 y, \hat{y}=2 x
y=\frac{\hat{x}}{3}, x=\frac{\hat{y}}{2}
Given equation is x^{2}+y^{2} \leq 1
\begin{aligned} & \left(\frac{\hat{y}}{2}\right)^{2}+\left(\frac{\hat{x}}{3}\right)^{2} \leq 1 \\ & \frac{(\hat{y})^{2}}{(2)^{2}}+\frac{(\hat{\mathrm{x}})^{2}}{(3)^{2}} \leq 1 \text { equation of ellipse } \end{aligned}
Area of ellipse =\pi a b
\begin{aligned} & =\pi \times 3 \times 2 \\ & =6 \pi \\ & \pi=3.14 \\ & \text { area }=6 \times 3.14 \\ & =18.84 \end{aligned}


Question 3
A vector field
B(x,y,z)=x\hat{i}+y\hat{j}-2z\hat{k}
is defined over a conical region having height h=2, base radius r=3 and axis along z, as shown in the figure. The base of the cone lies in the x-yplane and is centered at the origin.
If n denotes the unit outward normal to the curved surface S of the cone, the value of the integral
\int _S B\cdot n \; dS
equals _________ . (Answer in integer)

A
0
B
1
C
2
D
3
GATE ME 2023   Engineering Mathematics
Question 3 Explanation: 
\int_{\mathrm{B}} \bar{B} \cdot n d S It is closed surface then apply gauss divergence theorem
\int_{s} \bar{B} \cdot \hat{n} d S=\iiint_{\text {Div }} \cdot \vec{B} d v \quad \text {...(i) }
\vec{B}=x \hat{i}+y \hat{j}-22 \hat{k}
\text { Diversion } \vec{B}=1+1-2=0
Put in (i)
\int_{S} \vec{B} \cdot \hat{n} d S=\iiint 0 d v =0
Question 4
The figure shows the plot of a function over the interval [-4, 4]. Which one of the options given CORRECTLY identifies the function?

A
|2-x|
B
|2-|x||
C
|2+|x||
D
2-|x|
GATE ME 2023   Engineering Mathematics
Question 4 Explanation: 
(a) Graph of y = 2 - x

(b) Graph of y = |2 - x|

(c) Graph of y = |2 - |x||

Question 5
Given \int_{-\infty }^{\infty }e^{-x^2}dx=\sqrt{\pi}
If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.
A
\sqrt{\pi a}
B
\sqrt{\frac{\pi}{a}}
C
b \sqrt{\pi a}
D
b \sqrt{\frac{\pi}{a}}
GATE ME 2022 SET-2   Engineering Mathematics
Question 5 Explanation: 
\begin{aligned} &\text{ Let }(x+b)=t\\ &\Rightarrow \; dx=dt\\ &\text{When ,} x=-\infty ;t=-\infty \\ &\int_{-\infty }^{-\infty }e^{-n(x+b)^2}dx=\int_{-\infty }^{-\infty }e^{-at^2}dt\\ &\text{Let, }at^2=y^2\Rightarrow t=\frac{y}{\sqrt{a}}\\ &2at\;dt=3y\;dy\\ &dt=\frac{ydy}{at}=\frac{ydy}{a\frac{y}{\sqrt{a}}}=\frac{y}{\sqrt{a}}\\ &\int_{-\infty }^{-\infty }e^{-at^2}dt=\int_{-\infty }^{-\infty }e^{-y^2}\cdot \frac{dy}{\sqrt{a}}=\sqrt{\frac{\pi}{a}} \end{aligned}




There are 5 questions to complete.

14 thoughts on “Calculus”

  1. The question posted have minor mistakes in it, in answers it show up something else other than what mentioned in question. Please look into it ,otherwise its a good platform to practise

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