Question 1 |

The smallest perimeter that a rectangle with area of 4 square units can have is
______ units.
(Answer in integer)

4 | |

6 | |

8 | |

10 |

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.

\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.

18.84 | |

22.25 | |

62.15 | |

45.25 |

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}

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)

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)

0 | |

1 | |

2 | |

3 |

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

\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?

|2-x| | |

|2-|x|| | |

|2+|x|| | |

2-|x| |

Question 4 Explanation:

(a) Graph of y = 2 - x

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

(c) 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 ___.

If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.

\sqrt{\pi a} | |

\sqrt{\frac{\pi}{a}} | |

b \sqrt{\pi a} | |

b \sqrt{\frac{\pi}{a}} |

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.

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

Thank you for your suggestions.

Can you please share the questions numbers for the above suggestions.

question number 10

question no 9 mein b vector, wrong type ho gaya hai

in que 10 x->1 not zero

Thank You Jashwanth Reddy Earla,

We have updated the question.

que 35 f(x)= 2x^3-3x^2

Thank You Jashwanth Reddy Earla,

We have updated the question.

Question 101- Answer is maximum(x,-x)

Thank You Ajay,

We have updated the answer.

Error in statement of Q.40

error in que 40

Please correct question number 6 , Function have extra Y

Q11. First Solutions calculation is wrong, it should be 3π²+(sigma)π/0=P