Question 1 |

Strokes theorem connects

a line integral and a surface integral | |

a surface integral and a volume integral | |

a line integral and a volume integral | |

gradient of a function and its surface integral |

Question 1 Explanation:

A line integral and a surface integral is related by stroke's theorem

Question 2 |

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is

0.0036 | |

0.1937 | |

0.2234 | |

0.3874 |

Question 2 Explanation:

Probability of defective item

P=0.1

Probability of non-defective item

Q=1-p=1-0.1=0.9

Probability that exactly 2 of the chosen items are defective

=^{10} C_{2}(P)^{2}(Q)^{8}

=^{10} C_{2}(0.1)^{2}(0.9)^{8}=0.1937

P=0.1

Probability of non-defective item

Q=1-p=1-0.1=0.9

Probability that exactly 2 of the chosen items are defective

=^{10} C_{2}(P)^{2}(Q)^{8}

=^{10} C_{2}(0.1)^{2}(0.9)^{8}=0.1937

Question 3 |

\int_{-a}^{a}(\sin ^{6} x + \sin ^{7} x)dx
is equal to

2\int_{0}^{a}(\sin ^{6} x)dx | |

2\int_{0}^{a}(\sin ^{7} x)dx | |

2\int_{0}^{a}(\sin ^{6} x + \sin ^{7} x)dx | |

zero |

Question 3 Explanation:

I=\int_{-a}^{a}\left(\sin ^{6} x+\sin ^{7} x \right) dx

= 2 \int_{0}^{a} \sin ^{6} x d x+0

( because \int_{0}^{a} \sin ^{7}x\cdot d x=0)

= 2 \int_{0}^{a} \sin ^{6} x d x+0

( because \int_{0}^{a} \sin ^{7}x\cdot d x=0)

Question 4 |

A is a 3 x 4 real matrix and Ax=b is an inconsistent system of equations. The highest possible rank of A is

1 | |

2 | |

3 | |

4 |

Question 4 Explanation:

C =[A: B]_{3 \times 5}

\therefore \rho\left[C_{3 \times 5}\right] \leq \min \{3,5\}

\because The system is inconsistent

\rho(A) \lt \rho(C)

\therefore \rho(A)\lt 3

Hence maximum possible rank of

A=2

\therefore \rho\left[C_{3 \times 5}\right] \leq \min \{3,5\}

\because The system is inconsistent

\rho(A) \lt \rho(C)

\therefore \rho(A)\lt 3

Hence maximum possible rank of

A=2

Question 5 |

Changing the order of the integration in the double integral I=\int_{0}^{8}\int_{\frac{x}{4}}^{2}f(x,y)dydx
leads to I=\int_{r}^{s}\int_{p}^{q}f(x,y)dxdy. What is q?

4y | |

16y^{2} | |

x | |

8 |

Question 5 Explanation:

\text { When } \quad I=\int_{0}^{8} \int_{x / 4}^{2} f(x \cdot y) dydx

I=\int_{0}^{2} \int_{0}^{4 y} f(x)dydx

I=\int_{0}^{2} \int_{0}^{4 y} f(x)dydx

There are 5 questions to complete.