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Answers by NastyBall

0 votes
1
ISI2019-MMA-19
Let $G =\{a_1,a_2, \dots ,a_{12}\}$ be an Abelian group of order $12$ . Then the order of the element $ ( \prod_{i=1}^{12} a_i)$ is $1$ $2$ $6$ $12$
Let $G =\{a_1,a_2, \dots ,a_{12}\}$ be an Abelian group of order $12$ . Then the order of the element $ ( \prod_{i=1}^{12} a_i)$ is$1$$2$$6$$12$
0 votes
3
ISI2018-MMA-21
The angle between the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 4x$ is $\pi /2$ $\pi /3$ $\pi /4$ $\pi /6$
The angle between the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 4x$ is$\pi /2$$\pi /3$$\pi /4$$\pi /6$
0 votes
4
ISI2015-MMA-84
For positive real numbers $a_1, a_2, \cdots, a_{100}$, let $p=\sum_{i=1}^{100} a_i \text{ and } q=\sum_{1 \leq i < j \leq 100} a_ia_j.$ Then $q=\frac{p^2}{2}$ $q^2 \geq \frac{p^2}{2}$ $q< \frac{p^2}{2}$ none of the above
For positive real numbers $a_1, a_2, \cdots, a_{100}$, let $$p=\sum_{i=1}^{100} a_i \text{ and } q=\sum_{1 \leq i < j \leq 100} a_ia_j.$$ Then $q=\frac{p^2}{2}$$q^2 \geq ...
0 votes
5
ISI2015-MMA-83
If $\alpha, \beta$ are complex numbers then the maximum value of $\dfrac{\alpha \overline{\beta}+\overline{\alpha}\beta}{\mid \alpha \beta \mid}$ is $2$ $1$ the expression may not always be a real number and hence maximum does not make sense none of the above
If $\alpha, \beta$ are complex numbers then the maximum value of $\dfrac{\alpha \overline{\beta}+\overline{\alpha}\beta}{\mid \alpha \beta \mid}$ is$2$$1$the expression m...
0 votes
6
ISI2015-MMA-76
Given that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$, the value of $ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+xy+y^2)} dxdy$ is $\sqrt{\pi/3}$ $\pi/\sqrt{3}$ $\sqrt{2 \pi/3}$ $2 \pi / \sqrt{3}$
Given that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$, the value of $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+xy+y^2)} dxdy$$ is$\sqrt{\pi/3}$$...
0 votes
7
ISI2015-MMA-75
The length of the curve $x=t^3$, $y=3t^2$ from $t=0$ to $t=4$ is $5 \sqrt{5}+1$ $8(5 \sqrt{5}+1)$ $5 \sqrt{5}-1$ $8(5 \sqrt{5}-1)$
The length of the curve $x=t^3$, $y=3t^2$ from $t=0$ to $t=4$ is$5 \sqrt{5}+1$$8(5 \sqrt{5}+1)$$5 \sqrt{5}-1$$8(5 \sqrt{5}-1)$
1 votes
11
ISI2015-MMA-27
Let $\cos ^6 \theta = a_6 \cos 6 \theta + a_5 \cos 5 \theta + a_4 \cos 4 \theta + a_3 \cos 3 \theta + a_2 \cos 2 \theta + a_1 \cos \theta +a_0$. Then $a_0$ is $0$ $1/32$ $15/32$ $10/32$
Let $\cos ^6 \theta = a_6 \cos 6 \theta + a_5 \cos 5 \theta + a_4 \cos 4 \theta + a_3 \cos 3 \theta + a_2 \cos 2 \theta + a_1 \cos \theta +a_0$. Then $a_0$ is$0$$1/32$$...
1 votes
12
ISI2015-MMA-26
$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$
$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to$\infty$$0$$\log_e 2$$1$
0 votes
14
ISI2015-MMA-20
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals$e^{-1}$$e^{-1/2}$$e^{-2}$$1$
0 votes
15
ISI2015-MMA-19
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is$2$$2e$$2 ...
0 votes
16
ISI2015-MMA-11
The number of positive integers which are less than or equal to $1000$ and are divisible by none of $17$, $19$ and $23$ equals $854$ $153$ $160$ none of the above
The number of positive integers which are less than or equal to $1000$ and are divisible by none of $17$, $19$ and $23$ equals$854$$153$$160$none of the above