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Let x be a continuous variable defined over the interval $$\left( { - \infty ,\,\infty } \right)$$  , and f(x) = e<sup>-x-e<sup>-x</sup></sup> . The integral $${\text{g}}\left( {\text{x}} \right) = \int {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $$    is equal to

Correct Answer: e<sup>-e<sup>-x</sup></sup>

Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this series are denoted by $${a_k}$$ , that is
$$x\left( t \right) = \sum\limits_{k = - \infty }^\infty {{a_k}{e^{jk{{2\pi } \over T}t}}} .$$
The same function x(t) can also be considered as a periodic function with period T' = 40. Let bk be the Fourier series coefficients when period is taken as T'. If $$\sum\limits_{k = - \infty }^\infty {\left| {{a_k}} \right|} = 16,$$    then $$\sum\limits_{k = - \infty }^\infty {\left| {{b_k}} \right|} $$  is equal to

If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as:
$${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$     where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$

For a function g(t), it is given that
$$\int\limits_{ - \infty }^{ + \infty } {g\left( t \right){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}} $$    for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ - \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ - \infty }^{ + \infty } {y\left( t \right)} dt$$       is. . . . . . . .

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

Consider the line integral $$I = \int_{\text{c}} {\left( {{{\text{x}}^2} + {\text{i}}{{\text{y}}^2}} \right){\text{dz,}}} $$    where z = x + iy. The line c is shown in the figure below
Complex Variable mcq question image
The value of $$I$$ is

Mean Value theorem is applicable to the and continuous in open interval (a, b) b) Functions continuous in closed interval only and having same value at point ‘a’ and ‘b’ c) Functions continuous in closed interval and differentiable in open interval (a, b) d) Functions differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with

Let δ(t) denote the delta function. The value of the integral $$\int\limits_{ - \infty }^\infty {\delta \left( t \right)} \cos \left( {{{3t} \over 2}} \right)dt$$     is

A normal random variable X has following probability density function $${{\text{f}}_{\text{x}}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {8\pi } }}{{\text{e}}^{ - \left\{ {\frac{{{{\left( {{\text{x}} - 1} \right)}^2}}}{8}} \right\}}},\, - \infty Then $$\int\limits_1^\infty {{{\text{f}}_{\text{x}}}\left( {\text{x}} \right){\text{dx}}} $$   is