Admin: Ankit Pandey

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Ankit Pandey ♦♦ 10
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Posts by Ankit Pandey

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Answer: A: Find the mass of the leminiscate $r^2 = a^2 cos 2\theta$ if the density at any p
... Solution: Given curve: $r^{2}=a^{2} \cos 2 \theta \quad$ is lemniscate. The density at any point is proportional to the square of dist. From the pole. Distance from the pole =r $\quad \therefore$ Density $\propto r^{2}$ $\therefore$ Density $=k . r^{2}$ The mass of the lemniscate is given by, ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate$\int^1_0 \int^{1-x}_0 \int^{1-x-y}_0 \frac{1}{(x + y + z +1)^3} dz \ d
... Solution: $\begin{aligned} \text { Let } 1 &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d x \ d y \ d z \\ &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d z \ d y \ d x \end{aligned}$ $\begin{aligned} &=\int_{0}^{1} \int_{0}^{1-x}\left ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Change the order of integration and evaluate $ \int^1_0 \int^\sqrt{2-x^2}_x \fra
... Solution: Let $I=\int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{x d y d x}{\sqrt{x^{2}+y^{2}}}$ Region of integration is: $\quad x \leq y \leq \sqrt{2-x^{2}}\\ 0 \leq x \leq 1$ Curves: $(i) y=x$ line $\quad(i i) x=0, x=1 \quad$ lines parallel to the **y** axis (iii) $y=\sqrt{2-x^{2}} \Rightarrow x^{ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Using DUIS prove that $\int^{\infty}_0 e^{-(x^2+\frac{a^2}{x^2})}dx=\frac{\sqrt\
... Solution: Let $I(\mathrm{a})=\int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x---(1)$ Taking 'a' as parameter diff. w.r.t. a, $\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x$ Apply D.U.I.S rule, $\frac{d I(a)}{d a}=\int_{0}^{\i ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Find the length of the cycloid from one cusp to the next where
... Solution: Given curve: Cycloid $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ ![enter image description here][1] The length of given curve is: $\mathrm{S}=\int_{\theta_{1}}^{\theta_{2}} \sqrt{\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2}} \mathrm{d} \theta$ $\begin{ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate Integral :
... Solution: let $\quad I=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x$ Taking 'a' as parameter, $\quad I(a)=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x---(1)$ differentiate w.r.t **a** $\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{1} \frac{x^{a}-1}{\log x} d x$ $\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate $\int^1_0 \int_0^{x^2} e^ \frac{y}{x} dy dx $
... Solution: $\begin{aligned} \text { let } \mathrm{I} &=\int_{0}^{1} \int_{0}^{x^{2}} e^{\frac{y}{x}} d y d x \\ &=\int_{0}^{1}\left[\frac{e^{y}}{\frac{1}{x}}\right]_{0}^{x^{2}} d x \end{aligned}$ $=\int_{0}^{1} \frac{\left(e^{x}-1\right)}{\frac{1}{x}} d x$ $=\int_{0}^{1} x \cdot e^{x} d x- ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Solve by the method of variation of parameters
... Solution: $\left(D^{2}-6 D+9\right) y=\frac{e^{3 x}}{x^{2}}$ For complementary solution, $f(D)=0$ $\therefore\left(D^{2}-6 D+9\right)=0$ Roots are: $D=3,3 \quad$ Real roots but repeatative. The complementary solution of given diff. eqn is, $\quad \therefore y_{c}=\left(c_{1}+x c_{2}\right) e^ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Change the order of integration and evaluate
... Solution: Let $\mathrm{V}=\iiint x^{2} d x \ d y\ d z$ Region of integration is volume bounded by the planes $x=0, y=0, z=0$ And $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ Put $x=a u, y=b v, z=c w$ $\therefore \quad d x \ d y \ d z=a b c$ du.dv.dw ![enter image description here][1] The inters ...
written 3 days ago by Ankit Pandey ♦♦ 10
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Answer: A: Evaluate $\int\int\int_v x^2 dx \ dy \ dz$ over the volume bounded by the plan
... Solution: Let $I=\int_{0}^{2} \int_{\sqrt{2 y}}^{2} \frac{x^{2}}{\sqrt{x^{4}-4 y^{2}}} d x \ d y$ $\begin{aligned} \text { Region of integration : } & \sqrt{2 y} \leq x \leq 2 \\ 0 & \leq y \leq 2 \end{aligned}$ $\begin{aligned} \text { Curves: } & \text { (i) } x=2, y=2, y=0 \text { ...
written 3 days ago by Ankit Pandey ♦♦ 10

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