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NIELIT 2016 MAR Scientist C - Section B: 7
Lakshman Patel RJIT
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The area under the curve $y(x)=3e^{-5x}$ from $x=0 \text{ to } x=\infty$ is
$\dfrac{3}{5}$
$\dfrac{-3}{5}$
${5}$
$\dfrac{5}{3}$
nielit2016mar-scientistc
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Lakshman Patel RJIT
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NIELIT 2016 MAR Scientist C - Section B: 5
If $f(x,y)=x^{3}y+e^{x},$ the partial derivatives, $\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}$ are $3x^{2}y+1, \: x^{3}+1$ $3x^{2}y+e^{x}, \: x^{3}$ $x^{3}y+xe^{x}, \: x^{3}+e^{x}$ $2x^{2}y+\dfrac{e^{x}}{x}$
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NIELIT 2016 MAR Scientist C - Section B: 6
If $y=f(x)$, in the interval $[a,b]$ is rotated about the $x$-axis, the Volume of the solid of revolution is $(fâ(x)=dy/dx)$ $\int_{a}^{b} \pi [f(x)]^{2} dx \\$ $\int_{a}^{b}[f(x)]^{3} dx \\$ $\int_{a}^{b} \pi [{f}'(x)]^{2} dx \\$ $\int_{a}^{b} \pi^{2} f(x)dx \\$
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NIELIT 2016 MAR Scientist C - Section B: 14
Find the area bounded by the curve $y=\sqrt{5-x^{2}}$ and $y=\mid x-1 \mid$ $\dfrac{2}{0}(2\sqrt{6}-\sqrt{3})-\dfrac{5}{2}$ $\dfrac{2}{3}(6\sqrt{6}+3\sqrt{3})+\dfrac{5}{2}$ $2(\sqrt{6}-\sqrt{3})-5$ $\dfrac{2}{3}(\sqrt{6}-\sqrt{3})+5$
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NIELIT 2016 MAR Scientist C - Section B: 15
The equation of the plane through the point $(-1,3,2)$ and perpendicular to each of the planes $x+2y+3z=5$ and $3x+3y+z=0$ is $7x-8y+3z+25=0$ $7x+8y+3z+25=0$ $7x-8y+3z-25=0$ $7x-8y-3z-25=0$
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