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2 years ago

If f'(x)=2/x and f(sqrt(e))=5, then f(e)=

1 answer:

3 0

$f'(x)=\dfrac2x$
$\implies f(x)=\displaystyle\frac2x\,\mathrm dx=2\ln|x|+C$

$f(\sqrt e)=5\implies 5=2\ln|\sqrt e|+C$
$\implies 5=\dfrac22\ln e+C$
$\implies C=4$
$\implies f(x)=2\ln|x|+4$
$\implies f(e)=2\ln|e|+4=6$

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A point x is 34m due to east of a point Y . The bearing of a flagpole from x and Y are N18°W And N40°E respectively .

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Answer:

The distance of flag post from Y is 38.13 m

Step-by-step explanation:

Consider point Y at the intersection of both lines as shown below. Now the point X lies 34 meter from point Y in east direction.

Now flag pole at point X lies at a bearing of N18°W. That is at point X from north, flag post makes an angle of 18° towards west.

Similarly flag pole at point Y lies at a bearing of N40°E. That is at point Y from north, flag post makes an angle of 40° towards east.

Consider ∆ AXY as right angle triangle. Therefore measure of angle FXY is,

$\angle AXY=\angle AXF+\angle FXY$

$\therefore 90 \degree=18\degree+\angle FXY$

$\angle FXY=72\degree$

Consider ∆ BYX as right angle triangle. Therefore measure of angle FYX is,

$\angle BYX=\angle BYF+\angle FYX$

$\therefore 90 \degree=40\degree+\angle FYX$

$\angle FYX=50\degree$

Refer attachment 1.

From diagram consider the triangle FYX. To find the third angle that is ∠YFX can be calculated by using angle sum property of triangle.

∠YFX+∠FYX+∠FXY=180°

∠YFX+50°+72°=180°

∠YFX=58°

Refer attachment 2.

Now the distance FY can be calculated using sin rule as follows,

$\frac{\sin X}{FY}=\frac{\sin F}{YX}=\frac{\sin Y}{FX}$

Substituting the values,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}=\frac{\sin 50}{FX}$

Simplifying first two terms,

$\frac{\sin 72}{FY}=\frac{\sin 58}{34}$

Cross multiplying,

$34\times\sin 72=FY\times \sin 58$

$\dfrac{34\times\sin 72}{\sin 58}=FY$

$FY=38.13$ m

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