Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be  ( y x ) 3 . If the curve also passes through the point ( α , 6 1 0 )  in the first quadrant, then a is equal to………….

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  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    2 weeks ago

    3 x f ( x ) d x = ( f ( x ) x ) 3 x 3 3 x f ( x ) d x = f 3 ( x ) , differentiating w.r.to x

    x 3 f ( x ) + 3 x 2 f 3 ( x ) x 3 = 3 f 2 ( x ) f ' ( x ) 3 y 2 d y d x = x 3 y = 3 y 3 x 3 x y d y d x = x 4 + 3 y 2  

    After solving we get  y 2 = x 4 3 + c x 2  also curve passes through (3, 3) Þ c = -2


    y 2 = x 4 3 2 x 2
    which passes through ( α , 6 1 0 ) α 4 6 α 2 3 = 3 6 0 α = 6  

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Raj Pandey

Since a is a odd natural number then | 1 3 y a d y | = 3 6 4 3 | ( y a + 1 a + 1 ) 1 3 | = 3 6 4 3 3 a + 1 a + 1 = 3 6 4 3  

 Þ a = 5

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Vishal Baghel

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

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Vishal Baghel

Given curve is y=cosx

y=sinx for 0xπ2

And yaxis

We know that sinx=cosx at x=π4and<π4<π2 i.e,  cosπ4=sinπ4=1/√2

So the point of intersection is at x=π4

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Vishal Baghel

The given equation of the lines are

2x+y=4(1)3x2y=6(2)x3y+5=0(3)

 Area of ?ABC=area(PCBQ)area(?APC)area(?AQB)

=14y(3)dx12y(1)dx24y(2)dx=14(x+5)3dx12(42x)dx24(3x62dx)

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Vishal Baghel

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is y0=5042(x2)

=y=52(x2)

Similarly equation of BC is

y5=3564(x4)=y=5+(x+4)y=9x

And equation of AC is y0=3062(x2)

y=34(x2)

 Area of ?ABC

area(?ABE)+area(BCDE)area(?ACD)

=24yABdx+46yBCdx26yBCdx=2452(x2)dx+46(9x)dx2634(x2)dx=52[x222x]24+[9xx22]4634[x222x]26

=52[(4222.4)(2222.2)]+[(9.6622)(9.4422)]34[(6222.6)(2222.2)]=52[(88)(24)]+[541836+8]34[18122+4]=5+86=7unit2

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