The integral ∫xxsinx + cosx2dx =&nbs

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If BAC = 90°, and ar(ABC) = 55sq. units, then the abscissa of the vertex C is :

  • 1 + 5

  • 2 + 5

  • 1 + 25

  • 25 - 1


2.

If 1 + 1 - 22 . 1 + 1 - 42 . 3 + 1 - 62 . 5 + ... + 1 - 202 . 19 = α - 220β thenan ordered pair α, β = ?

  • (11, 97)

  • (10, 103)

  • (11, 103)

  • (10, 97)


3.

Let P(3, 3) be a point on the hyperbola,x2a2 - y2b2 = 1 . If the normal to it at P intersects the x-axis at (9,0) and e is its eccentricity, then the ordered pair (a2, e2) is equal to :

  • 92, 3

  • 32, 2

  • 9, 3

  • 92, 2


4.

The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

  • 9

  • 5

  • 3

  • 7


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5.

Let x2a2 + y2b2 = 1 a > b be agiven ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function φt = 512 + t - t2 then a+ b2 is equal to

  • 145

  • 126

  • 116

  • 135


6.

Let  α and β be the roots of x2  3x + p = 0 and γ and δ be the roots of x2  6x + q = 0. If α, β, γ, δ  from a geometric progression. Then ratio (2q + p) : (2q  p) is

  • 3 : 1

  • 5 : 3

  • 9 : 7

  • 33 : 31


7.

If A =cosθisinθisinθcosθ, θ = π24 and A5 = abcd, where i =  - 1, then which one of the following isnot true ?

  • 0  a2 +b2  1

  • a2 - d2 = 0

  • a2 - b2 = 12

  • a2 - c2 = 1


8.

Let y = y(x) be the solution of the differential equation,  xy'  y = x2(xcosx + sinx), x > 0. If y(π) = π, then  y''π2 + yπ2 is equal to :

  • 1 + π2

  • 1 + π2 + π24

  • 2 + π2 +π24

  • 2 + π2


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9.

Let fx = x - 2 and gx = ffx, x  0, 4. Then03gx - fxdx = ?

  • 0

  • 12

  • 32

  • 1


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10.

The integral xxsinx + cosx2dx = ? Where Cis a constant of integration : 

  • tanx + xsecxxsinx + cosx + C

  • tanx - xsecxxsinx + cosx + C

  • secx + xtanxxsinx + cosx + C

  • secx - xsecxxsinx + cosx + C


B.

tanx - xsecxxsinx + cosx + C

x2dxxsinx + cosx2 = xcosx . xcosxxsinx + cosx2dx= xcosxxcosxxsinx + cosx2dx -  ddxxsecxxcosxxsinx + cosx2dxdx= xcosx - 1xsinx + cosx +  sec2xdx xcosxxsinx + cosx2dx =  - 1xsinx + cosx and ddxxsecx = secx + xsecxtanx = secx1 + xsinxcosx = sec2xxsinx + cosx=  - xcosxxsinx + cosx + sinxcosx + C = tanx - xsecxxsinx + cosx + C


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