If y2 = ax2 + bx + c, where a, b, c are constants, then y3d2

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The slope of the tangent at (x, y) to a curve passing through 1, π4 is given by yx - cos2yx, then the equation of the curve is

  • y = tan-1logex

  • y = xtan-1logxe

  • y = xtan-1logex

  • None of the above


2.

Let f(x) = ax +bcx +d. Then, fof(x) = x provided that

  • d = - a

  • d = a

  • a = b = c =  d = 1

  • a = b = 1


3.

If A = cos2αcosαsinαcosαsinαsin2α and B = cos2βcosβsinβcosβsinβsin2β are two matrices such that the product AB is null matrix, then α - β is

  • 0

  • multiple of π

  • an odd multiple of π2

  • None of these


4.

If A is a square matrix of order n x n, then adj (adj A) is equal to

  • AnA

  • An - 1A

  • An - 2A

  • An - 3A


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5.

If f(x) = sin-12x1 + x2, then f(x) is differentiable on

  • [- 1, 1]

  • R - {- 1, 1}

  • R - (- 1, 1)

  • None of these


6.

The function f (x) = e- x is

  • continuous everywhere but not differentiable at x = 0

  • continuous and differentiable everywhere

  • not continuous at x = 0

  • None of the above


7.

For the equations x + 2y + 3z = 1, 2x + y + 3z = 2 and 5x + 5y + 9z = 4

  • there is only one solution

  • there exists infinitely many solution

  • there is no solution

  • None of the above


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8.

If y2 = ax2 + bx + c, where a, b, c are constants, then y3d2ydx2 is equal to

  • a constant

  • a function of x

  • a function of y

  • a function of x and y both


A.

a constant

Given, y2 = ax2 + bx + cOn differentiating w.r.t. x, we get2ydydx = 2ax + bAgain differentiating w.r.t. x, we get2dydx2 +2yd2ydx2 = 2a d2ydx2 = a - dydx2 yd2ydx2 = a - 2ax + b2y2 yd2ydx2 = 4ay2 - 2ax + b24y2 4y3d2ydx2 = 4aax2 + bx + c - 4a2x2 + 4abx + b2 4y3d2ydx2 = 4ac - b2 y3d2ydx2 = 4ac - b24 = constant


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9.

If  the function f(x) = 2x3 - 9ax2 + 12a2x + 1 attains its maximum and minimum at p and q respectively such that p2  = q, then a equals 

  • 0

  • 1

  • 2

  • None of these


10.

On the interval [0, 1] the function x25(1 - x)75 takes its maximum value at the point

  • 0

  • 1/4

  • 1/2

  • 1/3


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