If y = (1 + x)(1 + x2)(1 + x4)...(1 + x2n) then the value of 

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

51.

The point in the interval [0, 2], where f(x) = ex sin(x) has maximum slope, is

  • π2

  • π

  • 3π2


52.

The minimum value of f(x) = ex4 - x3 + x2

  • e

  • - e

  • 1

  • - 1


53.

In which of the following functions, Rolle's theorem is applicable

  • f(x) = f(x) = x in - 2 x 2

  • fx = tanx in 0  x  π

  • fx = 1 + x - 223 in 1  x  3

  • fx = xx - 22 in 0  x  2


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54.

If y = (1 + x)(1 + x2)(1 + x4)...(1 + x2n) then the value of dydxx = 0 is

  • 0

  • - 1

  • 1

  • 2


C.

1

y = 1 +x1 +x2(1 +x4) ... (a + x2n)

Taking log on both sides

logy = log1 +x +log1 +x2 +log1 +x4 + ... + log1 +x2nOn differentiating w. r. t. xdydx = y11 +x + 2x1 + x2 + 4x31 + x4 + ...At x = 0, dydx = 1Since at x = 0, y = 1


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55.

The function f(x) = seclogx + 1 + x2

  • odd

  • even

  • neither odd nor even

  • constant


56.

The domain of the function f(x) = cos-11 - x2 is

  • (- 3, 3)

  • [- 3, 3]

  • - , - 3  3, 

  • (- , - 3]  [3, )


57.

If the line ax + by + c = 0 is a tangent to the curve xy = 4, then 

  • a < 0, b > 0

  • a  0, b > 0

  • a < 0, b < 0

  • a  0, b < 0


58.

If the normal to the curve y = f(x) at the point (3, 4) make an angle 3π/4 with the positive x-axis, then f'(3) is

  • 1

  • - 1

  • 34

  • 34


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59.

The equation of normal of x+ y2 - 2x + 4y - 5 = 0 at (2, 1) is

  • y = 3x - 5

  • 2y = 3x - 4

  • y = 3x + 4

  • y = x + 1


60.

The area of the region bounded by y2 = x and y = x is

  • 13sq. unit

  • 16 sq. units

  • 23 sq. units

  • 1 sq. units


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