111pqrpqr + 1 is equal to from Mathematics JEE Ye

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The minimum value of the function max (x, x) is equal to

  • 0

  • 1

  • 2

  • 1/2


2.

If f(x) = 2x + 42x, then f'(2) is equal to

  • 0

  • - 1

  • 1

  • 2


3.

Let f(x) = 2x3 - 9ax2 + 12a2x + 1, where a > 0. The minimum of f is attained at a point q and the maximum is attained at a point p. If p = q, then a is equal to

  • 1

  • 3

  • 2

  • 0


4.

For all rest numbers x and y, it is known as the real valued function f satisfies f(x) + f(y) = f(x + y). If f(1) = 7, then r = 1100fr is equal to

  • 7 x 51 x 102

  • 6 x 50 x 102

  • 7 x 50 x 102

  • 7 x 50 x 101


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5.

If f(x) = xx2x312x3x2026x, then f'(x) is equal to

  • x3 + 6x2

  • 6x3

  • 3x

  • 6x2


6.

The difference between the maximum and minimum value of of the function fx = 0xt2 + t + 1dt on [2, 3] is

  • 39/6

  • 49/6

  • 59/6

  • 69/6


7.

If a and b are the non-zero distinct roots of x2 + ax + b = 0, then the minimum value of x2 + ax + b is

  • 2/3

  • 9/4

  • - 9/4

  • - 2/3


8.

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(3x) g(x), where g is differentiable.The f'(x) is equal to

  • 3f(x)

  • g(0)

  • f(x)g(0)

  • 3g(x)


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9.

The roots of the equation x - 1111x - 1111x - 1 = 0 are

  • 1, 2

  • - 1, 2

  • - 1, - 2

  • 1, - 2


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10.

111pqrpqr + 1 is equal to

  • q - p

  • q + p

  • q

  • p


A.

q - p

We have,111pqrpqr + 1On applying, C1  C1 - C2, C2  C2 - C3, we get= 001p - qq - rrp - qq - r - 1r + 1On taking common (p - q) from C1, we get= p - q0011q - rr1q - r - 1r + 1= p - q . 1q - r - 1 - q + r= q - p


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