The function f(ax) = ax + b is strictly increasing for all real x

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 Multiple Choice QuestionsMultiple Choice Questions

671.

Let y = 3x - 13x + 1sinx + loge1 + x, x > - 1. Then, at x = 0, dydx equals

  • 1

  • 0

  • - 1

  • - 2


672.

If f is a real-valued differentiable function such that f(x)f' (x) < 0 for all real x, then

  • f(x) must be an increasing function

  • f(x) must be a decreasing function

  • f(x) must be an increasing function

  • f(x) must be a decreasing function


673.

Rolle's theorem is applicable in the interval [- 2, 2] for the function

  • f(x) = x3

  • f(x) = 4x4

  • f(x) = 2x3 + 3

  • f(x) = πx


674.

The solution of 25d2ydx2 - 10dydx + y = 0, y(0) = 1, y(1) = 2e15 is

  • y = e5x + e- 5x

  • y = 1 + xe5x

  • y = 1 + xex5

  • y = 1 + xe- x5


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675.

If f(x) and g(x) are twice differentiable functions on (0, 3) satisfying f"(x) = g''(c), f'(1) = 4g'(D) = 6, f(2) = 3, g(2) = 9, then f(1) - g(1) is

  • 4

  • - 4

  • 0

  • - 2


676.

For function f(x) = ecosx, Rolle's theorem is

  • applicable, when π2  x  3π2

  • applicable, when 0  x  π2

  • applicable, when 0  x  π

  • applicable, when π4  x  π2


677.

f(x) = 0,            x = 0x - 3,     x > 0 the function f(x) is

  • increasing when x  0

  • strictly increasing when x > 0

  • strictly increasing at x = 0

  • not continuous at x = 0 and so it is not increasing when x > 0


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678.

The function f(ax) = ax + b is strictly increasing for all real x, if

  • a > 0

  • a < 0

  • a = 0

  •  0


A.

a > 0

Given, f(x) = ax + b

On differentiating w.r.t. x, we get

f'(x) = a

For strictly increasing, f'(x) > 0

                                        a > 0


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679.

If y = 2x- 2x2 + 3x - 5, then for x = 2 and x = 0.1, value of y is

  • 2.002

  • 1.9

  • 0

  • 0.9


680.

If the function

fx = x2 - A + 2x + Ax - 2,       for x  22,                                         for x = 2

is continuous at x = 2, then

  • A = 0

  • A = 1

  • A = - 1

  • A = 2


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