1 + ∆nfa is equal to from Mathematics Relat

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 Multiple Choice QuestionsMultiple Choice Questions

251.

Let f(x) = ax(a > 0) be written as f(x) = f1(x) + f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) + f1(x – y) equals :

  • 2f1(x)f1(y)

  • 2f1(x + y)f1(x - y)

  • 2f1(x + y)f2(x - y)

  • 2f1(x)f2(y)


252.

If the function f : R - 1, - 1  A defined by fx = x21 - x2, is surjective, then A is equal to :

  • R - (- 1, 0)

  • R - [- 1, 0)

  • - {- 1}

  • [0, )


253.

Let k = 10fa + k = 16210 - 1, where function satisfies f(x + y) = f(x)f(y) for all natural numbers x, y and f(1) = 2 . Then the natural number ‘a’ is

  • 16

  • 22

  • 20

  • 25


254.

If f(1) = 10, f(2) = 14, then using Newton's forward formula f(1.3) is equal to 

  • 12.2

  • 11.2

  • 10.2

  • 15.2


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255.

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

  • d = - a

  • d = a

  • a = b = c =  d = 1

  • a = b = 1


256.

The positive root of x2 - 78.8 = 0 after first approximation by Newton Raphson method assuming initial approximation to the root is 14, is

  • 9.821

  • 9.814

  • 9.715

  • 9.915


257.

In the usual notation the value of  is equal to

  •  -

  •  + 

  •  - 

  • None of the above


258.

The value of f(4) - f(3) is

  • f2 + 2f1 + 3f1

  • f3 + 2f2 + 3f1

  • f2 + 2f1 + 3f0

  • None of these


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259.

1 + nfa is equal to

  • f(a + h)

  • f(a + 2h)

  • f(a + nh)

  • f(a + (n - 1)h)


C.

f(a + nh)

1 + nfa = Enfa                     = fa + nh


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260.

Simplify the Boolean function (x · y) + [(x + y') - y]'.

  • 0

  • 1

  • x + y

  • xy


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