Let f(x) = a^x(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)

If the function f : R $- \left\{1, - 1\right\} \to \mathrm{A}$ defined by $\mathrm{f}\left(\mathrm{x}\right) = \frac{{\mathrm{x}}^2}{1 - {\mathrm{x}}^2}$, is surjective, then A is equal to :

• R - (- 1, 0)

• R - [- 1, 0)

• - {- 1}

• $[0, \infty )$

Let $\sum _{\mathrm{k} = 1}^0\mathrm{f}\left(\mathrm{a} + \mathrm{k}\right) = 16\left(2^{10} - 1\right)$, 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

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

Let f(x) = $\frac{\mathrm{ax} +\hspace{0.17em}\mathrm{b}}{\mathrm{cx} +\hspace{0.17em}\mathrm{d}}$. Then, fof(x) = x provided that

• d = - a

• d = a

• a = b = c =  d = 1

• a = b = 1

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

In the usual notation the value of $∆\nabla$ is equal to

• $∆ -\hspace{0.17em}\nabla$

• $∆ + \nabla$

• $\nabla - ∆$

• None of the above

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

• $∆\mathrm{f}\left(2\right) + {∆}^2\mathrm{f}\left(1\right) + {∆}^3\mathrm{f}\left(1\right)$

• $∆\mathrm{f}\left(3\right) + {∆}^2\mathrm{f}\left(2\right) + {∆}^3\mathrm{f}\left(1\right)$

• $∆\mathrm{f}\left(2\right) + {∆}^2\mathrm{f}\left(1\right) + {∆}^3\mathrm{f}\left(0\right)$

• None of these

${\left(1 + ∆\right)}^{\mathrm{n}}\mathrm{f}\left(\mathrm{a}\right)$ is equal to

• f(a + h)

• f(a + 2h)

• f(a + nh)

• f(a + (n - 1)h)

260.

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

• 0

• 1

• x + y

• xy