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)

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

• 0

• 1

• x + y

• xy

For the circuit show below, the Boolean polynomial is

• $\left(~ \mathrm{p} \vee \mathrm{q}\right) \vee \left(\mathrm{p} \vee ~ \mathrm{q}\right)$

• $\left(~ \mathrm{p} \wedge \mathrm{q}\right) \vee \left(\mathrm{p} \wedge \mathrm{q}\right)$

• $\left(~ \mathrm{p} \wedge ~ \mathrm{q}\right) \vee \left(\mathrm{q} \wedge \mathrm{p}\right)$

• $\left(~ \mathrm{p} \wedge \mathrm{q}\right) \vee \left(\mathrm{p} \wedge ~ \mathrm{q}\right)$

In a Boolean Algebra B, for all x, y in B, $\mathrm{x} \wedge \left(\mathrm{x} \vee \mathrm{y}\right)$ is equal to

• y

• x

• 1

• 0

The value of $\left(1 + ∆\right)\left(1 - \nabla \right)$ is

• 0

• - 1

• 1

• None of the above