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BPSC Lecturer ME Held on July 2016 (Advt. 35/2014)

Option 1 : no real roots between 3 and 4

CT 1: Engineering Mathematics

487

10 Questions
5 Marks
15 Mins

**Concept:**

Algebraic functions are continuous in nature. If an algebraic function, f(x) changes its sign between two numbers then,

- There must exist a value of x for which f(x) becomes zero in between those two numbers.
- That value of x is called the root of equation f(x) = 0

__Calculation:__

**Given:**

f(x) = x3 – 2x – 8

We can start solving by finding the values of x, where f(x) is changing sign.

f'(x) = 3x^{2} - 2, equating it to zero we get,

3x2 - 2 = 0

\(x = \; \pm \sqrt {\frac{2}{3}}\)

\(f\left( {\sqrt {\frac{2}{3}} } \right) = \; - 9.089\) and \(f\left( {-\sqrt {\frac{2}{3}} } \right) = \; - 6.911\)

At both these values of x, f(x) is negative so there will be only one real root and rest 2 roots of cubic equation will be imaginary.

Note: Imaginary roots always exist in pairs.

So, to guess the position of real root, we can check values of f(x) at few points,

f(0) = - 8, f(1) = - 9, f(2) = - 4, f(3) = 13, f(4) = 48

From here, we can see that,

**The sign of the function does not change between 3 and 4, so there is no real roots between 3 and 4.**- The sign of the function does not change between 1 and 2, so there are no real roots between 1 and 2.
- The sign of the function changes between 2 and 3, so there is at least one real root between 2 and 3.