Question

If α and β are the root of the quadratic equation ax² + bx + c =

0, then$f(x) = x - \lbrack x\rbrack,f(x)$ =

• A. $\frac{1}{2}$
• B. $f(x) = \sin\left( \frac{\pi x}{n - 1} \right) + \cos\left( \frac{\pi x}{n} \right),$
• C. $n \in Z,$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{2}$

Answer 2

Q ax² + bx + c = 0 has roots α and β then

i.e., cx² + bx + a = 0 has roots $\frac { 1 } { \alpha }$ and $\frac { 1 } { \beta }$.

⇒ c = c

=$\lim _ { x \rightarrow \frac { 1 } { \alpha } }$

=$\lim _ { x \rightarrow \frac { 1 } { \alpha } }$

=$\lim _ { x \rightarrow \frac { 1 } { \alpha } }$ $\left| \frac { \sin \left( \frac { c x ^ { 2 } + b x + a } { 2 } \right) } { ( 1 - \alpha x ) } \right|$

= $\lim _ { x \rightarrow \frac { 1 } { \alpha } }$ $\left| \frac { \sin \left\{ \frac { c } { 2 } \left( x - \frac { 1 } { \alpha } \right) \left( x - \frac { 1 } { \beta } \right) \right\} } { - \alpha \left( x - \frac { 1 } { \alpha } \right) } \right|$

=

= $\left. 1 \cdot \frac { \frac { c } { 2 } \left( \frac { 1 } { \alpha } - \frac { 1 } { \beta } \right) } { - \alpha } \right\rvert \,$ = $\left| \frac { \mathrm { c } } { 2 \alpha } \left( \frac { 1 } { \alpha } - \frac { 1 } { \beta } \right) \right|$ .