Question

If $Q\left(x\right)=m\left[\frac{(x-l)}{(m-l)}\right]+l\left[\frac{(x-m)}{(l-m)}\right]$, then $Q\left(l\right)+Q\left(m\right)$ is equal to

• A. $Q\left(lm\right)$
• B. $Q(l+m)$
• C. $2(l-m)$
• D. None of these

Answer 1

The correct option is B

$Q(l+m)$

Explanation for the correct option.

Given that, $Q\left(x\right)=m\left[\frac{(x-l)}{(m-l)}\right]+l\left[\frac{(x-m)}{(l-m)}\right]$

Substituting $x=l$ & $x=m$ in $Q\left(x\right)$ we have:

$\begin{array}{rcl}Q\left(l\right)+Q\left(m\right)& =& m\left[\frac{\left(l-l\right)}{\left(m-l\right)}\right]+l\left[\frac{\left(l-m\right)}{\left(l-m\right)}\right]+m\left[\frac{\left(m-l\right)}{\left(m-l\right)}\right]+l\left[\frac{\left(m-m\right)}{\left(l-m\right)}\right]\\ & =& 0+l+m+0\left[\because l-l=m-m=0\right]\\ & =& l+m\end{array}$

Hence, option B is correct.