Question

If p(x)=4x square +x-2 then value of p(2)-p(-1)+p(1/2)

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:p(x) = {4x}^{2} + x - 2$

Now, we have to find the value of

$\rm :\longmapsto\:p(2) -p( - 1) + p\bigg[\dfrac{1}{2} \bigg]$

Consider,

$\red{\rm :\longmapsto\:p(2)}$

$\rm \: = \: {4(2)}^{2} + 2 - 2$

$\rm \: = \:16$

$\bf\implies \:\boxed{ \tt{ \: p(2) = 16 \: }}$

Now, Consider

$\red{\rm :\longmapsto\:p( - 1)}$

$\rm \: = \: {4( - 1)}^{2} - 1 - 2$

$\rm \: = \: 4 - 1 - 2$

$\rm \: = \: 1$

$\bf\implies \:\boxed{ \tt{ \: p( - 1) = 1 \: }}$

Now, Consider,

$\red{\rm :\longmapsto\:p\bigg[\dfrac{1}{2} \bigg]}$

$\rm \: = \:4 \times {\bigg[\dfrac{1}{2} \bigg]}^{2} + \bigg[\dfrac{1}{2} \bigg] - 2$

$\rm \: = \:4 \times \dfrac{1}{4} + \dfrac{1}{2} - 2$

$\rm \: = \:1 + \dfrac{1}{2} - 2$

$\rm \: = \:\dfrac{1}{2} - 1$

$\rm \: = \:\dfrac{1 - 2}{2}$

$\rm \: = \: - \:\dfrac{1}{2}$

$\bf\implies \:\boxed{ \tt{ \: p\bigg[\dfrac{1}{2} \bigg] = \: - \: \dfrac{1}{2} }}$

So, Now Consider,

$\rm :\longmapsto\:p(2) - p( - 1) + p\bigg[\dfrac{1}{2} \bigg]$

$\rm \: = \:16 - 1 - \dfrac{1}{2}$

$\rm \: = \:15 - \dfrac{1}{2}$

$\rm \: = \: \dfrac{30 - 1}{2}$

$\rm \: = \: \dfrac{29}{2}$

Hence,

$\purple{\bf\implies \:\boxed{ \tt{ \: \:p(2) + p( - 1) + p\bigg[\dfrac{1}{2} \bigg] = \: \dfrac{29}{2} \: }}}$