Find the value of the expression
$(x^{5} + x^{2} + x) at x = 1.$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 3
$At x = 1,$

$x^{5} + x^{2} + x = 1^{5} + 1^{2} + 1 = 1 + 1 + 1 = 3$

Suggest Corrections

Similar questions

(x^{2} + x + 1) at x = - 1.

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 4 x}{x^{2}} f o r x < 0 k f o r x = 0 \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} f o r x > 0 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ at x = 0

y = | x^{2} - | x | | a t x = - 2.

{tan}^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}

{tan}^{- 1} \frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}

a t x = 0

f (x) = 1 + x + x^{2} + - - + x^{50} a t x = 1

Join BYJU'S Learning Program