The set of real values of x- for which hx-1-2x2-4x4-6x6-100x100 is concave downward is

H(x)=1+2x^2+4x^4+6x^6+ ⋯+100x^100 h'(x)=0+2^2x+4^2x^3+6^2x^5+⋯+100^2x^99 h”(x)=2^2+4^2(3x^2)+6^2(5x^4)+⋯+100^2(99x^98) So, h”(x) is always positive for all real ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sign of Quadratic Expression

The set of re...

Question

The set of real values of $x$ , for which $h (x) = 1 + 2 x^{2} + 4 x^{4} + 6 x^{6} + \dots + 100 x^{100}$ is concave downward is

x \in ϕ

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

x \in R^{-}

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

x \in W

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

x \in R^{+}

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

Open in App

Solution

The correct option is A $x \in ϕ$
$h (x) = 1 + 2 x^{2} + 4 x^{4} + 6 x^{6} + \dots + 100 x^{100}$
$h^{'} (x) = 0 + 2^{2} x + 4^{2} x^{3} + 6^{2} x^{5} + \dots + 100^{2} x^{99}$
$h^{''} (x) = 2^{2} + 4^{2} (3 x^{2}) + 6^{2} (5 x^{4}) + \dots + 100^{2} (99 x^{98})$
So, $h^{''} (x)$ is always positive for all real values of $x$
Hence $h (x)$ is always concave upward

Suggest Corrections

Similar questions

Q. If

f (x) = 1 + 2 x^{2} + 4 x^{4} + 6 x^{6} + . . . + 100 x^{100}

is a polynomial in a real variable x, then

f (x)

has?

Q. The set of real values of

x

, for which

h (x) = 1 + 2 x^{2} + 4 x^{4} + 6 x^{6} + \dots + 100 x^{100}

is concave downward is

Q. Exhaustive set of values of

a

for which the function

f (x) = x^{4} + a x^{3} + \frac{3 x^{2}}{2} + 1

will be concave upwards along the entire real line is

Q. Which of the following is correct for the quadratic polynomial

f (x) = 2 x^{2} + 8 x - 12

Q. The set of real values of

x

satisfying

{log}_{1 / 2} (x^{2} - 6 x + 12) \geq - 2

Join BYJU'S Learning Program

The set of real values of x, for which h(x)=1+2x2+4x4+6x6+⋯+100x100 is concave downward is

The correct option is A x∈ϕh(x)=1+2x2+4x4+6x6+⋯+100x100 h′(x)=0+22x+42x3+62x5+⋯+1002x99 h′′(x)=22+42(3x2)+62(5x4)+⋯+1002(99x98) So, h′′(x) is always positive for all real values of x Hence h(x) is always concave upward

The set of real values of $x$ , for which $h (x) = 1 + 2 x^{2} + 4 x^{4} + 6 x^{6} + \dots + 100 x^{100}$ is concave downward is