The number of real solutions of x7 + 5x3 + 3x + 1 = 0 is equal to………..

f ( x ) = x 7 + 5 x 3 + 3 x + 1 f ' ( x ) = 7 x 6 + 1 5 x 4 + 3 > 0 ∀ x ∈ R ∴ f ( x ) <!- [if gte vml 1]> for x ∞ -<!- [if gte vml 1]> <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->, f (x) ∞ <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->x = 0, f (x) = 1 ∴ f ( x ) = 0 has only one real root.

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Maths NCERT Exemplar Solutions Class 11th Chapter Seven Maths Class 11th Ncert Solutions Maths class 11th

The number of real solutions of x⁷ + 5x³ + 3x + 1 = 0 is equal to………..

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 weeks ago

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Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 1

Detailed Solution:

$f (x) = x^{7} + 5 x^{3} + 3 x + 1$

$f' (x) = 7 x^{6} + 15 x^{4} + 3 > 0 \forall x \in R$

$∴ f (x)$ is increasing

for x $\infty$ - , f (x) $\infty$

x = 0, f (x) = 1

$∴ f (x)$ = 0 has only one real root.

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