The number of distinct real roots of the equation x7-7 x-2=0 is

Question

The number of distinct real roots of the equation x7-7 x-2=0 is (A) 5 (B) 7 (C) 1 (D) 3

Tardigrade Admin · Accepted Answer

x7-7 x-2=0 x7-7 x=2 f(x)=x7-7 x (odd) y=2 f(x)=x(x2-71 / 3)(x4+x2 ⋅ 71 / 3+72 / 3) f prime(x)=7(x6-1)=7(x2-1)(x4+x2+1) f prime(x)=0 ⇒ x=± 1 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2f2af15470ebd802b99e2bc8addc94c4-.png" /> f(x) = 2 has 3 real distinct solution.

Q. The number of distinct real roots of the equation $x^{7} - 7 x - 2 = 0$ is

5

7

1

3

$x^{7} - 7 x - 2 = 0$
$x^{7} - 7 x = 2$
$f (x) = x^{7} - 7 x$ (odd) & $y = 2$
$f (x) = x (x^{2} - 7^{1/3}) (x^{4} + x^{2} \cdot 7^{1/3} + 7^{2/3})$
$f^{'} (x) = 7 (x^{6} - 1) = 7 (x^{2} - 1) (x^{4} + x^{2} + 1)$
$f^{'} (x) = 0 \Rightarrow x = \pm 1$

$f (x) = 2$ has $3$ real distinct solution.

Q. The number of distinct real roots of the equation x7−7x−2=0 is

5

7

1

3

x7−7x−2=0 x7−7x=2 f(x)=x7−7x (odd) & y=2 f(x)=x(x2−71/3)(x4+x2⋅71/3+72/3) f′(x)=7(x6−1)=7(x2−1)(x4+x2+1) f′(x)=0⇒x=±1 f(x)=2 has 3 real distinct solution.

Q. The number of distinct real roots of the equation $x^{7} - 7 x - 2 = 0$ is

$x^{7} - 7 x - 2 = 0$
$x^{7} - 7 x = 2$
$f (x) = x^{7} - 7 x$ (odd) & $y = 2$
$f (x) = x (x^{2} - 7^{1/3}) (x^{4} + x^{2} \cdot 7^{1/3} + 7^{2/3})$
$f^{'} (x) = 7 (x^{6} - 1) = 7 (x^{2} - 1) (x^{4} + x^{2} + 1)$
$f^{'} (x) = 0 \Rightarrow x = \pm 1$

$f (x) = 2$ has $3$ real distinct solution.