The number of real roots of (x - 1) (x - 2) (x - 3) (x - 4) = 3 is

Question

The number of real roots of (x - 1) (x - 2) (x - 3) (x - 4) = 3 is (A) 2 (B) 0 (C) 4 (D) none of these

Tardigrade Admin · Accepted Answer

beginarrayl (x+(1/x))3+(x+(1/x))=0 ⇒(x+(1/x))[(x+(1/x))2+1] endarray Since, (x+(1/x))2+1>0 Therefore, (x+(1/x))=0 ⇒ x 2+1=0 Therefore, number of real roots are zero.

Q. The number of real roots of
$(x - 1) (x - 2) (x - 3) (x - 4) = 3$ is

$2$

$0$

$4$

none of these

$(x + \frac{1}{x})^{3} + (x + \frac{1}{x}) = 0 \Rightarrow (x + \frac{1}{x}) [(x + \frac{1}{x})^{2} + 1]$
Since, $(x + \frac{1}{x})^{2} + 1 > 0$
Therefore, $(x + \frac{1}{x}) = 0$
$\Rightarrow x^{2} + 1 = 0$
Therefore, number of real roots are zero.

Q. The number of real roots of (x−1)(x−2)(x−3)(x−4)=3 is

2

0

4

none of these

(x+x1​)3+(x+x1​)=0⇒(x+x1​)[(x+x1​)2+1]​ Since, (x+x1​)2+1>0 Therefore, (x+x1​)=0 ⇒x2+1=0 Therefore, number of real roots are zero.

Q. The number of real roots of
$(x - 1) (x - 2) (x - 3) (x - 4) = 3$ is

$2$

$0$

$4$

$(x + \frac{1}{x})^{3} + (x + \frac{1}{x}) = 0 \Rightarrow (x + \frac{1}{x}) [(x + \frac{1}{x})^{2} + 1]$
Since, $(x + \frac{1}{x})^{2} + 1 > 0$
Therefore, $(x + \frac{1}{x}) = 0$
$\Rightarrow x^{2} + 1 = 0$
Therefore, number of real roots are zero.