The number of real roots of the equation 5 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.Rearranging,we get $|t - 1| = t^2 - 2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.Rearranging,we get $|t - 1| = t^2 - 2t - 5$.Let $g(t) = |t - 1|$ and $f(t) = t^2 - 2t - 5$.We look for intersections where $t > 0$.Case $1$: $t \ge 1$,then $t - 1 = t^2 - 2t - 5 \Rightarrow t^2 - 3t - 4 = 0 \Rightarrow (t - 4)(t + 1) = 0$. Since $t > 0$,$t = 4$. This gives $2^x = 4 \Rightarrow x = 2$.Case $2$: $0 Thus,there is only $1$ real root.

The number of real roots of the equation $5 + |2^x - 1| = 2^x(2^x - 2)$ is

Similar Questions

Solution of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

If $a(x+y) = b(x-y) = 2ab$,then the value of $2(x^2 + y^2)$ is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $35x^{2} - 53x + 20 = 0$
$II.$ $56y^{2} - 97y + 42 = 0$

The least integer $k$ which makes the roots of the equation ${x^2} + 5x + k = 0$ imaginary is

Vedclass Test Series

Exam Paper Generator

Online Exam Module