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 solutions where $t > 0$.Case $1$: $t \ge 1$.$t - 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - 3t - 4 = 0$ $\Rightarrow (t - 4)(t + 1) = 0$.Since $t \ge 1$,we have $t = 4$. Thus $2^x = 4 \Rightarrow x = 2$.Case $2$: $0 $-(t - 1) = t^2 - 2t - 5$ $\Rightarrow -t + 1 = t^2 - 2t - 5$ $\Rightarrow t^2 - t - 6 = 0$ $\Rightarrow (t - 3)(t + 2) = 0$.Neither $t = 3$ nor $t = -2$ satisfy $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

Which of the following is not a root of $f(x) = x^3 - 11x^2 + 36x - 36$?

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation $(-2+\sqrt{3})(|\sqrt{x}-3|) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0$, $x \ge 0$, then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

Let $\alpha, \alpha^2$ be the roots of $x^2 + x + 1 = 0$. Then the equation whose roots are $\alpha^{31}, \alpha^{62}$ is:

$\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 3 x^2 - 10 x - 24 = 0$. If $\alpha > \beta > \gamma$ and $\alpha^3 + 3 \beta^2 - 10 \gamma - 24 = 11 k$,then $k = $

Let $f(x)$ be a quadratic expression such that $f(0)+f(1)=0$. If $f(-2)=0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module