The number of integers satisfying the inequal | vedclass

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

Question

(B) Let $t = \sqrt{\log_3(x) - 1}$. Since the square root is defined,$\log_3(x) - 1 \ge 0$,so $\log_3(x) \ge 1$,which implies $x \ge 3$. Also,$t \ge 0

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $t = \sqrt{\log_3(x) - 1}$. Since the square root is defined,$\log_3(x) - 1 \ge 0$,so $\log_3(x) \ge 1$,which implies $x \ge 3$. Also,$t \ge 0$.Then $\log_3(x) = t^2 + 1$.The inequality becomes $t + \frac{\frac{1}{2} \cdot 3 \log_3(x)}{-1} + 2 > 0$.Substituting $\log_3(x) = t^2 + 1$:$t - \frac{3}{2}(t^2 + 1) + 2 > 0$$t - \frac{3}{2}t^2 - \frac{3}{2} + 2 > 0$$t - \frac{3}{2}t^2 + \frac{1}{2} > 0$Multiply by $-2$: $3t^2 - 2t - 1 Factoring the quadratic: $(3t + 1)(t - 1) This holds for $-\frac{1}{3} Substituting back $t = \sqrt{\log_3(x) - 1}$:$0 \le \sqrt{\log_3(x) - 1} $0 \le \log_3(x) - 1 $1 \le \log_3(x) $3^1 \le x $3 \le x The integers satisfying this are $3, 4, 5, 6, 7, 8$. There are $6$ such integers.

The number of integers satisfying the inequality $\sqrt{\log_3(x) - 1} + \frac{\frac{1}{2}\log_3(x^3)}{\log_3(\frac{1}{3})} + 2 > 0$ is

Similar Questions

The quadratic equation whose one root is $2 - \sqrt{3}$ will be

Number of rational roots of the equation $x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$ is equal to

If the roots of the equation $ax^2 + bx + c = 0$ are $l$ and $2l$,then:

Let $f(x) = x^2 - x + k - 2$,$k \in R$. Then the complete set of values of $k$ for which $y = |f(|x|)|$ is non-derivable at $5$ distinct points is:

For the equation $|x^2| + |x| - 6 = 0$,the roots are

Vedclass Test Series

Exam Paper Generator

Online Exam Module