Number of integers satisfying inequality, sqr | vedclass

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

Question

$\Rightarrow \log _{3} x=t^{2}+1$

$	herefore 2|t|-3\left(1+t^{2}ight)+4>0$

$	herefore 3 t^{2}-3|t|+|t|-1<0$

$\Rightarrow|\mathrm{t}

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\Rightarrow \log _{3} x=t^{2}+1$

$	herefore 2|t|-3\left(1+t^{2}ight)+4>0$

$	herefore 3 t^{2}-3|t|+|t|-1<0$

$\Rightarrow|\mathrm{t}| \in\left(-\frac{1}{3}, 1ight) \Rightarrow \mathrm{t}^{2} \in[0,1)$

$\Rightarrow x \in[3,9) \Rightarrow 6$ integers

Number of integers satisfying inequality, $\sqrt {{{\log }_3}(x) - 1} + \frac{{\frac{1}{2}{{\log }_3}\,{x^3}}}{{{{\log }_3}\,\frac{1}{3}}} + 2 > 0$ is

Similar Questions

Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $P_n=\alpha^n-\beta^n, n \in N$. Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to :

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is

The number of real solutions of the equation $\mathrm{x}|\mathrm{x}+5|+2|\mathrm{x}+7|-2=0$ is .....................

The integer $'k'$, for which the inequality $x^{2}-2(3 k-1) x+8 k^{2}-7>0$ is valid for every $x$ in $R ,$ is

The number of real roots of the equation $x | x |-5| x +2|+6=0$, is