Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 60873 by Kunal12588 last updated on 26/May/19

Commented by Prithwish sen last updated on 26/May/19

$i) \log_{2} ({4 x}^{2} - x - 1) - \log_{2} (x^{2} + 1) > 0$ $\frac{{4 x}^{2} - x - 1}{x^{2} + 1} > 1$ ${3 x}^{2} - x - 2 > 0$ $(3 x + 2) (x - 1) > 0$ $either or$ $3 x + 2 > 0 3 x + 2 < 0$ $x - 1 > 0 and x - 1 < 0$ $\Rightarrow x > 1 \Rightarrow x < - \frac{2}{3}$ $the soln . is$ $1 < x < - \frac{2}{3}$
Commented by maxmathsup by imad last updated on 27/May/19

$8) l e t D t h e s e t o f d e f i n i t i o n x \in D \Leftrightarrow 4 x^{2} - x - 1 > 0$ $Δ = 1 - 4 (- 1) = 5 \Rightarrow x_{1} = \frac{1 + \sqrt{5}}{8} a n d x_{2} = \frac{1 - \sqrt{5}}{8}$ $s o D =] - \infty, \frac{1 - \sqrt{5}}{8} [\cup] \frac{1 + \sqrt{5}}{8}, + \infty [$ $(e) \Leftrightarrow \frac{l n (4 x^{2} - x - 1)}{l n (x^{2} + 1)} > 1 \Rightarrow l n (4 x^{2} - x - 1) > l n (x^{2} + 1) \Rightarrow$ $4 x^{2} - x - 1 > x^{2} + 1 \Rightarrow 3 x^{2} - x - 2 > 0 \to Δ = 1 - 4 (- 6) = 25 \Rightarrow$ $x_{1} = \frac{1 + 5}{6} = 1 a n d x_{2} = \frac{1 - 5}{6} = - \frac{2}{3} \Rightarrow$ $S = (] - \infty, - \frac{2}{3} [\cup] 1, + \infty [) \cap (] - \infty, \frac{1 - \sqrt{5}}{8} [\cup] \frac{1 + \sqrt{5}}{8}, + \infty [) = . . . .$
Commented by Rasheed.Sindhi last updated on 27/May/19

$S i r p r i t h w i s h s e n$ $\log_{2} ({4 x}^{2} - x - 1) - \log_{2} (x^{2} + 1) > 0$ $\overset{?}{\Rightarrow} \frac{{4 x}^{2} - x - 1}{x^{2} + 1} > 1$
Commented by Prithwish sen last updated on 27/May/19

$∵ \log_{2} ({4 x}^{2} - x - 1) > \log_{2} (x^{2} + 1)$ $\Rightarrow ({4 x}^{2} - x - 1) > (x^{2} + 1)$
Commented by Kunal12588 last updated on 28/May/19

$w e l l t h e a n s w e r s a r e$ $8 . x \in (- \infty, \frac{- 14}{3}) \cup (4, \infty)$ $9 . I {d o n}^{'} t k n o w$ $10 . x \in {4} \cup [0, 1]$
	Answered by tanmay last updated on 27/May/19

	$9) {l o g}_{a} 2 \times {l o g}_{2} b ⩾ 1$ $\frac{l o g 2}{l o g a} \times \frac{l o g b}{l o g 2} ⩾ 1 [{l o g}_{a} b w h e n b > 0 a n d a > 0]$ $l o g b ⩾ l o g a$ $b ⩾ a$ $(x^{2} - x - 2) ⩾∣ 6 x ∣$ $n o w (x^{2} - x - 2) > 0$ $(x + 1) (x - 2) > 0$ $f (x) = (x + 1) (x - 2)$ $w h e n x > 2 f (x) > 0$ $w h e n x < - 1 f (x) > 0$ $w h e n 2 > x > - 1 f (x) < 0$ $s o x \notin (- 1, 2)$ $∣ 6 x ∣> 0$ $s o v a l u e o f x c a n n o t l i e b e t w e e n - 1 a n d 2$ $s o l v e$ $x^{2} - x - 2 ⩾∣ 6 x ∣$ $x^{2} - x - 2 - ∣ 6 x ∣⩾ 0$ $w h e n x > 2$ $x^{2} - x - 2 - 6 x ⩾ 0$ $x^{2} - 7 x - 2 ⩾ 0$ $x = \frac{7 + \sqrt{49 + 4 \times 1 \times 2}}{2} = 7.275$ $w h e n x < - 1$ $x^{2} - x - 2 + 6 x ⩾ 0$ $x^{2} + 5 x - 2 ⩾ 0$ $x = \frac{- 5 - \sqrt{25 + 8}}{2} = - 5.372$
	Answered by tanmay last updated on 27/May/19

	$10) {l o g}_{3} a = {l o g}_{9} b [a > 0 b > 0]$ $\frac{l o g a}{l o g 3} = \frac{l o g b}{l o g 9}$ $\frac{l o g a}{l o g 3} = \frac{l o g b}{2 l o g 3}$ ${l o g a}^{2} = l o g b$ $a^{2} = b$ ${(\sqrt{x} + ∣ \sqrt{x} - 1 ∣)}^{2} = 4 \sqrt{x} - 3 + 4 ∣ \sqrt{x} - 1 ∣$ $p = \sqrt{x} q =∣ \sqrt{x} - 1 ∣$ ${(p + q)}^{2} = 4 p - 3 + 4 q$ ${(p + q)}^{2} - 4 (p + q) + 3 = 0$ $(p + q - 3) (p + q - 1) = 0$ $p + q = 3 a n d p + q = 1$ $\sqrt{x} + ∣ \sqrt{x} - 1 ∣= 3 s o x = 4 (w h e n (\sqrt{x} - 1) > 0 x > 1)$ $b u t w h e n \sqrt{x} - 1 < 0 x < 1$ $t h e n \sqrt{x} + ∣ \sqrt{x} - 1 ∣$ $= \sqrt{x} - \sqrt{x} + 1$ $= 1 w h e n x < 1$ $s o x = 4 i s t h e s o l u t i o n$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum