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Question Number 204632 by mr W last updated on 23/Feb/24

	Answered by witcher3 last updated on 23/Feb/24

	$x > 1; x = \frac{1}{\sin (t)}; t \in] 0, \frac{π}{2} [$ $\Rightarrow \frac{1}{\sin (t)} + \frac{\frac{1}{\sin (t)}}{\sqrt{\frac{1}{\sin^{2} (t)} - 1}} = \frac{1}{\sin (t)} + \frac{1}{\cos (t)} = 2 \sqrt{2}$ $cauchy shwartz ({(\frac{1}{\sqrt{\sin (t)}})}^{2} + {(\frac{1}{\sqrt{\cos (t)}})}^{2}) ({(\sqrt{\sin (t)})}^{2} + (\sqrt{\cos^{2} (t)}) ⩾ {(1 + 1)}^{2}$ $\Rightarrow \frac{1}{\sin (t)} + \frac{1}{\cos (t)} ⩾ \frac{4}{\sqrt{2} \sin (t + \frac{π}{4})} ⩾ \frac{4}{\sqrt{2}} = 2 \sqrt{2}$ $equality if only if$ $\sin (t) = \cos (t) \Rightarrow t = \frac{π}{4} \Rightarrow x = \sqrt{2}$
	Answered by MM42 last updated on 23/Feb/24

	$l e t f (x) = x + \frac{x}{\sqrt{x^{2} - 1}} & D_{f} = (1, \infty)$ $f^{'} (x) = 1 - \frac{1}{(x^{2} - 1) \sqrt{x^{2} - 1}}$ $\Rightarrow {m i n}_{f} = (\sqrt{2}, 2 \sqrt{2})$
	Answered by Rasheed.Sindhi last updated on 24/Feb/24
	$Algebraic Solution 1+(1/( (√(x^2 −1)) ))=((2(√2))/x) ((1/( (√(x^2 −1)) )))^2 =(((2(√2))/x)−1)^2 (1/(x^2 −1))=(8/x^2 )+1−((4(√2))/x) (1/(x^2 −1))=((x^2 −4(√2) x+8)/x^2 ) x^4 −4(√2) x^3 +6x^2 +4(√2) x−8=0 x^2 −4(√2) x+6+((4(√2) )/x)+(1/x^2 )−(9/x^2 )=0 (x^2 +(1/x^2 ))−4(√2) (x−(1/x))+6−(9/x^2 )=0 (x^2 +(1/x^2 )−2)−4(√2) (x−(1/x))+8−(9/x^2 )=0 (x−(1/x))^2 −4(√2) (x−(1/x))+8−(9/x^2 )=0 (x−(1/x))^2 −4(√2) (x−(1/x))+(2(√2) )^2 =(9/x^2 ) (x−(1/x)−2(√2) )^2 =((3/x))^2 x−(1/x)−2(√2) =±(3/x) x^2 −1−2(√2) x=±3 x^2 −2(√2) x−1∓3=0 x^2 −2(√2) x−4=0 ∣ x^2 −2(√2) x+2=0 { ((x=((2(√2) ±(√(8+16)))/2))),((x=((2(√2) ±(√(8−8)))/2))) :} { ((x=((2(√2) ±2(√6))/2)=(√2) ±(√6) (invalid))),((x=((2(√2) )/2)=(√2) ✓ )) :}$
	$Algebraic Solution$ $1 + \frac{1}{\sqrt{x^{2} - 1}} = \frac{2 \sqrt{2}}{x}$ ${(\frac{1}{\sqrt{x^{2} - 1}})}^{2} = {(\frac{2 \sqrt{2}}{x} - 1)}^{2}$ $\frac{1}{x^{2} - 1} = \frac{8}{x^{2}} + 1 - \frac{4 \sqrt{2}}{x}$ $\frac{1}{x^{2} - 1} = \frac{x^{2} - 4 \sqrt{2} x + 8}{x^{2}}$ $x^{4} - 4 \sqrt{2} x^{3} + 6 x^{2} + 4 \sqrt{2} x - 8 = 0$ $x^{2} - 4 \sqrt{2} x + 6 + \frac{4 \sqrt{2}}{x} + \frac{1}{x^{2}} - \frac{9}{x^{2}} = 0$ $(x^{2} + \frac{1}{x^{2}}) - 4 \sqrt{2} (x - \frac{1}{x}) + 6 - \frac{9}{x^{2}} = 0$ $(x^{2} + \frac{1}{x^{2}} - 2) - 4 \sqrt{2} (x - \frac{1}{x}) + 8 - \frac{9}{x^{2}} = 0$ ${(x - \frac{1}{x})}^{2} - 4 \sqrt{2} (x - \frac{1}{x}) + 8 - \frac{9}{x^{2}} = 0$ ${(x - \frac{1}{x})}^{2} - 4 \sqrt{2} (x - \frac{1}{x}) + {(2 \sqrt{2})}^{2} = \frac{9}{x^{2}}$ ${(x - \frac{1}{x} - 2 \sqrt{2})}^{2} = {(\frac{3}{x})}^{2}$ $x - \frac{1}{x} - 2 \sqrt{2} = \pm \frac{3}{x}$ $x^{2} - 1 - 2 \sqrt{2} x = \pm 3$ $x^{2} - 2 \sqrt{2} x - 1 \mp 3 = 0$ $x^{2} - 2 \sqrt{2} x - 4 = 0 ∣ x^{2} - 2 \sqrt{2} x + 2 = 0$ ${\begin{cases} x = \frac{2 \sqrt{2} \pm \sqrt{8 + 16}}{2} \\ x = \frac{2 \sqrt{2} \pm \sqrt{8 - 8}}{2} \end{cases}$ ${\begin{cases} x = \frac{2 \sqrt{2} \pm 2 \sqrt{6}}{2} = \sqrt{2} \pm \sqrt{6} (i n v a l i d) \\ x = \frac{2 \sqrt{2}}{2} = \sqrt{2} ✓ \end{cases}$
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