Question-112320

Question Number 112320 by mohammad17 last updated on 07/Sep/20

Commented by mohammad17 last updated on 07/Sep/20

h e l p m e s i r i n Q 1

Answered by john santu last updated on 07/Sep/20

(1) l e t 2^{\cos t} = m \Rightarrow \frac{1}{m - 1} > \frac{1}{1 - \frac{1}{2} m}

\frac{1}{m - 1} > \frac{2}{2 - m} \Rightarrow \frac{1}{m - 1} - \frac{2}{2 - m} > 0

\Rightarrow \frac{1}{m - 1} + \frac{2}{m - 2} > 0

\Rightarrow \frac{m - 2 + 2 m - 2}{(m - 1) (m - 2)} > 0

\Rightarrow \frac{3 m - 4}{(m - 1) (m - 2)} > 0

\Rightarrow 1 < m < \frac{4}{3} \cup m > 2

\Rightarrow 1 < 2^{\cos t} < \frac{4}{3} \cup 2^{\cos t} > 2

\Rightarrow 0 < \cos t < 2 - \log_{2} (3) \cup \cos t > 1

f o r \cos t > 1, t \in R, t = \emptyset

∴ 0 < \cos t < 0.415

Commented by mohammad17 last updated on 07/Sep/20

t h a n k y o u s i r

Answered by 1549442205PVT last updated on 08/Sep/20

Put 2^{cost} = x . Then \frac{1}{x - 1} > \frac{1}{1 - \frac{1}{2} x}

\Leftrightarrow \frac{1}{x - 1} > \frac{2}{2 - x} \Leftrightarrow \frac{1}{x - 1} - \frac{2}{2 - x} > 0

\frac{2 - x - 2 (x - 1)}{(x - 1) (2 - x)} > 0 \Leftrightarrow \frac{4 - 3 x}{(x - 1) (2 - x)} > 0

\Leftrightarrow \frac{3 x - 4}{(x - 1) (x - 2)} > 0 \Leftrightarrow \frac{x - 4 / 3}{(x - 1) (x - 2)} > 0

\Leftrightarrow x \in (1; \frac{4}{3}) \cup (2; + \infty)

a) x \in (1, \frac{}{}) \Leftrightarrow 1 <^{c o s t} < \frac{}{}

\Leftrightarrow 0 < cost < \frac{\ln (4 / 3)}{\ln 2} \approx 0.4150 = \cos 65 ° 29^{'}

\Leftrightarrow t \in (- 90 ° + k 360 °; - 65 ° 28^{'} + k 360) \cup (65 ° 28^{'} + k 360 °; 90 ° + k 360)

b) x \in (2; + \infty) \Leftrightarrow 2^{cost} > 2 \Leftrightarrow cost > 1

inequality has no roots

Thus, given inequality has roots are :

t \in (- 90 ° + k 360 °; - 65 ° 28^{'} + k 360) \cup

(65 ° 28^{'} + k 360 °; 90 ° + k 360 °)