Skip to main content

Visit this link for 1 : 1 LIVE Classes.

Solve for $\left\{ {x,y,z} \right\}$,

$x+xy+xyz=12$
$y+yz+xyz=21$
$z+xz+xyz=30$

It may be possible to isolate x, y, z in terms of xyz (say t) as follows:

$x(1+y+yz)=12$ [From the 1st equation]

Let us find y+yz in terms of t from the 2nd equation.

$y+yz=21-xyz=21-t$

Now, this y+yz can be substituted in the 1st step above.

$x(1+21-t)=12$

$\therefore x = \frac{{12}}{{22 - t}}$

Also, $y(1+z+xz)=21$ [From the 2nd equation]

$z+xz=30-t$ [From the 3rd equation]

So, $y(1+30-t)=21$

$\therefore y = \frac{{21}}{{31 - t}}$

Putting these values of x and y in the 1st equation $x+xy+t=12$ or $x(1+y)+t=12$,

$\left( {\frac{{12}}{{22 - t}}} \right)\left( {1 + \frac{{21}}{{31 - t}}} \right) + t = 12$

$ \Rightarrow \left( {\frac{{12}}{{22 - t}}} \right)\left( {\frac{{52 - t}}{{31 - t}}} \right) + t = 12$

$ \Rightarrow 12(52 - t) + t(22 - t)(31 - t) = 12(22 - t)(31 - t)$

$ \Rightarrow 624 - 12t + 682t - 53{t^2} + {t^3} = 8184 - 636t + 12{t^2}$

$ \Rightarrow {t^3} - 65{t^2} + 1306t - 7560 = 0$

t=10 satisfies the above equation.

$ \Rightarrow {t^2}(t - 10) - 55t(t - 10) + 756(t - 10) = 0$

$ \Rightarrow (t - 10)({t^2} - 55t + 756) = 0$

$ \Rightarrow (t - 10)(t - 27)(t - 28) = 0$

t has three possible values: 10, 27, 28.

x and y have already been obtained in terms of t. z can also be found in terms of t in the same fashion.

$ z(1 + 12 - t) = 30$

$ \Rightarrow z = \frac{{30}}{{13 - t}}$

$\left\{ {x,y,z} \right\} \equiv \left\{ {\frac{{12}}{{22 - t}},\frac{{21}}{{31 - t}},\frac{{30}}{{13 - t}}} \right\}$

$ \equiv \left\{ {1,1,10} \right\},\left\{ { - \frac{{12}}{5},\frac{{21}}{4}, - \frac{{15}}{7}} \right\},\left\{ { - 2,7, - 2} \right\}$

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)