Question

If $ f(\sin \,2x) = \dfrac{{(2\tan x + {{\sec }^2}x)(1 + \cos 2x)}}{2} $ , then determine the range of $ f(t) $ . If range is $ [a,b] $ , then $ b = ? $

Answer 1

Hint: Here, we see that domain of ‘f’ is $ \sin 2x $ . Therefore, to find range of f(t) we first use formula of $ (1 + \cos 2x) $ and then writing remaining terms in term of basic T-ratio of $ \sin x\,\,and\,\,\cos x $ and simplifying in such a way that right hand side can be written in term of domain of f or as in term of $ \sin 2x. $
Use $ 2\sin x\cos x = \sin 2x $ and $ \dfrac{{1 + \cos 2x}}{2} = {\cos ^2}x $ .

Complete step-by-step answer:
From given problem we have,
 $ f(\sin 2x) = \dfrac{{\left( {2\tan x + {{\sec }^2}x} \right)(1 + \cos 2x)}}{2} $
Also, we know that $ \dfrac{{1 + \cos 2x}}{2} = {\cos ^2}x $
Or we can write it as
 $ 2{\cos ^2}x = 1 + \cos 2x $
Using this in the right hand side of the above formed equation. We have
 $
  f(\sin 2x) = \dfrac{{\left( {2\tan x + {{\sec }^2}x} \right)\left( {2{{\cos }^2}x} \right)}}{2} \\
 \Rightarrow f\left( {\sin 2x} \right) = \left( {2\tan x + {{\sec }^2}x} \right)\left( {{{\cos }^2}x} \right) \\
  $
Simplifying the right hand side in terms of $ \sin x\,\,and\,\,\cos x $ .
$ f(\sin 2x) = \left( {\dfrac{{2\sin x}}{{\cos \,x}} + \dfrac{1}{{\cos {\,^2}x}}} \right).\,{\cos ^2}x $
Taking the L.C.M. of above formed equation.
 $
  f(\sin 2x) = \left( {\dfrac{{2\sin x\cos x + 1}}{{{{\cos }^2}x}}} \right).{\cos ^2}x \\
 \Rightarrow f\left( {\sin 2x} \right) = 2\sin x.\cos x + 1 \\
\Rightarrow f\left( {\sin 2x} \right) = \sin 2x + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because \sin 2x = 2.\sin x.\cos x} \right\} \\
  $
Now, taking $ \sin 2x = t $ in above equation. We have,
 $ f\left( t \right) = t + 1 $
But, we know that for every sine function:
 $ \sin 2x\, \in [ - 1,1] $
Or we can write it as
 $ \Rightarrow - 1 \leqslant \sin 2x \leqslant 1 $
Adding 1 to the above equation. We have,
 $
  \Rightarrow - 1 + 1 \leqslant 1 + \sin 2x \leqslant 1 + 1 \\
\Rightarrow 0 \leqslant 1 + t \leqslant 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because \sin 2x = t} \right\} \\
\Rightarrow 0 \leqslant f(t) \leqslant 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because f(t) = 1 + t} \right\} \;
 $
Then, from above we see that range of ‘f(t)’ is [0 , 2]
But the range of ‘f (t)’ is [a, b].
Therefore, on comparing we have b = 2.
Hence, from above we see that required value of b is $ 2. $

Note: Domain is an independent set of those values for a given function which on substitution always gives real value of result but set of value of range is depending upon the value of the set of domains as it is necessary that all range must have domain.