1x=tanθ
$\textbf{If} $\mathbf{f(x)=sin\left(\frac{2x}{1+x^2}\right)+cos\left(\frac{4x}{1+x^2}\right)+1}$\\\\\\ \textbf{Then find Max. and Min. value of f(x)}.
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9 Answers
1708Actually sagnik I dont have any idea about solution . Can you show your process.
1max SHUD BE 17/8 ..
put cos(4x/..)= 1-2 sin^2(2x/...)
then u get a quad in sin(2x/..)
its max=17/8
11Yeah its 17/8.(calc error)
Put 2x/1+x^2=t
Thus,
f(x)= sin(t)+cos(2t)+1= sin(t)+1- 2sin^2t+1= -2sin^2(t)+sin(t)+2
This is a quadratic of sin(t). So, max value exists for sint=1/4 (<1)
i.e 17/8.
1min shud be obtained by
replacing sin(2x/1+x^2) by -sin1 in above quadratic
{for this, u need to analyse the graph of quadratic, and find range of
2x/1+x^2, which is [-1,1], and sin x is increasing in [-1,1]
hence min is at 2x/1+x^2= -1}
1708$Here I am just Describing Seoni Solution.\\\\ Let $\mathbf{\frac{2x}{1+x^2}= t}$. OR $\mathbf{tx^2-2x+t=0}$\\\\ If Given equation has real Roots then $\mathbf{D\geq 0}$\\\\ So $\mathbf{4-4t^2\geq0\Leftrightarrow -1\leq t\leq 1}$\\\\ So Here We have to find Max. and Min. value of\\\\ $\mathbf{f(t)=sint+2cost+1}$. Where $\mathbf{t\in [-1,1]}$\\\\ $\mathbf{f(t)=-2sin^2t+sint+2}$\\\\ $\mathbf{f(t)=\frac{17}{8}-2\left(sint-\frac{1}{4}\right)^2}$\\\\ So $\boxed{\boxed{\mathbf{f_{Max.}=\frac{17}{8}}$ at $\mathbf{sint=\frac{1}{4}}}}$\\\\ and $\mathbf{f_{Min.}=\frac{17}{8}-2.\left(-sin1-\frac{1}{4}\right)^2}$\\\\ OR $\boxed{\boxed{\mathbf{f_{Min.}=\frac{17}{8}-2.\left(sin1+\frac{1}{4}\right)^2}}}$ at $\mathbf{t=-1}$\\\\\\ bcz $\mathbf{f(t)=sint}$ is an Increasing function in $\mathbf{t\in [-1,1]}$