Question

If m and M are the minimum and maximum values of tan−1xtan-1x and the [] represents the greatest integer function then find the value of [−π2−m]+[M−π2]​

Answer 1

Answer:

-2

Step-by-step explanation:

As m > - π/2

     M < π/2

=>  [- π/2 - m] + [ M - π/2 ] = -1 -1 = -2

Answer 2

Answer:

Value of the given function is $M=5+\sqrt{20} \\ \,&m=5-\sqrt{20}$

Step-by-step explanation:

Given:

$F(x)=\frac{\tan ^{2} x+4 \tan x+9}{1+\tan ^{2} x}$, and $m and M$are minimum and maximum value.

To find: Value of $\pi^{2} -m+ M - \pi^{2}$

Solution:

So $\because \tan x=\frac{\sin x}{\cos x}$; so Replace given function in term of $\sin x$ and $\cos x$

$\begin{aligned}&\text { so } \\&F(x)=\frac{\frac{\sin ^{2} x}{\cos ^{2} x}+4 \frac{\sin x}{\cos x}+9}{1+\frac{\sin ^{2} x}{\cos ^{2} x}}=\frac{\sin ^{2} x+4 \sin x \cos x+9 \cos ^{2} x}{\cos ^{2} x+\sin ^{2} x} \\&F(x)=\frac{\sin ^{2} x+\cos ^{2} x+4 \sin x \cos x+8 \cos ^{2} x}{\cos ^{2} x+\sin ^{2} x} \\&\because \because \sin x+\cos ^{2} x=1\end{aligned}$

we get

$F(x)=1+4 \sin x \cos x+8 \cos ^{2} x \\ \because \quad \cos 2 x=2 \cos ^{2} x-1 \quad OR \\ 2 \cos ^{2} x=1+\cos 2 x \\ F(x)=1+2 \sin 2 x+4(\cos 2 x+1) \\ F(x)=1+2 \sin 2 x+4 \cos 2 x+4 \\ F(x)=5+2 \sin 2 x+4 \cos 2 x$

Now

${ so },\sqrt{a^{2}+b^{2}} & \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}} \\{ so }, \sqrt{4+16} & \leq 2 \sin 2 x+4 \cos 2 x \leq \sqrt{4+20}\\\end{aligned}$

Add '5'

$5-\sqrt{20} \leq 5+2 \sin 2 x+4 \cos 2 x \leq 5+\sqrt{20}$

so

$\begin{aligned}&M=5+\sqrt{20} \\&m=5-\sqrt{20}\end{aligned}$