Question Number 77024 by Sathvik last updated on 02/Jan/20
![The possible value of p for which graph of the function f(x)=2p^2 − 3ptan x+tan^2 x+1 does not lie below x-axis for all x∈(((−Π)/2),(Π/2)) is (a)0 (b)4 (c)3 (d)8](https://www.tinkutara.com/question/Q77024.png)
$$\mathrm{The}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\mathrm{graph}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2p}^{\mathrm{2}} − \\ $$$$\mathrm{3ptan}\:\mathrm{x}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}\:\mathrm{does}\:\mathrm{not}\:\mathrm{lie}\:\mathrm{below}\: \\ $$$$\mathrm{x}-\mathrm{axis}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\left(\frac{−\Pi}{\mathrm{2}},\frac{\Pi}{\mathrm{2}}\right)\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\mathrm{0}\:\:\:\:\:\:\left(\mathrm{b}\right)\mathrm{4}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\mathrm{3}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\mathrm{8} \\ $$
Answered by MJS last updated on 03/Jan/20
![well, just test them all... I found −2≤p≤2 ⇒ (a) is the right answer f(x): y=tan^2 x −3ptan x +2p^2 +1 f′(x): y=2tan^3 x −3ptan^2 x +2tan x −3p= =(tan^2 x +1)(2tan x −3p) ⇒ tan x =(3/2)p insert in f(x) y=1−(p^2 /4)=0 ⇒ p=±2 ⇒ −2≤p≤2](https://www.tinkutara.com/question/Q77059.png)
$$\mathrm{well},\:\mathrm{just}\:\mathrm{test}\:\mathrm{them}\:\mathrm{all}… \\ $$$$\mathrm{I}\:\mathrm{found}\:−\mathrm{2}\leqslant{p}\leqslant\mathrm{2} \\ $$$$\Rightarrow\:\left({a}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer} \\ $$$${f}\left({x}\right):\:{y}=\mathrm{tan}^{\mathrm{2}} \:{x}\:−\mathrm{3}{p}\mathrm{tan}\:{x}\:+\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}'\left({x}\right):\:{y}=\mathrm{2tan}^{\mathrm{3}} \:{x}\:−\mathrm{3}{p}\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{2tan}\:{x}\:−\mathrm{3}{p}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{1}\right)\left(\mathrm{2tan}\:{x}\:−\mathrm{3}{p}\right) \\ $$$$\Rightarrow\:\mathrm{tan}\:{x}\:=\frac{\mathrm{3}}{\mathrm{2}}{p} \\ $$$$\mathrm{insert}\:\mathrm{in}\:{f}\left({x}\right) \\ $$$${y}=\mathrm{1}−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0}\:\Rightarrow\:{p}=\pm\mathrm{2} \\ $$$$\Rightarrow\:−\mathrm{2}\leqslant{p}\leqslant\mathrm{2} \\ $$