MCQ
સમીકરણ $e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^{x}+1=0$ ના વાસ્તવિક બીજની સંખ્યા મેળવો.
- A$1$
- B$6$
- C$4$
- D$2$
$\Rightarrow\left(e^{3 x}-1\right)^{2}-e^{x}\left(e^{3 x}-1\right)=12 e^{2 x}$
$\left(e^{3 x}-1\right)^{2}\left(e^{x}-e^{-x}-e^{-2 x}\right)=12$
$\Rightarrow \underbrace{e^{x}-e^{-x}-e^{-2 x}}_{\text {increasing(let } f(x))}=\frac{12}{\underbrace{e^{3 x}-1}_{\text {decreasing }(l e t g(x))}}$
$\Rightarrow$ No. of real roots $=2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
વિધાન-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$