Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 72 Question 72 of 401 Medium NUMERICAL
If α = l i m x → 0 + ( e tan x − e x tan x − x ) \alpha =\underset{x\to {0}^{+}}{lim}(\frac{{e}^{\sqrt{\tan x}}-{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}) α = x → 0 + l im ( t a n x − x e t a n x − e x ) and β = l i m x → 0 ( 1 + sin x ) 1 2 cot x \beta =\underset{x\to 0}{lim}(1+\sin x{)}^{\frac{1}{2}\cot x} β = x → 0 l im ( 1 + sin x ) 2 1 c o t x are the roots of the quadratic equation a x 2 + b x − e = 0 a{x}^{2}+bx-\sqrt{e}=0 a x 2 + b x − e = 0 , then 12 log e ( a + b ) 12{\log }_{e}(a+b) 12 log e ( a + b ) is equal to \_\_\_\_\_\_\_\_\_.
Answer: 6
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α = l i m x → 0 + e tan x − e x ( tan x − x ) = l i m x → 0 e x ( e tan x − x − 1 ) ( tan x − x ) = 1 β = l i m x → 0 ( 1 + sin x ) 1 2 cot x = l i m x → 0 e ( sin x ) ( 1 2 cot x ) = l i m x → 0 e 1 2 cos x = e 1 / 2 \begin{gathered} \alpha =\underset{x\to {0}^{+}}{lim}\frac{{e}^{\sqrt{\tan x}}-{e}^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\ =\underset{x\to 0}{lim}\frac{{e}^{\sqrt{x}}({e}^{\sqrt{\tan x}-\sqrt{x}}-1)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\ \beta =\underset{x\to 0}{lim}(1+\sin x{)}^{\frac{1}{2}\cot x}=\underset{x\to 0}{lim}{e}^{(\sin x)(\frac{1}{2}\cot x)} \\ =\underset{x\to 0}{lim}{e}^{\frac{1}{2}\cos x}={e}^{1/2} \end{gathered} α = x → 0 + l im ( tan x − x ) e t a n x − e x = x → 0 l im ( tan x − x ) e x ( e t a n x − x − 1 ) = 1 β = x → 0 l im ( 1 + sin x ) 2 1 c o t x = x → 0 l im e ( s i n x ) ( 2 1 c o t x ) = x → 0 l im e 2 1 c o s x = e 1/2
= e = − e a ⇒ a = − 1 = − b a = 1 + e = b = e + 1 ⇒ 12 ln ( a + b ) = 12 ln ( e + 1 − 1 ) = 12 ln ( e 1 / 2 ) = 6 \begin{gathered} =\sqrt{e}=\frac{-\sqrt{e}}{a}\Rightarrow a=-1 \\ =\frac{-b}{a}=1+\sqrt{e} \\ \, =b=\sqrt{e}+1 \\ \Rightarrow 12\ln (a+b)=12\ln (\sqrt{e}+1-1)=12\ln ({e}^{1/2})=6 \end{gathered} = e = a − e ⇒ a = − 1 = a − b = 1 + e = b = e + 1 ⇒ 12 ln ( a + b ) = 12 ln ( e + 1 − 1 ) = 12 ln ( e 1/2 ) = 6 Answer the question to unlock