Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 64 Question 64 of 401 Medium NUMERICAL
For t > − 1 t>-1 t > − 1 , let α t {\alpha }_{t} α t and β t {\beta }_{t} β t be the roots of the equation
( ( t + 2 ) 1 / 7 − 1 ) x 2 + ( ( t + 2 ) 1 / 6 − 1 ) x + ( ( t + 2 ) 1 / 21 − 1 ) = 0 l i m t → − 1 + α t = a l i m t → − 1 + β t = b ((t+2{)}^{1/7}-1){x}^{2}+((t+2{)}^{1/6}-1)x+((t+2{)}^{1/21}-1)=0\underset{t\to -{1}^{+}}{lim}{\alpha }_{t}=a\underset{t\to -{1}^{+}}{lim}{\beta }_{t}=b (( t + 2 ) 1/7 − 1 ) x 2 + (( t + 2 ) 1/6 − 1 ) x + (( t + 2 ) 1/21 − 1 ) = 0 t → − 1 + l im α t = a t → − 1 + l im β t = b
then 72 ( a + b ) 2 72(a+b{)}^{2} 72 ( a + b ) 2 is equal to \_\_\_\_\_\_\_\_\_\_\_.
Answer: 98
Solutions · 1 Post your solution Official solution
a + b = l i m t → − 1 + ( α + β ) = l i m t → − 1 + − ( t + 2 ) 1 6 − 1 ( t + 2 ) 1 7 − 1 t + 2 = y a + b = l i m y → 1 + y 1 / 6 − 1 y 1 / 7 − 1 = 7 6 72 ( a + b ) 2 = 72 49 36 = 98 \begin{gathered} a+b=\underset{t\to -{1}^{+}}{lim}(\alpha +\beta )=\underset{t\to -{1}^{+}}{lim}-\frac{(t+2{)}^{\frac{1}{6}}-1}{(t+2{)}^{\frac{1}{7}}-1} \\ t+2=y \\ a+b=\underset{y\to {1}^{+}}{lim}\frac{{y}^{1/6}-1}{{y}^{1/7}-1}=\frac{7}{6} \\ 72(a+b{)}^{2}=72\frac{49}{36}=98 \end{gathered} a + b = t → − 1 + l im ( α + β ) = t → − 1 + l im − ( t + 2 ) 7 1 − 1 ( t + 2 ) 6 1 − 1 t + 2 = y a + b = y → 1 + l im y 1/7 − 1 y 1/6 − 1 = 6 7 72 ( a + b ) 2 = 72 36 49 = 98 Answer the question to unlock