Exams Joint Entrance Examination Mathematics Application of Derivatives Question 10 Question 10 of 212 Medium MCQ
Let a > 0 a>0 a > 0 . If the function f ( x ) = 6 x 3 − 45 a x 2 + 108 a 2 x + 1 f(x)=6{x}^{3}-45a{x}^{2}+108{a}^{2}x+1 f ( x ) = 6 x 3 − 45 a x 2 + 108 a 2 x + 1 attains its local maximum and minimum values at the points x 1 {x}_{1} x 1 and x 2 {x}_{2} x 2 respectively such that x 1 x 2 = 54 {x}_{1}{x}_{2}=54 x 1 x 2 = 54 , then a + x 1 + x 2 a+{x}_{1}+{x}_{2} a + x 1 + x 2 is equal to :
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f ( x ) = 6 x 3 − 45 a x 2 + 108 a 2 x + 1 f ′ ( x ) = 0 \begin{gathered} f(x)=6{x}^{3}-45a{x}^{2}+108{a}^{2}x+1 \\ {f}^{'}(x)=0 \end{gathered} f ( x ) = 6 x 3 − 45 a x 2 + 108 a 2 x + 1 f ′ ( x ) = 0
x 1 x 2 = 108 a 2 18 = 54 ⇒ a 2 = 9 ⇒ a = 3 a + x 1 + x 2 = 3 + 90 6 = 3 + 15 = 18 \begin{gathered} {x}_{1}{x}_{2}=\frac{108{a}^{2}}{18}=54 \\ \Rightarrow {a}^{2}=9\Rightarrow a=3 \\ a+{x}_{1}+{x}_{2}=3+\frac{90}{6}=3+15=18 \end{gathered} x 1 x 2 = 18 108 a 2 = 54 ⇒ a 2 = 9 ⇒ a = 3 a + x 1 + x 2 = 3 + 6 90 = 3 + 15 = 18 Answer the question to unlock