Exams Joint Entrance Examination Mathematics Application of Derivatives Question 9 Question 9 of 212 Medium MCQ
Let x = − 1 x=-1 x = − 1 and x = 2 x=2 x = 2 be the critical points of the function f ( x ) = x 3 + a x 2 + b log e ∣ x ∣ + 1 , x ≠ 0 f(x)={x}^{3}+a{x}^{2}+b{\log }_{e}|x|+1,x\ne 0 f ( x ) = x 3 + a x 2 + b log e ∣ x ∣ + 1 , x = 0 .
Let m m m and M respectively be the absolute minimum and the absolute maximum values of f f f in the interval [ − 2 , − 1 2 ] [-2,-\frac{1}{2}] [ − 2 , − 2 1 ] . Then ∣ M + m ∣ |M+m| ∣ M + m ∣ is equal to
( ( ( Take log e 2 = 0.7 ) : {\log }_{e}2=0.7): log e 2 = 0.7 ) :
Solutions · 1 Post your solution Official solution
3 − 2 a − b = 0 12 + 4 a + b 2 = 0 } a = − 9 2 b = 12 \left.\begin{gathered} 3-2 a-b=0 \\ 12+4 a+\frac{b}{2}=0 \end{gathered}\right\} \begin{aligned} & a=\frac{-9}{2} \\ & b=12\end{aligned} 3 − 2 a − b = 0 12 + 4 a + 2 b = 0 ⎭ ⎬ ⎫ a = 2 − 9 b = 12
\[ ∴ f ( x ) = x 3 − 9 2 x 2 + 12 ln ∣ x ∣ + 1 f ( − 1 ) = − 1 − 9 2 + 1 = − 9 2 = − 4.5 f ( − 2 ) = − 8 − 18 + 12 ln 2 + 1 = − 25 + 12 ln 2 = − 16.6 f ( − 1 2 ) = − 1 8 − 9 8 + 12 ln ( 1 2 ) + 1 = − 8.5 ∣ M + m ∣ = ∣ − 16.6 − 4.5 ∣ = 21.1 \begin{aligned}
& \therefore f(x)=x^3-\frac{9}{2} x^2+12 \ln |x|+1 \\
& f(-1)=-1-\frac{9}{2}+1=-\frac{9}{2}=-4.5 \\
& f(-2)=-8-18+12 \ln 2+1 \\
& \quad=-25+12 \ln 2=-16.6 \\
& f\left(-\frac{1}{2}\right)=-\frac{1}{8}-\frac{9}{8}+12 \ln \left(\frac{1}{2}\right)+1=-8.5 \\
& |M+m|=|-16.6-4.5|=21.1
\end{aligned} ∴ f ( x ) = x 3 − 2 9 x 2 + 12 ln ∣ x ∣ + 1 f ( − 1 ) = − 1 − 2 9 + 1 = − 2 9 = − 4.5 f ( − 2 ) = − 8 − 18 + 12 ln 2 + 1 = − 25 + 12 ln 2 = − 16.6 f ( − 2 1 ) = − 8 1 − 8 9 + 12 ln ( 2 1 ) + 1 = − 8.5 ∣ M + m ∣ = ∣ − 16.6 − 4.5∣ = 21.1 \] Answer the question to unlock