Question

Match List I with List II

LIST I		LIST II
A.	The maximum value of the function \(f(x)=25x-\frac{5x^2}{2}+7\) in [-1,6] is	I.	24
B.	The minimum value of the function \(f(x)=2x^3-15x^2+36x+1\) in [1,5] is	II.	\(\frac{1}{16}\)
C.	The maximum value of the function \(f(x)=\frac{x}{2}-x^2\) in [0,1] is	III.	\(\frac{139}{2}\)
D.	The least value of the function \(f(x)=\frac{9}{x+3}+x\) in [-7,1], \(x\ne-3\) is	IV.	\(-\frac{37}{4}\)

Choose the correct answer from the options given below:

• A. А-I, В-III, C-II, D-IV
• B. A-III, B-I, С-IV, D-II
• C. A-II, B-IV, С-III, D-I
• D. A-III, B-I, С-II, D-IV

Answer 1

The Correct Option is D

To solve the problem, we need to find the extrema (maximum or minimum values) of the given functions within the specified intervals and then match List I with List II accordingly.

• For A: The function \(f(x)=25x-\frac{5x^2}{2}+7\) is defined in the interval \([-1,6]\). To find the extrema, first calculate the derivative: \(f'(x)=25-5x\). Set \(f'(x)=0\) to find critical points: \(25-5x=0\Rightarrow x=5\). Evaluate \(f(x)\) at \(x=-1\), \(x=5\), and \(x=6\).
  \(f(-1)=25(-1)-\frac{5(-1)^2}{2}+7=24.5\)
  \(f(5)=25(5)-\frac{5(5)^2}{2}+7=\frac{139}{2}\)
  \(f(6)=25(6)-\frac{5(6)^2}{2}+7=-1\)
  The maximum value is \(\frac{139}{2}\), corresponding to III.
• For B: Consider \(f(x)=2x^3-15x^2+36x+1\) on \([1,5]\). Find the derivative: \(f'(x)=6x^2-30x+36\). Solve \(f'(x)=0\):
  \(6x^2-30x+36=0\Rightarrow x^2-5x+6=0 \Rightarrow (x-2)(x-3)=0\Rightarrow x=2,3\). Check \(f\) at \(x=1,2,3,5\):
  \(f(1)=24\)
  \(f(2)=25\)
  \(f(3)=19\)
  \(f(5)=26\)
  The minimum value is 19, corresponding to I.
• For C: The function \(f(x)=\frac{x}{2}-x^2\) in \([0,1]\) gives \(f'(x)=\frac{1}{2}-2x\). Set \(f'(x)=0\):
  \(\frac{1}{2}-2x=0\Rightarrow x=\frac{1}{4}\). Evaluate \(f(x)\) at \(x=0,\frac{1}{4},1\):
  \(f(0)=0\)
  \(f\left(\frac{1}{4}\right)=\frac{1}{16}\)
  \(f(1)=-\frac{1}{2}\)
  The maximum value is \(\frac{1}{16}\), corresponding to II.
• For D: Evaluate \(f(x)=\frac{9}{x+3}+x\) over \([-7,1]\), avoiding \(x=-3\). Calculate the derivative: \(f'(x)=1-\frac{9}{(x+3)^2}\). Solve \(f'(x)=0\):
  \(1-\frac{9}{(x+3)^2}=0\Rightarrow (x+3)^2=9\Rightarrow x+3=\pm 3\Rightarrow x=0,-6\). Verify \(f(x)\) at \(x=-7,-6,0,1\):
  \(f(-7)=-1.5\)
  \(f(-6)=-\frac{37}{4}\)
  \(f(0)=3\)
  \(f(1)=4\)
  The least value is \(-\frac{37}{4}\), corresponding to IV.

Therefore, the matching is: A-III, B-I, C-II, D-IV.