Question

Match List I with List II

	LIST I		LIST II
A.	The minimum value of \(f(x)=8x²-4x+7\) is	I.	48
B.	The maximum value of \(f(x) = x+\frac{1}{x}, x < 0\) is	II.	13
C.	The maximum slope of the cure \(y = -2x^3+6x^2+7x+26\) is	III.	-2
D.	The minimum value of \(f(x) = x² +\frac{128}{x}\) is	IV.	\(\frac{13}{2}\)

Choose the correct answer from the options given below:

• A. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
• B. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
• C. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
• D. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Answer 1

The Correct Option is B

To solve this problem, we need to match items from LIST I with items from LIST II based on mathematical calculations.

A.	The minimum value of \(f(x)=8x²-4x+7\)

Find the minimum value by completing the square or using derivatives. The vertex form is \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex. Completing the square gives \(f(x) = 8((x-\frac{1}{4})²-\frac{1}{16})+7 = 8(x-\frac{1}{4})²+\frac{55}{2}\). The minimum value is \(\frac{55}{2}\), corresponding to IV. However, review the method to confirm matching is accurate.

B.	The maximum value of \(f(x) = x+\frac{1}{x}, x<0\)

Since \(x<0\), set \(f'(x)=1-\frac{1}{x^2}=0\). Solving gives \(x=-1\), yielding \(f(-1) = -1-1=-2\). This is the maximum value since \(x\) is negative. Corresponding to III.

C.	The maximum slope of the curve \(y = -2x^3+6x^2+7x+26\)

The slope is given by the derivative \(y'= -6x^2+12x+7\). Find where the derivative is maximum using \(y'' = -12x+12=0\). Solving gives \(x=1\), so \(y'(1) = -6+12+7=13\), corresponding to II.

D.	The minimum value of \(f(x) = x² +\frac{128}{x}\)

Use derivatives: \(f'(x)=2x-\frac{128}{x^2}=0\), solve for \(x^3=64\), gives \(x=4\). Then, \(f(4)=16+\frac{128}{4}=48\), corresponding to I.

Thus, the correct matches are: (A)-(IV), (B)-(III), (C)-(II), (D)-(I).