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CAT 1993 — QA Question 32

Basics (Functions)Easy
Passage / Data

Answer the following questions based on the information given below:

ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C.

The maximum possible value of y = min (1/2 – 3x2/4, 5x2/4) for the range 0 < x < 1 is

Answer & solution

  • A

    1/3

  • B

    1/2

  • C

    5/27

  • 5/16

Solution

So maximum possible value will be at the point of intersection of the two graphs.

∴ 12-3x24=5x24⇒x2=14

Hence, required maximum value = 5x24=54×14=516.

Hence, option (d).

CAT 1993 QA Q32: The maximum possible value of y = min (1/2 &ndash; 3x 2 /4, 5x 2 /4) for the range 0 < x < 1 is — Solution | TheCATExam