AP ICET 2019 Shift 1 Paper

Question : 122 of 200

Marks: +1, -0

The smallest value of x for which

({12}/x+36)(4-x^2)

= 0 is

Solution:

({12}/{x}+36)(4-x^{2})=0

⇒ {12}/{x}(1+3 x)(x+2)(x-2)=0

x={-1}/3,-2

or 2
The least value of

x = –2.

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