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UPTET Paper 2 Math and Science Model Test 2

Section: Mathematics and Science

Question : 142 of 150

Marks: +1, -0

The minimum value of (x – 2) (x – 9) is

Solution:

Concept:

A quadratic expression can be written in perfect square form to find its minimum value.

Explanation:

First, expand the product:

(x-2)(x-9) = x^2 - 11x + 18

Complete the square:

x^2 - 11x = (x - \frac{11}{2})^2 - (\frac{11}{2})^2 = (x - \frac{11}{2})^2 - \frac{121}{4}

Add the constant 18:

(x - \frac{11}{2})^2 - \frac{121}{4} + 18 = (x - \frac{11}{2})^2 - \frac{121}{4} + \frac{72}{4}

Combine the fractions:

(x - \frac{11}{2})^2 - \frac{49}{4}

A square term is always

\ge 0

, so its minimum value is 0.

Thus the minimum of the whole expression is

0 - \frac{49}{4} = -\frac{49}{4}

Answer:

Option D:

-\frac{49}{4}

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