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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 39-40

Consider the following for the next two (02) items that follow:

2x^{2} + 3x - \alpha = 0

has roots

-2

and

\beta

while the equation

x^{2} - 3mx + 2m^{2} = 0

has both roots positive, where

\alpha > 0

and

\beta > 0

Question : 40 of 120

Marks: +1, -0

{\rm{\beta }}\sqrt {\rm{m}}

Solution:

Concept:

a, b, c

are in GP, then

b^2 = a \cdot c

Explanation:

Given that

\beta

2

2m

are in GP.

Using the GP property:

2^2 = \beta \cdot (2m)

4 = 2\beta m

Dividing by

2

gives

\beta m = 2

From the first equation (provided earlier), we know

\beta = \frac{1}{2}

Substitute

\beta = \frac{1}{2}

into

\beta m = 2

\frac{1}{2} \cdot m = 2

⇒

m = 4

Now compute

\beta \sqrt{m}

\beta \sqrt{m} = \frac{1}{2} \cdot \sqrt{4} = \frac{1}{2} \cdot 2 = 1

Answer:

The value of

\beta \sqrt{m}

1

, which corresponds to option A.

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