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If a = 0.125 then what is value of 4a24a+1+3a\sqrt{4a^2-4a+1}+3a ?

A1.500

B1.125

C1.250

D1.225

Answer:

B. 1.125

Read Explanation:

Solution:

Given:

a = 0.125

Concept used:

(A - B)2 = A2 - 2AB + B2

Calculation:

Now,

4a24a+1\sqrt{4a^2-4a+1}

(14a+4a2)⇒\sqrt{(1-4a+4a^2)}

(12a)2⇒\sqrt{(1-2a)^2}

(12a)⇒(1-2a)

So,

4a24a+1+3a\sqrt{4a^2-4a+1}+3a

⇒ (1 - 2a) + 3a

⇒ 1 + a

⇒ 1.125

∴ The required value of  4a24a+1+3a\sqrt{4a^2-4a+1}+3a is 1.125.


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