Test Index

Manipal Entrance Test 2014 Math Paper

© examsnet.com

Question : 12 of 80

Marks: +1, -0

If

a \cos^{3} \alpha + 3 a \cos \alpha \sin^{2} \alpha = m

a \sin^{3} \alpha + 3 a \cos^{2} \alpha \sin \alpha = n

(m+n)^{\frac{2}{3}} + (m-n)^{\frac{2}{3}}

is equal to

$2 a^{3}$
$2 a^{\frac{1}{3}}$
$2 a^{\frac{2}{3}}$
$2 a^{3}$

Solution:

Given relations are

a \cos^{3} \alpha + 3 a \cos \alpha \sin^{2} \alpha = m

....(i)

a \sin^{3} \alpha + 3 a \cos^{2} \alpha \sin \alpha = n

.....(ii)

On adding Eq. (i) and (ii) we get

(m+n) = a \cos^{3} \alpha + 3 a \cos \alpha \sin^{2} \alpha

+ 3 a \cos^{2} \alpha \sin \alpha + a \sin^{3} \alpha

= a (\cos \alpha + \sin \alpha)^{3}

Similarly on subtracting Eq. (ii) from Eq (i) we get m

m - n = a (\cos \alpha - \sin \alpha)^{3}

Now,

(m+n)^{\frac{2}{3}} + (m-n)^{\frac{2}{3}}

= a^{\frac{2}{3}} \{ (\cos \alpha + \sin \alpha)^{2} + (\cos \alpha - \sin \alpha)^{2} \}

= a^{\frac{2}{3}} \{ 2 (\cos^{2} \alpha + \sin^{2} \alpha) \} = 2 a^{\frac{2}{3}}

© examsnet.com

Go to Question:

Prev Question

Next Question