AP ICET 2017 Shift 2 Paper

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Question : 80 of 200

Marks: +1, -0

(4+√{15})^{\;\;{3}/{2}}+(4-√{15})^{\;\;{3}/{2}}=p√{10}⇒p=

$2√2$
$7/2$
7
3

Solution:

4+√{15}={5}/{2}+{3}/{2}+2√{{5}/{2}}⋅√{{3}/{2}}=(√{{5}/{2}}+√{;{3}/{2}})^{2}

Similarly,

4-√15=(√{5/2}-√{3/2})^2

∴

(4+√{15})^{3/2}+(4-√{15})^{3/2}

=(√{{5}/{2}}+√{{3}/{2}})^{3}+(√{{5}/{2}}-√{{3}/{2}})^{3}

Using

a^3+b^3=(a+b)^3–3ab(a+b)

Let

a=√{\;\;{5}/{2}}+√{{3}/{2}}

and

b=√{\;\;{5}/{2}}-√{{3}/{2}}

a+b=2√{\;\;{5}/{2}}=√{10}

ab={5}/{2}-{3}/{2}=1

[Using

(A+B)(A–B)=A^2–B^2

]

∴(√{{5}/{2}}+√{{3}/{2}})^{3}+(√{{5}/{2}}-√{{3}/{2}})^{3}

=(√{10})^{3}-3×1×√{10}

=10√{10}-3√{10}=7√{10}

Now,

7√{10}=p√{10}

∴{p}=7

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