Basics & LogarithmshardFree

Basics & Logarithms — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

x = \sqrt{3-\sqrt{5}}

and

y = \sqrt{3+\sqrt{5}}

. If the value of the expression

x - y + 2x^2y + 2xy^2 - x^4y + xy^4

can be expressed in the form

\sqrt{p}+\sqrt{q}

(where

p, q \in \mathbb{N}

), then the value of

(p+q)

is:

450

160

\sqrt{160}

610

correct

Solution

Step 1: Compute the basic symmetric quantities

x^2 = 3-\sqrt{5}, \quad y^2 = 3+\sqrt{5}

x^2+y^2 = 6, \quad x^2y^2 = (3-\sqrt{5})(3+\sqrt{5}) = 4 \implies xy = 2

(x+y)^2 = x^2+y^2+2xy = 6+4 = 10 \implies x+y = \sqrt{10}

(x-y)^2 = x^2+y^2-2xy = 6-4 = 2 \implies x-y = -\sqrt{2}

(Here

x-y < 0

since

y > x

.) Step 2: Regroup the expression

x - y + 2x^2y + 2xy^2 - x^4y + xy^4 = (x-y) + 2xy(x+y) + xy(y^3-x^3)

Step 3: Evaluate each group

y^3 - x^3 = (y-x)(y^2+xy+x^2) = \sqrt{2}\,(6+2) = 8\sqrt{2}

(x-y) + 2xy(x+y) + xy(y^3-x^3) = -\sqrt{2} + 4\sqrt{10} + 2(8\sqrt{2}) = 15\sqrt{2} + 4\sqrt{10}

Step 4: Express in the required form

15\sqrt{2} = \sqrt{450}, \quad 4\sqrt{10} = \sqrt{160}

Therefore

p = 450

q = 160

, and

p+q = 610

. Answer: (4)

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