Basics & LogarithmshardFree

Basics & Logarithms — JEE Maths practice question

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

Question
If x=4−23x = \sqrt{4-2\sqrt{3}} and y=9−45y = \sqrt{9-4\sqrt{5}}, then the value of (5 x−3 y)2(\sqrt{5}\,x - \sqrt{3}\,y)^2 is equal to (a−bc)(a - b\sqrt{c}), where a,b,ca, b, c are co-prime and cc is an odd integer. Then a+b+ca+b+c is equal to:
A3636correct
B3434
C3333
D1919
Solution
Step 1: Remove the radicals
4−23=3−23+1=(3−1)2  ⟹  x=3−14 - 2\sqrt{3} = 3 - 2\sqrt{3} + 1 = (\sqrt{3}-1)^2 \implies x = \sqrt{3}-1
9−45=5−45+4=(5−2)2  ⟹  y=5−29 - 4\sqrt{5} = 5 - 4\sqrt{5} + 4 = (\sqrt{5}-2)^2 \implies y = \sqrt{5}-2
 (Both principal square roots are positive since 3>1\sqrt{3} > 1 and 5>2\sqrt{5} > 2.) Step 2: Form 5 x−3 y\sqrt{5}\,x - \sqrt{3}\,y
5 x=15−5,3 y=15−23\sqrt{5}\,x = \sqrt{15}-\sqrt{5}, \quad \sqrt{3}\,y = \sqrt{15}-2\sqrt{3}
5 x−3 y=23−5\sqrt{5}\,x - \sqrt{3}\,y = 2\sqrt{3} - \sqrt{5}
 Step 3: Square the result
(23−5)2=12−415+5=17−415(2\sqrt{3}-\sqrt{5})^2 = 12 - 4\sqrt{15} + 5 = 17 - 4\sqrt{15}
 Step 4: Identify the constants a=17a = 17, b=4b = 4, c=15c = 15, which are pairwise co-prime with cc odd.
a+b+c=17+4+15=36a+b+c = 17+4+15 = 36
 Answer: (1)
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