Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
If 11+21=ab+cd\sqrt{11+\sqrt{21}} = \sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{c}{d}}, where a,b,c,da, b, c, d are natural numbers with gcd⁡(a,b)=gcd⁡(c,d)=1\gcd(a,b) = \gcd(c,d) = 1, then a+b+c+da+b+c+d is equal to:
A21\sqrt{21}
B26\sqrt{26}
C2626correct
D2121
Solution
Step 1: Express the radical with a doubled surd
11+21=22+2212=22+2212\sqrt{11+\sqrt{21}} = \sqrt{\frac{22+2\sqrt{21}}{2}} = \frac{\sqrt{22+2\sqrt{21}}}{\sqrt{2}}
 Step 2: Denest the inner radical Seeking m+n=22m+n = 22 and mn=21mn = 21 gives m=21m = 21, n=1n = 1:
22+221=(21+1)2  ⟹  22+221=21+122 + 2\sqrt{21} = (\sqrt{21}+1)^2 \implies \sqrt{22+2\sqrt{21}} = \sqrt{21}+1
 Step 3: Write in the required form
11+21=21+12=212+12\sqrt{11+\sqrt{21}} = \frac{\sqrt{21}+1}{\sqrt{2}} = \sqrt{\frac{21}{2}} + \sqrt{\frac{1}{2}}
 Step 4: Identify the constants a=21a = 21, b=2b = 2, c=1c = 1, d=2d = 2, with gcd⁡(21,2)=1\gcd(21,2) = 1 and gcd⁡(1,2)=1\gcd(1,2) = 1.
a+b+c+d=21+2+1+2=26a+b+c+d = 21+2+1+2 = 26
 Answer: (3)
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