JEE Main 2021MathematicsBasic of MathematicsEasyMCQ

JEE Main 2021 — Basic of Mathematics Question with Solution

JEE Main 2021 (16 Mar Shift 1)

Question

If for x∈0,π2,log10sinx+log10cosx=-1 and log10sinx+cosx=12log10n-1,n>0, then the value of n is equal to : 

Choose an option

Show full solutionCorrect option: B
Correct answer
B12

Step-by-step explanation

Given x∈0,π2

log10sinx+log10cosx=-1

⇒log10sinx.cosx=-1

⇒sinx.cosx=110   ...(1)

log10sinx+cosx=12log10n-1

⇒sinx+cosx=10log10n-12=10log10n-log1010=n10

By squaring we get, 

sin2x+cos2x+2sinxcosx=n10

1+2sinx.cosx=n10

⇒1+15=n10⇒n=12

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About this question

This is a previous-year question from JEE Main 2021, covering the Basic of Mathematics chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.