JEE Main 2022MathematicsApplication of DerivativesEasyMCQ

JEE Main 2022 — Application of Derivatives Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

If the angle made by the tangent at the point x0,y0 on the curve x=12t+sintcost,
y=121+sint2,0<t<π2, with the positive x-axis is π3, then y0 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
C27

Step-by-step explanation

Given, 

Curve x=12t+sintcost

y=121+sint2

Now dydx=dydtdxdt

tanπ3=ddt121+sint2ddt12t+sintcost

3=12×21+sintcost121+cos2t

3=21+sint·cost2cos2t

3=1+sintcost=cost2+sint22cos2t2-sin2t2

3=tant2+π4

π3=t2+π4 ⇒t2=π12 ⇒ t=π6

Now putting t=π6 in y=121+sint2

as x0,y0 lie on curve

So y0=121+sinπ62

=12322=27

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About this question

This is a previous-year question from JEE Main 2022, covering the Application of Derivatives 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.