JEE Main 2021MathematicsDefinite IntegrationHardMCQ

JEE Main 2021 — Definite Integration Question with Solution

JEE Main 2021 (20 Jul Shift 1)

Question

Let a be a positive real number such that ∫0aex-xdx=10e-9 where, [x] is the greatest integer less than or equal to x. Then, a is equal to:

Choose an option

Show full solutionCorrect option: B
Correct answer
B10+loge2

Step-by-step explanation

Given:

a>0

Let n≤a<n+1, n∈W

Then,

a=a+a

Here, a=n
Now,

∫0aex-xdx=10e-9

⇒∫0nexdx+∫naex-xdx=10e-9

⇒n∫01exdx+∫naex-ndx=10e-9

⇒ne-1+ea-n-1=10e-9

⇒ne-n+ea-n-1=10e-9

⇒ne-n-1+eae-n=10e-9

On comparing, we get

n=10

and,

-n-1+eae-n=-9

⇒-10-1+ea-10=-9

⇒ea-10=2

⇒a-10logee=loge2

⇒a=10+loge2

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About this question

This is a previous-year question from JEE Main 2021, covering the Definite Integration 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.