Verified Solution • 2025–26 Session moderate

QJEEM18Mabf6a: JEE Main 2018 Mathematics - Class 12 JEEM18Mabf6a Solution: If        and then

If    I1=∫01e−xcos⁡2x dx;{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;

   I2=∫01e−x2cos⁡2x dx{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx  and

I3=∫01e−x3dx;{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx; then

Multiple Choice Options

  • A I2 > I3 > I1
  • B I2 > I1 > I3
  • C I3 > I2 > I1
  • D I3 > I1 > I2

Step-by-Step Verified Solution

Given,

I1=∫01e−xcos⁡2x dx;{I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;

I2=∫01e−x2cos⁡2x dx{I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx and

I3=∫01e−x3dx{I_3} = \int_0^1 {{e^{ - {x^3}}}} dx

For x ∈\in (0, 1)

⇒\Rightarrow   x > x2 or −- x < −- x2

and x2 > x3 or −- x2 < −- x3

∴\therefore e−x2{e^{ - {x^2}}} < e−x3{e^{ - {x^3}}} and e−x{e^{ - x}} < e−x2{e^{ - {x^2}}}

⇒\Rightarrow {e^{ - x}} &lt; {e^{ - {x^2}}} &lt; {e^{ - {x^3}}}

⇒\Rightarrow {e^{ - {x^3}}} &gt; {e^{ - {x^2}}} &gt; {e^{ - x}}

⇒\Rightarrow {I_3} &gt; {I_2} &gt; {I_1}

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