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Application of Integration question

2012 · Shift 1 · Q33
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  5. /2012 · Shift 1 · Q33

Application of Integration question

2012 · Shift 1 · Q33

JEE AdvancedMathematicsApplication of IntegrationMultiple correct+4 / −1
Let SSS be the area of the region enclosed by y=e−x2y = {e^{ - {x^2}}}y=e−x2, y=0y=0y=0, x=0x=0x=0, and x=1x=1x=1; then
  1. A
    S≥1eS \ge {1 \over e}S≥e1​
  2. B
    S≥1−1eS \ge 1 - {1 \over e}S≥1−e1​
  3. C
    S≤14(1+1e)S \le {1 \over 4}\left( {1 + {1 \over {\sqrt e }}} \right)S≤41​(1+e​1​)
  4. D
    S≤12+1e(1−12)S \le {1 \over {\sqrt 2 }} + {1 \over {\sqrt e }}\left( {1 - {1 \over {\sqrt 2 }}} \right)S≤2​1​+e​1​(1−2​1​)
View written solutionFree

Correct answer: A, D, B

  1. Interpret the area

The required area is

S=∫01e−x2 dx.S=\int_0^1 e^{-x^2}\,dx.S=∫01​e−x2dx.

We must check which given inequalities are true.

  1. Useful fact about the function

Let f(x)=e−x2,0≤x≤1.f(x)=e^{-x^2}, \quad 0\le x\le 1.f(x)=e−x2,0≤x≤1. Then f(x)f(x)f(x) is decreasing on [0,1][0,1][0,1] because

f′(x)=−2xe−x2≤0.f'(x)=-2x e^{-x^2}\le 0.f′(x)=−2xe−x2≤0.

Also,

f(0)=1,f(1)=1e.f(0)=1, \qquad f(1)=\frac1e.f(0)=1,f(1)=e1​.
  1. Check option A: S≥1eS\ge \frac1eS≥e1​

Since e−x2e^{-x^2}e−x2 is decreasing on [0,1][0,1][0,1], its minimum value on this interval is at x=1x=1x=1, namely 1e\frac1ee1​. Hence,

e−x2≥1efor all x∈[0,1].e^{-x^2}\ge \frac1e \quad \text{for all } x\in[0,1].e−x2≥e1​for all x∈[0,1].

Integrating from 000 to 111,

S=∫01e−x2 dx≥∫011e dx=1e.S=\int_0^1 e^{-x^2}\,dx \ge \int_0^1 \frac1e\,dx=\frac1e.S=∫01​e−x2dx≥∫01​e1​dx=e1​.

So A is correct.

  1. Check option B: S≥1−1eS\ge 1-\frac1eS≥1−e1​

We use the inequality e−t≥1−tfor all t≥0.e^{-t}\ge 1-t \quad \text{for all } t\ge 0.e−t≥1−tfor all t≥0. Put t=x2t=x^2t=x2. Then

Therefore,

S=∫01e−x2 dx≥∫01(1−x2) dx=[x−x33]01=1−13=23.S=\int_0^1 e^{-x^2}\,dx \ge \int_0^1 (1-x^2)\,dx =\left[x-\frac{x^3}{3}\right]_0^1=1-\frac13=\frac23.S=∫01​e−x2dx≥∫01​(1−x2)dx=[x−3x3​]01​=1−31​=32​.

Now compare:

23>1−1e\frac23 > 1-\frac1e32​>1−e1​

because 1/e≈0.36791/e\approx 0.36791/e≈0.3679, so

and 23≈0.6667\frac23\approx 0.666732​≈0.6667. Thus

So B is correct.

  1. Check option C: S≤14(1+1e)S\le \frac14\left(1+\frac1{\sqrt e}\right)S≤41​(1+e​1​)

We estimate the RHS numerically:

14(1+1e)≈14(1+0.6065)≈0.4016.\frac14\left(1+\frac1{\sqrt e}\right) \approx \frac14(1+0.6065) \approx 0.4016.41​(1+e​1​)≈41​(1+0.6065)≈0.4016.

But

S≥23≈0.6667,S\ge \frac23 \approx 0.6667,S≥32​≈0.6667,

from option B's proof. Hence

S≤0.4016S \le 0.4016S≤0.4016

is impossible. So C is false.

  1. Check option D: S≤12+1e(1−12)S\le \frac1{\sqrt2}+\frac1{\sqrt e}\left(1-\frac1{\sqrt2}\right)S≤2​1​+e​1​(1−2​1​)

Let a=12.a=\frac1{\sqrt2}.a=2​1​. Since f(x)=e−x2f(x)=e^{-x^2}f(x)=e−x2 is decreasing,

S=∫01e−x2 dx=∫0ae−x2 dx+∫a1e−x2 dx.S=\int_0^1 e^{-x^2}\,dx =\int_0^a e^{-x^2}\,dx+\int_a^1 e^{-x^2}\,dx.S=∫01​e−x2dx=∫0a​e−x2dx+∫a1​e−x2dx.

Now on [0,a][0,a][0,a], we have e−x2≤1,e^{-x^2}\le 1,e−x2≤1, so

∫0ae−x2 dx≤a=12.\int_0^a e^{-x^2}\,dx \le a=\frac1{\sqrt2}.∫0a​e−x2dx≤a=2​1​.

On [a,1][a,1][a,1], since the function is decreasing,

e−x2≤e−a2=e−1/2=1e.e^{-x^2}\le e^{-a^2}=e^{-1/2}=\frac1{\sqrt e}.e−x2≤e−a2=e−1/2=e​1​.

Hence,

∫a1e−x2 dx≤1e(1−a)=1e(1−12).\int_a^1 e^{-x^2}\,dx \le \frac1{\sqrt e}(1-a) =\frac1{\sqrt e}\left(1-\frac1{\sqrt2}\right).∫a1​e−x2dx≤e​1​(1−a)=e​1​(1−2​1​).

Adding,

S≤12+1e(1−12).S\le \frac1{\sqrt2}+\frac1{\sqrt e}\left(1-\frac1{\sqrt2}\right).S≤2​1​+e​1​(1−2​1​).

So D is correct.

  1. Final conclusion

Correct options are:

A,  B,  D\boxed{A,\;B,\;D}A,B,D​
  1. Comparison with stored correct answer

Stored correct answer: A,D,BA, D, BA,D,B

This is the same set as A,B,DA, B, DA,B,D. So the derived answer agrees with the stored answer.

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