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Definite Integration question

2004 · Shift 0 · Q87
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Definite Integration question

2004 · Shift 0 · Q87

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If f(x)=ex1+ex,I1=∫f(−a)f(a)xg{x(1−x)}dxf\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx}f(x)=1+exex​,I1​=f(−a)∫f(a)​xg{x(1−x)}dx and I2=∫f(−a)f(a)g{x(1−x)}dx,{I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,I2​=f(−a)∫f(a)​g{x(1−x)}dx, then the value of I2I1{{{I_2}} \over {{I_1}}}I1​I2​​ is
  1. A
    111
  2. B
    −3-3−3
  3. C
    −1-1−1
  4. D
    222
View written solutionFree

Correct answer: D

  1. Find the limits using the given function

Given f(x)=ex1+ex.f(x)=\frac{e^x}{1+e^x}.f(x)=1+exex​.

First compute: f(−a)=e−a1+e−a=11+ea,f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{1}{1+e^a},f(−a)=1+e−ae−a​=1+ea1​, f(a)=ea1+ea.f(a)=\frac{e^a}{1+e^a}.f(a)=1+eaea​.

Now observe that f(−a)+f(a)=11+ea+ea1+ea=1.f(-a)+f(a)=\frac{1}{1+e^a}+\frac{e^a}{1+e^a}=1.f(−a)+f(a)=1+ea1​+1+eaea​=1.

So if we denote α=f(−a),β=f(a),\alpha=f(-a),\qquad \beta=f(a),α=f(−a),β=f(a), then α+β=1.\alpha+\beta=1.α+β=1.

Hence the interval of integration is symmetric about 12\frac1221​.

  1. Write the integrals

We have I1=∫αβx g{x(1−x)} dx,I_1=\int_{\alpha}^{\beta} x\,g\{x(1-x)\}\,dx,I1​=∫αβ​xg{x(1−x)}dx, I2=∫αβg{x(1−x)} dx.I_2=\int_{\alpha}^{\beta} g\{x(1-x)\}\,dx.I2​=∫αβ​g{x(1−x)}dx.

We need to evaluate I2I1.\frac{I_2}{I_1}.I1​I2​​.

  1. Use the symmetry substitution

Consider in I1I_1I1​ the substitution x↦1−x.x\mapsto 1-x.x↦1−x. Then dx↦−dxdx\mapsto -dxdx↦−dx, and since α+β=1\alpha+\beta=1α+β=1, the limits interchange properly: x=α⇒1−x=β,x=β⇒1−x=α.x=\alpha \Rightarrow 1-x=\beta, \qquad x=\beta \Rightarrow 1-x=\alpha.x=α⇒1−x=β,x=β⇒1−x=α.

So

=\int_{\alpha}^{\beta} (1-x)g(x(1-x))\,dx.$$ This is because $$x(1-x)=(1-x)x,$$ so the argument of $g$ remains unchanged. --- 4. **Add the two expressions for $I_1$** Now, $$I_1=\int_{\alpha}^{\beta} xg(x(1-x))\,dx,$$ and also $$I_1=\int_{\alpha}^{\beta} (1-x)g(x(1-x))\,dx.$$ Adding, $$2I_1=\int_{\alpha}^{\beta} [x+(1-x)]g(x(1-x))\,dx.$$ Thus, $$2I_1=\int_{\alpha}^{\beta} g(x(1-x))\,dx=I_2.$$ Therefore, $$I_2=2I_1.$$ So, $$\frac{I_2}{I_1}=2.$$ --- 5. **Match with the options** The correct option is $$\boxed{2}.$$ So **Option D** is correct.
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