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Definite Integration question

2024 · 31 Jan · Shift 2 · Q40
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  5. /2024 · 31 Jan · Shift 2 · Q40

Definite Integration question

2024 · 31 Jan · Shift 2 · Q40

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f,g:(0,∞)→Rf, g:(0, \infty) \rightarrow \mathbb{R}f,g:(0,∞)→R be two functions defined by f(x)=∫−xx(∣t∣−t2)e−t2dtf(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d tf(x)=−x∫x​(∣t∣−t2)e−t2dt and g(x)=∫0x2t1/2e−tdtg(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d tg(x)=0∫x2​t1/2e−tdt. Then, the value of 9(f(log⁡e9)+g(log⁡e9))9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)9(f(loge​9​)+g(loge​9​)) is equal to :
  1. A
    10
  2. B
    9
  3. C
    8
  4. D
    6
View written solutionFree

Correct answer: C

  1. Evaluate f(x)f(x)f(x).

Given

f(x)=∫−xx(∣t∣−t2)e−t2 dt.f(x)=\int_{-x}^{x} (|t|-t^2)e^{-t^2}\,dt.f(x)=∫−xx​(∣t∣−t2)e−t2dt.

For symmetric limits, split the integrand into even/odd parts.

  • ∣t∣|t|∣t∣ is even,
  • t2t^2t2 is even,
  • hence (∣t∣−t2)(|t|-t^2)(∣t∣−t2) is even,
  • and e−t2e^{-t^2}e−t2 is also even.

So the whole integrand is even. Therefore,

f(x)=2∫0x(t−t2)e−t2 dt.f(x)=2\int_0^x (t-t^2)e^{-t^2}\,dt.f(x)=2∫0x​(t−t2)e−t2dt.

Thus,

f(x)=2(∫0xte−t2 dt−∫0xt2e−t2 dt).f(x)=2\left(\int_0^x t e^{-t^2}\,dt-\int_0^x t^2 e^{-t^2}\,dt\right).f(x)=2(∫0x​te−t2dt−∫0x​t2e−t2dt).

Now compute these terms.

For the first integral,

∫0xte−t2 dt=−12e−t2∣0x=12(1−e−x2).\int_0^x t e^{-t^2}\,dt = -\frac12 e^{-t^2}\Big|_0^x = \frac12(1-e^{-x^2}).∫0x​te−t2dt=−21​e−t2​0x​=21​(1−e−x2).

So,

2∫0xte−t2 dt=1−e−x2.2\int_0^x t e^{-t^2}\,dt = 1-e^{-x^2}.2∫0x​te−t2dt=1−e−x2.

For the second integral, use integration by parts through the identity

ddt(te−t2)=e−t2−2t2e−t2.\frac{d}{dt}(t e^{-t^2})=e^{-t^2}-2t^2 e^{-t^2}.dtd​(te−t2)=e−t2−2t2e−t2.

Hence,

2t2e−t2=e−t2−ddt(te−t2),2t^2 e^{-t^2}=e^{-t^2}-\frac{d}{dt}(t e^{-t^2}),2t2e−t2=e−t2−dtd​(te−t2),

so

∫0xt2e−t2 dt=12∫0xe−t2 dt−12 te−t2∣0x=12∫0xe−t2 dt−12xe−x2.\int_0^x t^2 e^{-t^2}\,dt=\frac12\int_0^x e^{-t^2}\,dt-\frac12\, t e^{-t^2}\Big|_0^x =\frac12\int_0^x e^{-t^2}\,dt-\frac12 x e^{-x^2}.∫0x​t2e−t2dt=21​∫0x​e−t2dt−21​te−t2​0x​=21​∫0x​e−t2dt−21​xe−x2.

Therefore,

2∫0xt2e−t2 dt=∫0xe−t2 dt−xe−x2.2\int_0^x t^2 e^{-t^2}\,dt=\int_0^x e^{-t^2}\,dt-x e^{-x^2}.2∫0x​t2e−t2dt=∫0x​e−t2dt−xe−x2.

So,

f(x)=(1−e−x2)−(∫0xe−t2 dt−xe−x2)=1−e−x2−∫0xe−t2 dt+xe−x2.f(x)= (1-e^{-x^2})-\left(\int_0^x e^{-t^2}\,dt-x e^{-x^2}\right) =1-e^{-x^2}-\int_0^x e^{-t^2}\,dt+x e^{-x^2}.f(x)=(1−e−x2)−(∫0x​e−t2dt−xe−x2)=1−e−x2−∫0x​e−t2dt+xe−x2.
  1. Evaluate g(x)g(x)g(x).

Given

g(x)=∫0x2t1/2e−t dt.g(x)=\int_0^{x^2} t^{1/2}e^{-t}\,dt.g(x)=∫0x2​t1/2e−tdt.

Put t=u2t=u^2t=u2, so dt=2u dudt=2u\,dudt=2udu, t1/2=ut^{1/2}=ut1/2=u, and limits become 000 to xxx:

g(x)=∫0xu e−u2(2u) du=2∫0xu2e−u2 du.g(x)=\int_0^x u\,e^{-u^2}(2u)\,du=2\int_0^x u^2 e^{-u^2}\,du.g(x)=∫0x​ue−u2(2u)du=2∫0x​u2e−u2du.

From above,

2∫0xu2e−u2 du=∫0xe−u2 du−xe−x2.2\int_0^x u^2 e^{-u^2}\,du = \int_0^x e^{-u^2}\,du - x e^{-x^2}.2∫0x​u2e−u2du=∫0x​e−u2du−xe−x2.

Hence,

g(x)=∫0xe−u2 du−xe−x2.g(x)=\int_0^x e^{-u^2}\,du - x e^{-x^2}.g(x)=∫0x​e−u2du−xe−x2.
  1. Add f(x)f(x)f(x) and g(x)g(x)g(x).

Using the expressions obtained,

f(x)+g(x)=(1−e−x2−∫0xe−t2 dt+xe−x2)+(∫0xe−t2 dt−xe−x2).f(x)+g(x)=\left(1-e^{-x^2}-\int_0^x e^{-t^2}\,dt+x e^{-x^2}\right) +\left(\int_0^x e^{-t^2}\,dt-x e^{-x^2}\right).f(x)+g(x)=(1−e−x2−∫0x​e−t2dt+xe−x2)+(∫0x​e−t2dt−xe−x2).

The integral terms cancel and the xe−x2x e^{-x^2}xe−x2 terms also cancel. Thus,

f(x)+g(x)=1−e−x2.f(x)+g(x)=1-e^{-x^2}.f(x)+g(x)=1−e−x2.
  1. Substitute x=ln⁡9x=\sqrt{\ln 9}x=ln9​.

Then

x2=ln⁡9⇒e−x2=e−ln⁡9=19.x^2=\ln 9 \quad\Rightarrow\quad e^{-x^2}=e^{-\ln 9}=\frac19.x2=ln9⇒e−x2=e−ln9=91​.

So,

f(ln⁡9)+g(ln⁡9)=1−19=89.f\left(\sqrt{\ln 9}\right)+g\left(\sqrt{\ln 9}\right)=1-\frac19=\frac89.f(ln9​)+g(ln9​)=1−91​=98​.

Therefore,

9(f(ln⁡9)+g(ln⁡9))=9⋅89=8.9\left(f\left(\sqrt{\ln 9}\right)+g\left(\sqrt{\ln 9}\right)\right) =9\cdot \frac89=8.9(f(ln9​)+g(ln9​))=9⋅98​=8.
  1. Check options.
  • A: 101010 ❌
  • B: 999 ❌
  • C: 888 ✅
  • D: 666 ❌

So the correct option is C.

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