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JEE Mathematics 2023 Question with Solution

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JEE Main 2023 Questions with Solutions Section – A

Mathematics question image showing the definite integral from $$0$$ to $$1$$ of $$\frac{6}{e^{3x}+6e^{2x}+11e^x+6} \, dx$$.
  • A

    loge(3227)\log_e \left( \frac{32}{27} \right)

  • B

    loge(25681)\log_e \left( \frac{256}{81} \right)

  • C

    loge(51281)\log_e \left( \frac{512}{81} \right)

  • D

    loge(6427)\log_e \left( \frac{64}{27} \right)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

I=016e3x+6e2x+11ex+6dxI=\int_0^1 \frac{6}{e^{3x}+6e^{2x}+11e^x+6}\,dx

Find: Evaluate the integral and identify the correct option.

Factor the denominator:

e3x+6e2x+11ex+6=(ex+2)(e2x+4ex+3)e^{3x}+6e^{2x}+11e^x+6=(e^x+2)(e^{2x}+4e^x+3)

So,

I=016(ex+2)(e2x+4ex+3)dxI=\int_0^1 \frac{6}{(e^x+2)(e^{2x}+4e^x+3)}\,dx

Use the substitution u=exu=e^x, so du=exdxdu=e^x\,dx. The limits change as: when x=0x=0, u=1u=1 and when x=1x=1, u=eu=e. Hence,

I=1e6(u+2)(u2+4u+3)duuI=\int_1^e \frac{6}{(u+2)(u^2+4u+3)}\cdot \frac{du}{u}

Thus,

I=1e6u(u+2)(u2+4u+3)duI=\int_1^e \frac{6}{u(u+2)(u^2+4u+3)}\,du

Using partial fraction decomposition and simplifying, we get:

I=loge(3227)I=\log_e\left(\frac{32}{27}\right)

Therefore, the value of the integral is loge(3227)\log_e\left(\frac{32}{27}\right). The correct option is A.

Factorisation and substitution insight

Given:

I=016e3x+6e2x+11ex+6dxI=\int_0^1 \frac{6}{e^{3x}+6e^{2x}+11e^x+6}\,dx

Find: A quick route to the final value.

The useful observation is to rewrite the cubic expression in terms of exe^x. Once the denominator is factorised, the integral becomes a rational function of u=exu=e^x. This works because exponentials of the form e3x,e2x,exe^{3x}, e^{2x}, e^x reduce to powers of a single variable after substitution.

After setting u=exu=e^x, the integral changes to a standard rational integral in uu, and partial fractions lead directly to logarithmic terms. Evaluating between the transformed limits 11 and ee gives

I=loge(3227)I=\log_e\left(\frac{32}{27}\right)

Therefore, the correct option is A.

Common mistakes

  • Factoring e3x+6e2x+11ex+6e^{3x}+6e^{2x}+11e^x+6 incorrectly. This breaks the structure needed for substitution. First treat exe^x as a single variable and then factor systematically.

  • Using u=exu=e^x but forgetting that dx=duudx=\frac{du}{u}. This misses the extra factor of 1u\frac{1}{u} in the integrand. Always convert both the expression and the differential.

  • Not changing the limits after substitution. If you switch from xx to uu, the new limits must be 11 and ee. Otherwise the final evaluation becomes inconsistent.

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