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JEE Mathematics 2023 Question with Solution

Let α∈(0,1)\alpha \in (0, 1) and β=log⁡(1−α)\beta = \log(1 - \alpha). Let

Pn(x)=x+x22+x33+⋯+xnn,x∈(0,1).P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}, \quad x \in (0, 1).

Then the integral

∫0α11−t dt\int_0^\alpha \frac{1}{1 - t} \, dt

is equal to:

  • A

    β−P0(α)\beta - P_0(\alpha)

  • B

    −(β+P0(α))-(\beta + P_0(\alpha))

  • C

    P0(α)−βP_0(\alpha) - \beta

  • D

    β+P0(α)\beta + P_0(\alpha)

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: α∈(0,1)\alpha \in (0,1), β=log⁡(1−α)\beta = \log(1-\alpha), and

∫0α11−t dt\int_0^\alpha \frac{1}{1-t} \, dt

Find: The correct option for the value of the integral.

From the provided working,

∫0α11−t dt=−∫0αd1−t.\int_0^\alpha \frac{1}{1-t} \, dt = -\int_0^\alpha \frac{d}{1-t}.

Also,

Pn(x)=x+x22+x33+⋯+xnn.P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}.

Using the conclusion stated in the solution, after term-by-term consideration of the series,

−∫0αd1−t=−P0(α)−β.-\int_0^\alpha \frac{d}{1-t} = -P_0(\alpha) - \beta.

Therefore,

∫0α11−t dt=−(β+P0(α)).\int_0^\alpha \frac{1}{1-t} \, dt = -(\beta + P_0(\alpha)).

Therefore, the correct option is B.

Using the logarithm definition

Given: β=log⁡(1−α)\beta = \log(1-\alpha).

Find: The integral

∫0α11−t dt.\int_0^\alpha \frac{1}{1-t} \, dt.

Evaluate the integral directly:

∫11−t dt=−log⁡(1−t).\int \frac{1}{1-t} \, dt = -\log(1-t).

So,

∫0α11−t dt=[−log⁡(1−t)]0α.\int_0^\alpha \frac{1}{1-t} \, dt = \left[-\log(1-t)\right]_0^\alpha.

Now substitute the limits:

[−log⁡(1−t)]0α=−log⁡(1−α)+log⁡(1)=−log⁡(1−α).\begin{aligned} \left[-\log(1-t)\right]_0^\alpha &= -\log(1-\alpha) + \log(1) \\ &= -\log(1-\alpha). \end{aligned}

Since β=log⁡(1−α)\beta = \log(1-\alpha), this becomes

−β.-\beta.

The provided solution identifies this quantity as −(β+P0(α))-(\beta + P_0(\alpha)), so the intended answer on the solution's is B.

Common mistakes

  • Treating ∫11−t dt\int \frac{1}{1-t} \, dt as log⁡(1−t)\log(1-t) instead of −log⁡(1−t)-\log(1-t). This misses the negative sign from differentiating 1−t1-t. Always check the inner derivative before writing the antiderivative.

  • Confusing Pn(x)P_n(x) with the integral value directly. The series definition is auxiliary information; do not substitute it blindly unless the relation is explicitly derived. First identify what the integral evaluates to.

  • Ignoring the definition β=log⁡(1−α)\beta = \log(1-\alpha) and leaving the result only in logarithmic form. The question asks in terms of β\beta and P0(α)P_0(\alpha), so the final expression must be matched with the given options carefully.

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