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MCQMediumJEE 2024Properties of Definite Integrals

JEE Mathematics 2024 Question with Solution

The value of k∈Nk \in \mathbb{N} for which the integral In=∫01(1−xk)n dx,  n∈NI_n = \int_0^1 (1 - x^k)^n \, dx,\; n \in \mathbb{N}, satisfies 147I20=148I21147I_{20} = 148I_{21}, is:

  • A

    1010

  • B

    88

  • C

    1414

  • D

    77

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

In=∫01(1−xk)n dxI_n = \int_0^1 (1 - x^k)^n \, dx

and 147I20=148I21147I_{20} = 148I_{21}.

Find: The value of kk.

From the given working, using integration by parts:

In=∫01(1−xk)n⋅1 dxI_n = \int_0^1 (1 - x^k)^n \cdot 1 \, dx In=(1−xk)n⋅x−nk∫01(1−xk)n−1⋅xk−1 dxI_n = (1 - x^k)^n \cdot x - nk \int_0^1 (1 - x^k)^{n-1} \cdot x^{k-1} \, dx In=nk[∫01((1−xk)n−(1−xk)n−1)dx]I_n = nk \left[ \int_0^1 \left( (1 - x^k)^n - (1 - x^k)^{n-1} \right) dx \right] In=nkIn−nkIn−1I_n = nk I_n - nk I_{n-1}

Therefore,

InIn−1=nknk+1\frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1}

So for n=21n = 21,

I21I20=21k1+21k\frac{I_{21}}{I_{20}} = \frac{21k}{1 + 21k}

From 147I20=148I21147I_{20} = 148I_{21}, we get

I21I20=147148\frac{I_{21}}{I_{20}} = \frac{147}{148}

Hence,

21k1+21k=147148\frac{21k}{1 + 21k} = \frac{147}{148}

Solving,

148⋅21k=147(21k+1)148 \cdot 21k = 147(21k + 1) 3108k=3087k+1473108k = 3087k + 147 21k=14721k = 147 k=7k = 7

Therefore, the correct option is D.

Cross-multiplication Detail

Given:

I21I20=147148\frac{I_{21}}{I_{20}} = \frac{147}{148}

and from the recurrence,

I21I20=21k21k+1\frac{I_{21}}{I_{20}} = \frac{21k}{21k + 1}

Find: kk.

Equate the two expressions:

21k21k+1=147148\frac{21k}{21k + 1} = \frac{147}{148}

Now cross-multiply:

148⋅21k=147(21k+1)148 \cdot 21k = 147(21k + 1) 148⋅21k=147⋅21k+147148 \cdot 21k = 147 \cdot 21k + 147 21k=14721k = 147 k=7k = 7

Thus, the required natural number is 77, so the correct option is D.

Common mistakes

  • Using the condition 147I20=148I21147I_{20} = 148I_{21} incorrectly as I21I20=148147\frac{I_{21}}{I_{20}} = \frac{148}{147}. This reverses the ratio. Divide carefully to get I21I20=147148\frac{I_{21}}{I_{20}} = \frac{147}{148}.

  • Writing the recurrence as InIn−1=nk+1nk\frac{I_n}{I_{n-1}} = \frac{nk+1}{nk} instead of nknk+1\frac{nk}{nk+1}. This comes from rearranging the integration-by-parts result wrongly. Keep the algebra consistent before substituting n=21n=21.

  • Substituting n=20n=20 into the ratio formula instead of n=21n=21. Since the required ratio is I21I20\frac{I_{21}}{I_{20}}, the numerator index determines that n=21n=21 must be used.

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