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Definite Integration question

2021 · 1 Sep · Shift 2 · Q30
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  5. /2021 · 1 Sep · Shift 2 · Q30

Definite Integration question

2021 · 1 Sep · Shift 2 · Q30

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let Jn,m=∫012xnxm−1dx{J_{n,m}} = \int\limits_0^{{1 \over 2}} {{{{x^n}} \over {{x^m} - 1}}dx}Jn,m​=0∫21​​xm−1xn​dx, ∀\forall∀ n > m and n, m ∈\in∈ N. Consider a matrix A=[aij]3×3A = {[{a_{ij}}]_{3 \times 3}}A=[aij​]3×3​ where aij={j6+i,3−ji+3,3,i≤j0,i>j{a_{ij}} = \left\{ {\begin{matrix} {{j_{6 + i,3}} - {j_{i + 3,3}},} & {i \le j} \\ {0,} & {i \gt j} \\ \end{matrix} } \right.aij​={j6+i,3​−ji+3,3​,0,​i≤ji>j​. Then ∣adjA−1∣\left| {adj{A^{ - 1}}} \right|​adjA−1​ is :
  1. A
    (15)2 ×\times× 242
  2. B
    (15)2 ×\times× 234
  3. C
    (105)2 ×\times× 238
  4. D
    (105)2 ×\times× 236
View written solutionFree

Correct answer: C

  1. Given integral

We have

Jn,m=∫01/2xnxm−1 dx,n>m.J_{n,m} = \int_0^{1/2} \frac{x^n}{x^m-1}\,dx, \qquad n>m.Jn,m​=∫01/2​xm−1xn​dx,n>m.

Also,

aij={J6+i,3−Ji+3,3,i≤j,0,i>j.a_{ij}= \begin{cases} J_{6+i,3}-J_{i+3,3}, & i\le j,\\ 0,& i>j. \end{cases}aij​={J6+i,3​−Ji+3,3​,0,​i≤j,i>j.​

So for each fixed row index iii, all entries with j≥ij\ge ij≥i are equal. Hence AAA is upper triangular.

  1. Find a general expression for Jn,3−Jn−3,3J_{n,3}-J_{n-3,3}Jn,3​−Jn−3,3​

Observe:

Jn,3−Jn−3,3=∫01/2(xnx3−1−xn−3x3−1)dxJ_{n,3}-J_{n-3,3} = \int_0^{1/2} \left(\frac{x^n}{x^3-1}-\frac{x^{n-3}}{x^3-1}\right)dxJn,3​−Jn−3,3​=∫01/2​(x3−1xn​−x3−1xn−3​)dx =∫01/2xn−3(x3−1)x3−1 dx=∫01/2xn−3 dx.= \int_0^{1/2} \frac{x^{n-3}(x^3-1)}{x^3-1}\,dx = \int_0^{1/2} x^{n-3}\,dx.=∫01/2​x3−1xn−3(x3−1)​dx=∫01/2​xn−3dx.

Therefore,

Jn,3−Jn−3,3=∫01/2xn−3dx.J_{n,3}-J_{n-3,3} = \int_0^{1/2} x^{n-3}dx.Jn,3​−Jn−3,3​=∫01/2​xn−3dx.

Now in our matrix,

J6+i,3−Ji+3,3J_{6+i,3}-J_{i+3,3}J6+i,3​−Ji+3,3​

corresponds to taking n=6+in=6+in=6+i, so

J6+i,3−Ji+3,3=∫01/2xi+3dx.J_{6+i,3}-J_{i+3,3} = \int_0^{1/2} x^{i+3}dx.J6+i,3​−Ji+3,3​=∫01/2​xi+3dx.

Thus,

aij={∫01/2xi+3dx,i≤j,0,i>j.a_{ij}= \begin{cases} \displaystyle \int_0^{1/2} x^{i+3}dx, & i\le j,\\ 0,& i>j. \end{cases}aij​=⎩⎨⎧​∫01/2​xi+3dx,0,​i≤j,i>j.​

Compute:

∫01/2xi+3dx=[xi+4i+4]01/2=1i+4⋅12i+4.\int_0^{1/2} x^{i+3}dx = \left[\frac{x^{i+4}}{i+4}\right]_0^{1/2} = \frac{1}{i+4}\cdot \frac{1}{2^{i+4}}.∫01/2​xi+3dx=[i+4xi+4​]01/2​=i+41​⋅2i+41​.

So diagonal entries are:

a11=15⋅25,a22=16⋅26,a33=17⋅27.a_{11}=\frac{1}{5\cdot 2^5}, \qquad a_{22}=\frac{1}{6\cdot 2^6}, \qquad a_{33}=\frac{1}{7\cdot 2^7}.a11​=5⋅251​,a22​=6⋅261​,a33​=7⋅271​.

Hence

A=(15⋅2515⋅2515⋅25016⋅2616⋅260017⋅27).A= \begin{pmatrix} \frac1{5\cdot 2^5} & \frac1{5\cdot 2^5} & \frac1{5\cdot 2^5}\\[4pt] 0 & \frac1{6\cdot 2^6} & \frac1{6\cdot 2^6}\\[4pt] 0 & 0 & \frac1{7\cdot 2^7} \end{pmatrix}.A=​5⋅251​00​5⋅251​6⋅261​0​5⋅251​6⋅261​7⋅271​​​.
  1. Find ∣A∣|A|∣A∣

Since AAA is upper triangular,

∣A∣=a11a22a33=15⋅25⋅16⋅26⋅17⋅27.|A| = a_{11}a_{22}a_{33} = \frac1{5\cdot 2^5}\cdot \frac1{6\cdot 2^6}\cdot \frac1{7\cdot 2^7}.∣A∣=a11​a22​a33​=5⋅251​⋅6⋅261​⋅7⋅271​.

Now,

5⋅6⋅7=210,25+6+7=218.5\cdot 6\cdot 7 = 210, \qquad 2^{5+6+7}=2^{18}.5⋅6⋅7=210,25+6+7=218.

So

∣A∣=1210⋅218.|A| = \frac{1}{210\cdot 2^{18}}.∣A∣=210⋅2181​.
  1. Use determinant formula for adjugate

For an invertible n×nn\times nn×n matrix MMM,

∣adj⁡(M)∣=∣M∣n−1.|\operatorname{adj}(M)| = |M|^{n-1}.∣adj(M)∣=∣M∣n−1.

Here the matrix is A−1A^{-1}A−1 of order 3×33\times 33×3, so

∣adj⁡(A−1)∣=∣A−1∣2.\left|\operatorname{adj}(A^{-1})\right| = |A^{-1}|^{2}.​adj(A−1)​=∣A−1∣2.

But

∣A−1∣=1∣A∣=210⋅218.|A^{-1}| = \frac1{|A|} = 210\cdot 2^{18}.∣A−1∣=∣A∣1​=210⋅218.

Therefore,

∣adj⁡(A−1)∣=(210⋅218)2.\left|\operatorname{adj}(A^{-1})\right| = (210\cdot 2^{18})^2.​adj(A−1)​=(210⋅218)2.

Now,

210=2⋅105,210 = 2\cdot 105,210=2⋅105,

so

210⋅218=105⋅219.210\cdot 2^{18} = 105\cdot 2^{19}.210⋅218=105⋅219.

Hence

∣adj⁡(A−1)∣=(105⋅219)2=(105)2⋅238.\left|\operatorname{adj}(A^{-1})\right| = (105\cdot 2^{19})^2 = (105)^2\cdot 2^{38}.​adj(A−1)​=(105⋅219)2=(105)2⋅238.
  1. Match with options

This is

(105)2×238,(105)^2\times 2^{38},(105)2×238,

which matches Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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