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Definite Integration question

2021 · 27 Jul · Shift 1 · Q42
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Definite Integration question

2021 · 27 Jul · Shift 1 · Q42

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let the domain of the function f(x)=log⁡4(log⁡5(log⁡3(18x−x2−77)))f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)f(x)=log4​(log5​(log3​(18x−x2−77))) be (a, b). Then the value of the integral ∫absin⁡3x(sin⁡3x+sin⁡3(a+b−x)dx\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dxa∫b​(sin3x+sin3(a+b−x)sin3x​dx is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1

  1. Find the domain of f(x)=log⁡4(log⁡5(log⁡3(18x−x2−77))).f(x)=\log_4\left(\log_5\left(\log_3(18x-x^2-77)\right)\right).f(x)=log4​(log5​(log3​(18x−x2−77))).

    For the expression to be defined, we proceed from inside outward.

    Step 1: Condition for log⁡3(18x−x2−77)\log_3(18x-x^2-77)log3​(18x−x2−77)

    We need 18x−x2−77>0.18x-x^2-77>0.18x−x2−77>0. Rearranging, −x2+18x−77>0-x^2+18x-77>0−x2+18x−77>0 x2−18x+77<0.x^2-18x+77<0.x2−18x+77<0. Factorize: x2−18x+77=(x−7)(x−11).x^2-18x+77=(x-7)(x-11).x2−18x+77=(x−7)(x−11). So, (x−7)(x−11)<0  ⟹  7<x<11.(x-7)(x-11)<0 \implies 7<x<11.(x−7)(x−11)<0⟹7<x<11.

    Step 2: Condition for log⁡5(log⁡3(18x−x2−77))\log_5\left(\log_3(18x-x^2-77)\right)log5​(log3​(18x−x2−77))

    Since this is a logarithm, its argument must be positive: log⁡3(18x−x2−77)>0.\log_3(18x-x^2-77)>0.log3​(18x−x2−77)>0. Because base 3>13>13>1, this means 18x−x2−77>1.18x-x^2-77>1.18x−x2−77>1. So, −x2+18x−78>0-x^2+18x-78>0−x2+18x−78>0 x2−18x+78<0.x^2-18x+78<0.x2−18x+78<0. Factorize: x2−18x+78=(x−6)(x−13).x^2-18x+78=(x-6)(x-13).x2−18x+78=(x−6)(x−13). Hence, 6<x<13.6<x<13.6<x<13.

    Combining with Step 1 gives still 7<x<11.7<x<11.7<x<11.

    Step 3: Condition for outermost log⁡4\log_4log4​

    We need log⁡5(log⁡3(18x−x2−77))>0.\log_5\left(\log_3(18x-x^2-77)\right)>0.log5​(log3​(18x−x2−77))>0. Since base 5>15>15>1, this implies log⁡3(18x−x2−77)>1.\log_3(18x-x^2-77)>1.log3​(18x−x2−77)>1. Again base 3>13>13>1, so 18x−x2−77>3.18x-x^2-77>3.18x−x2−77>3. Thus, −x2+18x−80>0-x^2+18x-80>0−x2+18x−80>0 x2−18x+80<0.x^2-18x+80<0.x2−18x+80<0. Factorize: x2−18x+80=(x−8)(x−10).x^2-18x+80=(x-8)(x-10).x2−18x+80=(x−8)(x−10). Therefore, 8<x<10.8<x<10.8<x<10.

    So the domain is (a,b)=(8,10).(a,b)=(8,10).(a,b)=(8,10).

  2. Evaluate the integral I=∫absin⁡3xsin⁡3x+sin⁡3(a+b−x) dx.I=\int_a^b \frac{\sin^3 x}{\sin^3 x+\sin^3(a+b-x)}\,dx.I=∫ab​sin3x+sin3(a+b−x)sin3x​dx.

    Since a=8a=8a=8 and b=10b=10b=10, we have a+b=18.a+b=18.a+b=18. So I=∫810sin⁡3xsin⁡3x+sin⁡3(18−x) dx.I=\int_8^{10} \frac{\sin^3 x}{\sin^3 x+\sin^3(18-x)}\,dx.I=∫810​sin3x+sin3(18−x)sin3x​dx.

  3. Use the symmetry substitution

    Let x↦a+b−x=18−x.x\mapsto a+b-x=18-x.x↦a+b−x=18−x. Then I=∫810sin⁡3(18−x)sin⁡3(18−x)+sin⁡3x dx.I=\int_8^{10} \frac{\sin^3(18-x)}{\sin^3(18-x)+\sin^3 x}\,dx.I=∫810​sin3(18−x)+sin3xsin3(18−x)​dx.

    Add this with the original integral: 2I=∫810[sin⁡3xsin⁡3x+sin⁡3(18−x)+sin⁡3(18−x)sin⁡3(18−x)+sin⁡3x]dx.2I=\int_8^{10} \left[\frac{\sin^3 x}{\sin^3 x+\sin^3(18-x)}+\frac{\sin^3(18-x)}{\sin^3(18-x)+\sin^3 x}\right]dx.2I=∫810​[sin3x+sin3(18−x)sin3x​+sin3(18−x)+sin3xsin3(18−x)​]dx.

    The bracket simplifies to 111, hence 2I=∫8101 dx=10−8=2.2I=\int_8^{10} 1\,dx=10-8=2.2I=∫810​1dx=10−8=2. Therefore, I=1.I=1.I=1.

  4. Final answer 1\boxed{1}1​

  5. Comparison with stored answer

    Stored correct answer = 111.

    Our derived answer matches it.

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