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Let \(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\fra…

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Let \(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\) If \(\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{\mathrm{b}}^{\frac{1}{13}}}-\frac{1}{{\mathrm{c}}^{\frac{1}{13}}}\right),\mathrm{b},\mathrm{c}\in \mathrm{ℕ},\) then \(3(b+c)\) is equal to

[JEE Main 2025, 23 Jan (Shift 1)]

a

40

b

39

c

22

d

26

✓ Correct answer: b)

39

Explanation

\(\mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}.\\ \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\frac{x-11}{x+15}{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{2}}.\\ \text{ Let }\frac{\mathrm{x}-11}{\mathrm{x}+15}=\mathrm{t}\Rightarrow \frac{26}{(\mathrm{x}+15{)}^{2}}\mathrm{dx}=\mathrm{dt}\\ I\left(\mathrm{x}\right)=\frac{1}{26}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{11/13}}=\frac{1}{26}\cdot \frac{{\mathrm{t}}^{2/13}}{2/13}+C\\ I\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{x-11}{x+15}\right)}^{2/13}+C\\ \text{Now,}\\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13}\\ =\frac{1}{4}\left(\frac{1}{{2}^{2/13}}-\frac{1}{{3}^{2/13}}\right)\\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{4}^{1/13}}-\frac{1}{{9}^{1/13}}\right)\\ \text{Compare with,}\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{{b}^{\frac{1}{13}}}-\frac{1}{{c}^{\frac{1}{13}}}\right)\\ \text{we get}b=4&c=9\\ \text{Now,}3\left(b+c\right)=3\left(4+9\right)=39\\\)

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