JEE Main 2026 28 January Evening Shift
Exam JEE Main
Year 2026
Date 28 January
Shift Evening Shift
Subject Mathematics
Chapter Indefinite Integration JEE Main 2026 · 28 January · Evening Shift · Mathematics · Indefinite Integration
Let f(x)=∫ d x x^ ((2)/(3)) +2 x^ ((1)/(2)) be such that...
Let
f ( x ) = ∫ d x x ( 2 3 ) + 2 x ( 1 2 ) f(x)=\int \frac{\mathrm{d} x}{x^{\left(\frac{2}{3}\right)}+2 x^{\left(\frac{1}{2}\right)}} f ( x ) = ∫ x ( 3 2 ) + 2 x ( 2 1 ) d x be such that
f ( 0 ) = − 26 + 24 log e ( 2 ) f(0)=-26+24 \log _{\mathrm{e}}(2) f ( 0 ) = − 26 + 24 log e ( 2 ) . If
f ( 1 ) = a + b log e ( 3 ) f(1)=\mathrm{a}+\mathrm{b} \log _{\mathrm{e}}(3) f ( 1 ) = a + b log e ( 3 ) , where
a , b ∈ Z \mathrm{a}, \mathrm{b} \in \mathbf{Z} a , b ∈ Z , then
a + b \mathrm{a}+\mathrm{b} a + b is equal to :
Step-by-step solution
Let
u = x 1 / 6 u = x^{1/6} u = x 1/6 , so
x = u 6 x = u^6 x = u 6 and
d x = 6 u 5 d u dx = 6u^5 du d x = 6 u 5 d u . Then
x 2 / 3 = u 4 x^{2/3} = u^4 x 2/3 = u 4 ,
x 1 / 2 = u 3 x^{1/2} = u^3 x 1/2 = u 3 .
f ( x ) = ∫ 6 u 5 d u u 4 + 2 u 3 = ∫ 6 u 2 d u u + 2 f(x) = \int \frac{6u^5 du}{u^4 + 2u^3} = \int \frac{6u^2 du}{u+2} f ( x ) = ∫ u 4 + 2 u 3 6 u 5 d u = ∫ u + 2 6 u 2 d u .
By polynomial division:
u 2 u + 2 = u − 2 + 4 u + 2 \frac{u^2}{u+2} = u - 2 + \frac{4}{u+2} u + 2 u 2 = u − 2 + u + 2 4 .
f ( x ) = 6 ( u 2 2 − 2 u + 4 ln ∣ u + 2 ∣ ) + C = 3 u 2 − 12 u + 24 ln ( u + 2 ) + C f(x) = 6\left(\frac{u^2}{2} - 2u + 4\ln|u+2|\right) + C = 3u^2 - 12u + 24\ln(u+2) + C f ( x ) = 6 ( 2 u 2 − 2 u + 4 ln ∣ u + 2∣ ) + C = 3 u 2 − 12 u + 24 ln ( u + 2 ) + C .
Substituting back:
f ( x ) = 3 x 1 / 3 − 12 x 1 / 6 + 24 ln ( x 1 / 6 + 2 ) + C f(x) = 3x^{1/3} - 12x^{1/6} + 24\ln(x^{1/6}+2) + C f ( x ) = 3 x 1/3 − 12 x 1/6 + 24 ln ( x 1/6 + 2 ) + C .
From
f ( 0 ) = − 26 + 24 ln ( 2 ) f(0) = -26 + 24\ln(2) f ( 0 ) = − 26 + 24 ln ( 2 ) :
24 ln ( 2 ) + C = − 26 + 24 ln ( 2 ) 24\ln(2) + C = -26 + 24\ln(2) 24 ln ( 2 ) + C = − 26 + 24 ln ( 2 ) , so
C = − 26 C = -26 C = − 26 .
At
x = 1 x = 1 x = 1 :
f ( 1 ) = 3 − 12 + 24 ln ( 3 ) − 26 = − 35 + 24 ln ( 3 ) f(1) = 3 - 12 + 24\ln(3) - 26 = -35 + 24\ln(3) f ( 1 ) = 3 − 12 + 24 ln ( 3 ) − 26 = − 35 + 24 ln ( 3 ) .
Thus
a = − 35 a = -35 a = − 35 ,
b = 24 b = 24 b = 24 , and
a + b = − 11 a + b = -11 a + b = − 11 .
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