Let f ( x ) = d x x ( 2 3 ) + 2 x ( 1 2 ) be such that f ( 0 ) = - 26 + 24 log e ( 2 ) . If f ( 1 ) = a + b log e ( 3 ) , where a , b, then a+b is equal to: [2026]

Question

Let $f (x)$ $= \int \frac{d x}{x^{(\frac{2}{3})} + 2 x^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 \log_{e} (2) .$ If $f (1) = a + b \log_{e} (3),$ where $a, b \in ℤ$ , then a+b is equal to: [2026]

Smart Achievers · Accepted Answer

(1)

$f (x) = \int \frac{d x}{x^{2 / 3} + 2 x^{1 / 2}}$

$Put x = t^{6} \Rightarrow d x = 6 t^{5} d t$

$= \int \frac{6 t^{5}}{t^{4} + 2 t^{3}} d t = 6 \int \frac{(t^{2} - 4) + 4}{t + 2} d t$

$= 6 [\int (t - 2) d t + 4 \int \frac{1}{t + 2} d t]$

$= 6 [\frac{t^{2}}{2} - 2 t + 4 l n (t + 2)] + C$

$= 3 x^{1 / 3} - 12 x^{1 / 6} + 24 l n (x^{1 / 6} + 2) + C$

$f (0) = 24 l n 2 + C = - 26 + 24 l n 2 (given)$

$\Rightarrow C = - 26$

$Now$

$f (1) = - 35 + 24 l n 3 = a + b l n 3 (as given in ques.)$

$\Rightarrow a = - 35, b = 24$

$\Rightarrow a + b = - 11$

Solution

Q.
Let $f (x)$ $= \int \frac{d x}{x^{(\frac{2}{3})} + 2 x^{(\frac{1}{2})}}$ be such that $f (0) = - 26 + 24 \log_{e} (2) .$ If $f (1) = a + b \log_{e} (3),$ where $a, b \in ℤ$ , then a+b is equal to: [2026]

1 -11

2 -26

3 -18

4 -5

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