यदि fleft(frac{x-4}{x-2}right)=2x+1,x in R-{1 | vedclass

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(A) दिया गया है $f\left(\frac{x-4}{x-2}\right)=2x+1$. माना $y = \frac{x-4}{x-2}$. तब $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$. अतः,$\frac{2}{x-2} =

Vedclass · Accepted Answer

$5x-4 \log |x-1|+c$,जहाँ $c$ समाकलन स्थिरांक है।

Vedclass · Answer

(A) दिया गया है $f\left(\frac{x-4}{x-2}\right)=2x+1$. माना $y = \frac{x-4}{x-2}$. तब $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$. अतः,$\frac{2}{x-2} = 1-y$,जिसका अर्थ है $x-2 = \frac{2}{1-y} = \frac{-2}{y-1}$. इस प्रकार,$x = 2 - \frac{2}{y-1} = \frac{2y-2-2}{y-1} = \frac{2y-4}{y-1}$. अब,$f(y) = 2x+1 = 2\left(\frac{2y-4}{y-1}\right) + 1 = \frac{4y-8+y-1}{y-1} = \frac{5y-9}{y-1} = \frac{5(y-1)-4}{y-1} = 5 - \frac{4}{y-1}$. इसलिए,$f(x) = 5 - \frac{4}{x-1}$. अब,$\int f(x) dx = \int \left(5 - \frac{4}{x-1}\right) dx = 5x - 4 \log |x-1| + c$.

यदि $f\left(\frac{x-4}{x-2}\right)=2x+1$,$x \in R-\{1, 2\}$ है,तो $\int f(x) dx$ का मान ज्ञात कीजिए।

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