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TS EAMCET 18-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 68 of 160

Marks: +1, -0

a f(x)+b f\left(\frac{1}{x}\right)=x+1

\frac{d}{dx}(x^{2} f(x))=2 x^{2}+2 x+\frac{1}{3}

a-b=

Solution: 👈: Video Solution

a f(x)+b f\left(\frac{1}{x}\right)=x+1

Substitute

x

\frac{1}{x}

a f\left(\frac{1}{x}\right)+b f(x)=\frac{1}{x}+1

From Eq. (i)

\times a-

Eq. (ii)

b_{\text{e}}

a^2 f(x)+a b f\left(\frac{1}{x}\right) =a x+a

a b f\left(\frac{1}{x}\right) +b^2 f(x)=\frac{b}{x}+b

(a^2-b^2) f(x)=a x+(a-b)-\frac{b}{x}

\Rightarrow x^2 f(x)=\frac{1}{a^2-b^2} [a x^3+(a-b) x^2-b x]

\frac{d}{dx} (x^2 f(x)) =\frac{1}{a^2-b^2} [3 a x^2+2(a-b) x-b]

According to the question,

\frac{3a}{a^2-b^2}=2, \frac{a-b}{a^2-b^2}=1,-\frac{b}{a^2-b^2}=\frac{1}{3}

\Rightarrow \frac{3a}{2}=a-b=-3 b=a^2-b^2

\Rightarrow 3a=2a-2b \Rightarrow a=-2b

\Rightarrow \text{Also } \frac{a-b}{a^2-b^2}=1 \text{ and } \frac{1}{a+b}=1

\Rightarrow a+b=1 \Rightarrow -2a+b=1

\Rightarrow b=-1 \Rightarrow a=2

\therefore a-b=2-(-1)=3

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