Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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If for Non-zero X, Af(X) + Bf ( 1 X ) = 1 X − 5 , Where a ≠ B, Then Find F(X). - Mathematics

If for non-zero xaf(x) + bf \[\left( \frac{1}{x} \right) = \frac{1}{x} - 5\] , where a ≠ b, then find f(x).

 
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Solution

Given :

\[af\left( x \right) + bf\left( \frac{1}{x} \right) = \frac{1}{x} - 5\] ...(i)

\[\Rightarrow af \left( \frac{1}{x} \right) + bf\left( x \right) = \frac{1}{\frac{1}{x}} - 5\]
\[\Rightarrow af\left( \frac{1}{x} \right)+ bf\left( x \right) = x - 5\]  ...(ii)

On adding equations (i) and (ii), we get:  

\[af\left( x \right) + bf\left( x \right) + bf\left( \frac{1}{x} \right) + af\left( \frac{1}{x} \right) = \frac{1}{x} - 5 + x - 5\]

\[\Rightarrow \left( a + b \right)f\left( x \right) + \left( a + b \right)f\left( \frac{1}{x} \right) = \frac{1}{x} + x - 10\]

\[\Rightarrow f\left( x \right) + f\left( \frac{1}{x} \right) = \frac{1}{\left( a + b \right)}\left[ \frac{1}{x} + x - 10 \right]\]     ...(iii) 

On subtracting (ii) from (i), we get:

\[af\left( x \right) - bf\left( x \right) + bf\left( \frac{1}{x} \right) - af\left( \frac{1}{x} \right) = \frac{1}{x} - 5 - x + 5\]

\[\Rightarrow \left( a - b \right)f\left( x \right) - f\left( \frac{1}{x} \right)\left( a - b \right) = \frac{1}{x} - x\]

\[\Rightarrow f\left( x \right) - f\left( \frac{1}{x} \right) = \frac{1}{\left( a - b \right)}\left[ \frac{1}{x} - x \right]\]        ...(iv)

On adding equations (iii) and (iv), we get:

\[2f\left( x \right) = \frac{1}{a + b}\left[ \frac{1}{x} + x - 10 \right] + \frac{1}{a - b}\left[ \frac{1}{x} - x \right]\]

\[\Rightarrow 2f\left( x \right) = \frac{\left( a - b \right)\left[ \frac{1}{x} + x - 10 \right] + \left( a + b \right)\left[ \frac{1}{x} - x \right]}{\left( a + b \right)\left( a - b \right)}\]

 \[\Rightarrow 2f\left( x \right) = \frac{\frac{a}{x} + ax - 10a - \frac{b}{x} - bx + 10b + \frac{a}{x} - ax + \frac{b}{x} - bx}{a^2 - b^2}\]

\[\Rightarrow 2f\left( x \right) = \frac{\frac{2a}{x} - 10a + 10b - 2bx}{a^2 - b^2}\]

\[\Rightarrow f\left( x \right) = \frac{1}{a^2 - b^2} \times \frac{1}{2}\left[ \frac{2a}{x} - 10a + 10b - 2bx \right]\]

\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - 5a + 5b - bx \right]\]

Therefore , 

\[f\left( x \right) = \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx - 5a + 5b \right]\]

\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5\left( a - b \right)}{a^2 - b^2}\]

\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5\left( a - b \right)}{\left( a - b \right)\left( a + b \right)}\]

\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5}{\left( a + b \right)}\]

Hence,

\[f\left( x \right) = \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5}{\left( a + b \right)}\]

 

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.2 | Q 11 | Page 12
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