Verified Solution • 2025–26 Session moderate

QJEEM21Ma618d: JEE Main 2021 Mathematics - Class 11 JEEM21Ma618d Solution: If a + = 1, b + = 2 and , then the value of the expression is…

If a + α\alpha = 1, b + β\beta = 2 and af(x)+αf(1x)=bx+βx,x≠0af(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x},x \ne 0 , then the value of the expression f(x)+f(1x)x+1x{{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} is __________.

Step-by-Step Verified Solution

af(x)+αf(1x)=bx+βxaf(x) + \alpha f\left( {{1 \over x}} \right) = bx + {\beta \over x} ........(i)

Replace xx with 1x{1 \over x}

af(1x)+af(x)=bx+βxaf\left( {{1 \over x}} \right) + af(x) = {b \over x} + \beta x ..... (ii)

(i) + (ii)

(a+α)[f(x)+f(1x)]=(x+1x)(b+β)(a + \alpha )\left[ {f(x) + f\left( {{1 \over x}} \right)} \right] = \left( {x + {1 \over x}} \right)(b + \beta )

f(x)+f(1x)x+1x=β+ba+α=21=2{{f(x) + f\left( {{1 \over x}} \right)} \over {x + {1 \over x}}} = {{\beta + b} \over {a + \alpha }} = {2 \over 1} = 2

Related Learning Resources & Exam Preparation