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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
f(x)=ax+bcx+df(x) = \dfrac{ax+b}{cx+d} for all x∈R−{−dc}x \in \mathbb{R}-\left\{-\dfrac{d}{c}\right\}. If f(5)=5f(5) = 5, f(13)=13f(13) = 13, and f(f(x))=xf(f(x)) = x for all xx, then the range of f(x)=R−{x}f(x) = \mathbb{R}-\{x\}. Then xx.
Solution
Answer: 9
Step 1: Use f(f(x))=xf(f(x)) = x to get d=−ad = -a
f(x)=ax+bcx−af(x) = \frac{ax+b}{cx-a}
 Step 2: Set up and solve the system from fixed points f(5)=5f(5) = 5: 5a+b=5(5c−a)⇒10a+b=25c5a+b = 5(5c-a) \Rightarrow 10a+b = 25c. f(13)=13f(13) = 13: 13a+b=13(13c−a)⇒26a+b=169c13a+b = 13(13c-a) \Rightarrow 26a+b = 169c. Subtracting: 16a=144c⇒a=9c16a = 144c \Rightarrow a = 9c. Then b=25c−90c=−65cb = 25c-90c = -65c. Step 3: Find the excluded value in the range
f(x)=9cx−65ccx−9c=9x−65x−9f(x) = \frac{9cx-65c}{cx-9c} = \frac{9x-65}{x-9}
 Setting y=9x−65x−9y = \dfrac{9x-65}{x-9} and solving for xx: x=9y−65y−9x = \dfrac{9y-65}{y-9}, undefined when y=9y = 9. Range =R−{9}= \mathbb{R}-\{9\}, so x=9x = 9. Answer: 9
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