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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Functions

BITSAT 2016 · Mathematics · Functions

Let f(x)=(a x+b)/(c x+d), then fof (x)=x, provided that:

Let f(x)=ax+bcx+d,f(x)=\frac{a x+b}{c x+d}, then fof (x)=x,(x)=x, provided that:

Options

  1. Ad=−a\mathrm{d}=-\mathrm{a}
  2. Bd=a\mathrm{d}=\mathrm{a}
  3. Ca=b=1\quad a=b=1
  4. Da=b=c=d=1\quad a=b=c=d=1

Correct answer

A. d=−a\mathrm{d}=-\mathrm{a}

Step-by-step solution

f(x)=ax+bcx+df(x)=\frac{a x+b}{c x+d} fof (x)=a{ax+bcx+d}+bc{ax+bcx+d}+d⇒a2x+ab+bcx+bdacx+bc+cdx+d2=x⇒(ac+dc)x2+(bc+d2−bc−a2)x−ab−bd=0,∀x∈R⇒(a+d)c=0,d2−a2=0(x)=\frac{a\left\{\frac{a x+b}{c x+d}\right\}+b}{c\left\{\frac{a x+b}{c x+d}\right\}+d} \Rightarrow \frac{a^{2} x+a b+b c x+b d}{a c x+b c+c d x+d^{2}}=x\Rightarrow(a c+d c) x^{2}+\left(b c+d^{2}-b c-a^{2}\right) x-a b-b d=0, \quad \forall x \in \mathbb{R}\Rightarrow(a+d) c=0, d^{2}-a^{2}=0 and (a+d)b=0⇒a+d=0(a+d) b=0\Rightarrow a+d=0
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