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JEE Advanced 2011 · Relations and Functions (Class 12)

Let f:(0,1)→Rf : (0, 1) \to \mathbf{R} be defined by f(x)=b−x1−bxf(x) = \frac{b - x}{1 - bx}, where bb is a constant such that 0<b<10 < b < 1. Then

Answers: A. ff is not invertible on (0,1)(0, 1) · B. f≠f−1f \neq f^{-1} on (0,1)(0, 1) and f′(b)=1f′(0)f'(b) = \frac{1}{f'(0)}

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Written solution

f(x)=b−x1−bx,0<b<1f(x) = \dfrac{b-x}{1-bx},\quad 0<b<1

f:(0,1)→Rf:(0,1)\to\mathbf{R}

check (1)–(4): invertible, f−1\text{check }(1)\text{--}(4):\ \text{invertible},\ f^{-1}

f′(b), f−1’s differentiabilityf'(b),\ f^{-1}\text{'s differentiability}

1. f(f(x))=?1.\ f(f(x)) = ?

2. find f′(x), f′(0), f′(b)2.\ \text{find } f'(x),\ f'(0),\ f'(b)

3. domain of f−13.\ \text{domain of } f^{-1}

f(f(x))=b−b−x1−bx1−b⋅b−x1−bxf(f(x)) = \dfrac{b - \frac{b-x}{1-bx}}{1 - b\cdot\frac{b-x}{1-bx}}

=x(1−b2)1−b2=x= \dfrac{x(1-b^2)}{1-b^2} = x

f(f(x))=x⇒f=f−1f(f(x)) = x \Rightarrow f = f^{-1}

⇒f invertible; (1), (2) FALSE\Rightarrow f \text{ invertible};\ (1),\ (2)\ \text{FALSE}

f′(x)=(−1)(1−bx)−(b−x)(−b)(1−bx)2f'(x) = \dfrac{(-1)(1-bx) - (b-x)(-b)}{(1-bx)^2}

=b2−1(1−bx)2= \dfrac{b^2-1}{(1-bx)^2}

f′(0)=b2−1f'(0) = b^2-1

f′(b)=b2−1(1−b2)2=−11−b2f'(b) = \dfrac{b^2-1}{(1-b^2)^2} = -\dfrac{1}{1-b^2}

1f′(0)=1b2−1=−11−b2=f′(b)\dfrac{1}{f'(0)} = \dfrac{1}{b^2-1} = -\dfrac{1}{1-b^2} = f'(b)

⇒f′(b)=1f′(0) ⇒ (3) TRUE\Rightarrow f'(b) = \dfrac{1}{f'(0)}\ \Rightarrow\ (3)\ \text{TRUE}

f decreasing on (0,1): f(0)=b, lim⁡x→1−f(x)=−1f \text{ decreasing on } (0,1):\ f(0)=b,\ \lim_{x\to1^-}f(x)=-1

range(f)=(−1,b)=domain(f−1)≠(0,1)\text{range}(f) = (-1,b) = \text{domain}(f^{-1}) \ne (0,1)

⇒(4) FALSE\Rightarrow (4)\ \text{FALSE}

f=f−1 and f′(b)=1f′(0)f=f^{-1}\ \text{and}\ f'(b)=\dfrac{1}{f'(0)}

option (3)\boxed{\text{option }(3)}

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