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AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 63 of 160

Marks: +1, -0

f: R \rightarrow R

is a function such that

|f(x)-y| \leq \;\frac{1}{2}|x-y| \forall x, y \in R

f'(x) \geq \frac{1}{2} \;\; \forall x \in R, f(1)=\frac{1}{2}

1
0
2
infinite

Solution: 👈: Video Solution

|f(x)-y| \leq \;\frac{1}{2}|x-y|, \forall x, y \in R

=\lim\limits_{h \rightarrow 0}|f(x+h)-f(x)| \leq \;\frac{1}{2}|x+h-x|

=\lim\limits_{h \rightarrow 0}\;\frac{|f(x+h)-f(x)|}{h} \leq \;\frac{1}{2}

f'(x) \leq \;\frac{1}{2}

But

f'(x) \geq \;\frac{1}{2}

(given)

\Rightarrow \;\; f'(x)=\;\frac{1}{2} \Rightarrow f(x)=\;\frac{1}{2}x+c

\Rightarrow \;\; f(1)=\;\frac{1}{2} \times 1+c \Rightarrow \;\frac{1}{2}=\;\frac{1}{2}+c

\;\; c=0

\therefore \;\; f(x)=\;\frac{1}{2}x \;\text{ or }\; y=\;\frac{1}{2}x

Substitute

y=\;\frac{1}{2}x

in equation

y=x^2-2x-5

to determine the number of points of intersection

\;\frac{1}{2}x=x^2-2x-5

\Rightarrow 2x^2-5x-10=0

\Rightarrow x=\frac{5 \pm \sqrt{25+80}}{4}=\frac{5 \pm \sqrt{105}}{4}

\therefore

There are two point of intersection.

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