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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 66 of 160

Marks: +1, -0

f(x)

\begin{cases} 1+\cos x, & x\leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{cases}

a^2+b^2

Solution: 👈: Video Solution

\because f(x)

is continuous everywhere

\Rightarrow f(x)

is continuous at

x=0

and

x=2

\therefore \;\;

LHL at

(x=0)=

RHL at

(x=0)=f(0)

\lim\limits_{x \rightarrow 0^{-}}f(x)=\lim\limits_{x \rightarrow 0^{+}}f(x)=2

\;\; \lim\limits_{x \rightarrow 0^{-}}(1+\cos x) \;\; =\lim\limits_{x \rightarrow 0^{+}}(a-x)=2

\Rightarrow \;\; 1+1 \;\; =a-0=2

\Rightarrow \;\; a \;\; =2

and LHL at

(x=2)=

RHL at

(x=2)=f(2)

\;\lim\limits_{x \rightarrow 2^{-}}f(x)= \;\; \lim\limits_{x \rightarrow 2^{+}}f(x)=a-2

\Rightarrow\;\lim\limits_{x \rightarrow 2^{-}}(a-x)=\lim\limits_{x \rightarrow 2^{+}}(x^{2}-b^{2})=2-2

\;\text{ or }\;\;2-2 \;\; =2^{2}-b^{2}=0

\Rightarrow\;b^{2}=4

\therefore \;a^{2}+b^{2}=4+4=8

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