For every pair of continuous functions f, g:[ | vedclass

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Question

Consider $h(x)=f(x)-g(n) ext { Assume } \quad a < b$ $h(a)=\lambda-g(a)>0 $ $h(b)=f(b)-\lambda<0 $ $ ext { else if } a > b

Vedclass · Accepted Answer

$(A,D)$

Vedclass · Answer

Consider $h(x)=f(x)-g(n) ext { Assume } \quad a < b$ $h(a)=\lambda-g(a)>0 $ $h(b)=f(b)-\lambda<0 $ $ ext { else if } a > b h(a) < 0 ext { and } h(b) > 0 .$ By intermediate value theorem $\Rightarrow h(c)=0$ $\quad\quad.......(1)$ $(A)$ $\quad( f ( c ))^2+3 f ( c )=( g ( c ))^2+3 g ( c )$ $( f ( c )- g ( c ))( f ( c )+ g ( c )+3)=0$ So there exist a ' $c$ ' : $f(c)-g(c)$ from (1). Hence A is correct. $(D)$ Similarly $( f ( c ))^2=( g ( c ))^2$ $(f(c)-g(c))(f(c)+g(c))=0$ $\Rightarrow \quad( D )$ is correct. $B$ \& $C$ are wrong as by counter eg If $f(x)=g(x)=\lambda eq 0$, then $B ightarrow \lambda^2+\lambda=\lambda^2+3 \lambda$ is not possible. $C ightarrow \lambda^2+3 \lambda=\lambda^2+\lambda$ is not possible.

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