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Options

A. `(f(c ))^(2)+3f(c )=(g(c ))^(2)+3g(c )` for some c `in` [0,1]
B. `(f(c ))^(2)+f(c )=(g(c ))^(2)`+3g(c ) for some c `in` [0,1]
C. `(f(c ))^(2)+3f(c )=(g(c ))^(2)`+g(c ) for some c `in` [0,1]
D. `(f(c ))^(2)=(g(c ))^(2)` for some c `in` [0,1]

Answer

1,4

Explanation

Let f(x) and g(X) assume their maximum value at `x_(1)` and `x_(2)` respectively where `x_(1)lex_(2)`
`therefore f(x_(1))=g(x_(2))=lambda`
Now let h(x)=f(x)-g(x)
`therefore h(x_(1))=f(x_(1))-g(X_(1))`
`=lambda-g(x_(1))`
`gt0`
and `h(_(2)=f(x_(2))-g(x_(1))`
`=lambda-g(x)_(1))`
`lt0`
If `x_(1)gtx_(2)` then `h(x_(1)lt0` and `h(x_(2))gt0`
So by intermediate value theorem for some c in [0,1]
h(c )=0
From `f(c )^(2)+3f(c )+g(c )+3=0`
So there exist a such that f(c )-g(c )=0
Hence (a) is correct
simiarly `f(c )^(2)=g(c )^(2)`
Thus (4) is correct
2 and 3 are wrong by counter example
If `f(x) =g(x)=lambda ne 0` then
`lambda^(2)+lambda=lambda^(2)+3lambda` which is not possible
and `lambda^(2)+3lambda=lambda^(2)+lambda` which is not possible