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If `f: RrarrR` and `g: RrarrR` are defined by `f(x)=2x+3a n dg(x)=x^2+7,` then the value of `x` such that `g(f(x))=8` `1,2` `-1,2` `-1,-2` `1,-2`

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Verified by Experts

Since,
`f(x)=2x+3`
`g[f(x)]=8`
`=> g(2x+3)=8`
...
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