If f (x) = x ^ 2 + g ^ ' (1)x + g ^ " (2) and g(x) = f(1) x ^ 2 + x...

Given:

$f\left(x\right)$ = {x}^{2} + {g}^{'}\left(1\right)$x + {g}^{"}\left(2\right)$

$⇒ {f}^{'}\left(x\right) = 2x + {g}^{'}\left(1\right)</p><p>$⇒ {f}^{"}\left(x\right)$= 2$

Now,

$g\left(x\right)$ = f \left(1\right)${x}^{2} + x{f}^{'}\left(x\right) + {f}^{"}\left(x\right)</p><p>$⇒ g \left(x\right)$= f \left(1\right)${x}^{2} + x \left[2x + {g}^{'}\left(1\right)\right] + 2

$⇒ g \left(x\right)$ = \left[f\left(1\right) + 2\right]{x}^{2} + {g}^{'}\left(1\right)x + 2$</p><p>$⇒ {g}^{'}\left(x\right) = 2f\left(1\right)x + 4x + {g}^{'}\left(1\right) . . . . \left(1\right)

$⇒ {g}^{"}\left(x\right)$ = 2f \left(1\right)$+ 4 . . . \left(2\right)$

Put x = 1 in \left(1\right)$, then we get</p><p>$2f\left(1\right) + 4 = 0$</p><p>$⇒ f\left(1\right) = - 2$</p><p>$∴ {g}^{"}\left(2\right) = 2f\left(1\right) + 4 = 0$</p><p>Now,</p><p>$f\left(x\right) = {x}^{2} + {g}^{'}\left(1\right)x + {g}^{"}\left(2\right)$</p><p>Put$x = 1$, then we get</p><p>$f\left(1\right)$= 1 + {g}^{'}\left(1\right)$ + {g}^{"}\left(2\right)$</p><p>$⇒ {g}^{'}\left(1\right) = - 3$</p><p>So,</p><p>${f}^{'}\left(x\right) = 2x - 3$</p><p>$f\left(x\right)$= {x}^{2} - 3x + c$

Since, $f\left(1\right)$ = - 2 ⇒ c = 0$</p><p>So,</p><p>$f\left(x\right) = {x}^{2} - 3x$</p><p>And, </p><p>$g\left(x\right) = f\left(1\right){x}^{2} + x{f}^{'}\left(x\right)$+ {f}^{"}\left(x\right)$

$⇒ g\left(x\right) = - 2{x}^{2} + x\left(2x - 3\right) + 2$

$⇒ g\left(x\right) = - 3x + 2$

Hence, $f\left(4\right) - g\left(4\right) = 4 + 10 = 14$