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Let f(x) = x^2 + xg'(1) + g''(2) and g(x) = f(1)x^2 + xf'(x) + f''(x), find f(x) and g(x).

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Maths Class 12 Differential Calculus.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f(x) = {x}^{2} + xg'(1) + g''(2) - - - (1)$

So,

$\bf :\longmapsto\:f(1) = 1 + g'(1) + g''(2) - - - - (2)$

Also,

$\bf :\longmapsto\:f'(x) = 2x + g'(1) - - - - (3)$

Also,

$\bf :\longmapsto\:f''(x) = 2 - - - - (4)$

Now, Further given that

$\red{\rm :\longmapsto\:g(x) = f(1) {x}^{2} + xf'(x) + f''(x)}$

On substituting the values evaluated above in equation (2), (3) and (4), we get

$\red{\rm :\longmapsto\:g(x) = \bigg(1 + g'(1) + g''(2) \bigg) {x}^{2} + x(2x + g'(1)) + 2}$

$\red{\rm :\longmapsto\:g(x) = {x}^{2} + g'(1) {x}^{2} + g''(2){x}^{2} + 2 {x}^{2} + g'(1)x + 2}$

$\red{\rm :\longmapsto\:g(x) = {x}^{2}(3 + g'(1) + g''(2)) + xg'(1) + 2}$

So,

$\red{\rm :\longmapsto\:g'(x) =2x(3 + g'(1) + g''(2)) + g'(1)}$

Thus,

$\red{\rm :\longmapsto\:g'(1) =2(3 + g'(1) + g''(2)) + g'(1)}$

$\red{\rm :\longmapsto\:0=2(3 + g'(1) + g''(2))}$

$\red{\rm :\longmapsto \: 3 + g'(1) + g''(2) = 0 - - - - (5)}$

Also,

$\red{\rm :\longmapsto\:g''(x) =2(3 + g'(1) + g''(2))}$

Thus,

$\red{\rm :\longmapsto\:g''(2) =2(3 + g'(1) + g''(2))}$

$\red{\rm :\longmapsto\:g''(2) =6 + 2g'(1) + 2g''(2)}$

$\red{\rm :\longmapsto\:g''(2) =6 + 2g'(1) + 2g''(2)}$

$\red{\rm :\longmapsto\:6 + 2g'(1) + g''(2) = 0 - - - - (6)}$

On Subtracting equation (5) from equation (6), we get

$\red{\rm :\longmapsto\:g'(1) + 3 = 0}$

$\bf\implies \:g'(1) = - \: 3$

On substituting this value in equation (5), we get

$\red{\rm :\longmapsto \: 3 - 3 + g''(2) = 0}$

$\bf\implies \:g'(2) =\: 0$

Thus,

$\bf :\longmapsto\:f'(x) = 2x + g'(1) = 2x - 3$

So, on substituting the values of g'(1) and g''(2), in equation (1), we get

$\red{\rm :\longmapsto\: \red{\boxed{ \bf{ \: f(x) = {x}^{2} - 3x}}}}$

and

$\red{\rm :\longmapsto\: \red{\boxed{ \bf{ \: g(x) = - 3x + 2}}}}$