Question

Please solve the question in attached file​

Answer 1

Given Question :-

If f(x) is a quadratic expression which is positive for all real values of x and g(x) = f(x) - f'(x) + f''(x), then for all real values of x,

(a) g(x) > 0

(b) g(x) < 0

(c) g(x) = 0

(d) g(x) < - 1

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

Given that,

• f(x) is a quadratic expression which is positive for all real values of x.

Let assume that

$\rm :\longmapsto\:f(x) = {ax}^{2} + bx + c \: such \: that \: f(x) > 0$

$\rm \implies\:\boxed{\tt{ a > 0 \: \: and \: \: {b}^{2} - 4ac < 0 }}- - - (1)$

Now,

$\red{\rm :\longmapsto\:f'(x) = 2ax + b}$

$\red{\rm :\longmapsto\:f''(x) = 2a}$

Now, Consider

$\rm :\longmapsto\:g(x) = f(x) - f'(x) + f''(x)$

$\rm :\longmapsto\:g(x) = {ax}^{2} + bx + c - 2ax - b + 2a$

$\rm :\longmapsto\:g(x) = {ax}^{2} + (b - 2a)x + c - b + 2a$

Now, g(x) is a quadratic expression with a > 0

So, Discriminant of g(x), which is evaluated as

$\rm :\longmapsto\:D = {(b - 2a)}^{2} - 4a(2a - b + c)$

$\rm \:  =  \: {b}^{2} + {4a}^{2} - \cancel{4ab} - {8a}^{2} + \cancel{4ab} - 4ac$

$\rm \:  =  \: {b}^{2} - {4a}^{2} - 4ac$

$\rm \:  =  \: {b}^{2} - 4ac - {4a}^{2}$

$\rm \:   \: < 0$

$[ \because \: {b}^{2} - 4ac < 0 \: \: \: and \: \: \: {4a}^{2} > 0 \: ]$

So, for g(x) we have

↝ coefficient of x², a > 0 and Discriminant, D < 0

$\bf\implies \:g(x) > 0$

So, option (a) is correct.

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Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac