Quadratic Equation and Inequalities

To solve for the coefficient of $ x^3 $ in the function $ h(x) = f(x + 1) - g(x + 2) $, we start by expanding both $ f(x + 1) $ and $ g(x + 2) $.

First, expand $ f(x + 1) $:

$ f(x+1) = a_1 + 10(x+1) + a_2 (x+1)^2 + a_3 (x+1)^3 + (x+1)^4 $

From this expansion, the coefficient of $ x^3 $ in $ f(x+1) $ is $ a_3 + 4 $.

Next, expand $ g(x + 2) $:

$ g(x+2) = b_1 + 3(x+2) + b_2 (x+2)^2 + b_3 (x+2)^3 + (x+2)^4 $

Here, the coefficient of $ x^3 $ in $ g(x+2) $ is $ b_3 + 8 $.

Now, calculate the coefficient of $ x^3 $ in $ h(x) = f(x+1) - g(x+2) $:

$ \text{Coefficient of } x^3 \text{ in } h(x) = (a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4 $

Given that $ f(x) \neq g(x) $ for every $ x \in \mathbb{R} $, it follows that $ f(x) - g(x) \neq 0 $ for any $ x $. This means the equation:

$ (a_1 - b_1) + 7x + (a_2 - b_2)x^2 + (a_3 - b_3)x^3 = 0 $

cannot have any real roots. For this polynomial to have no real roots, the degree of the polynomial must be less than or equal to zero, implying that the coefficients of the highest degree terms must equate to zero. Thus, $ a_3 - b_3 = 0 $.

Substituting $ a_3 - b_3 = 0 $ into our earlier expression for the coefficient of $ x^3 $ in $ h(x) $ gives:

$ a_3 - b_3 - 4 = -4 $

Thus, the coefficient of $ x^3 $ in $ h(x) $ is $-4$.

Given, a quadratic equation $p(x)=0$ with real coefficients has purely imaginary roots.

Let $i \lambda$ and $-i \lambda$ are the roots of $p(x)=0$ where $i=\sqrt{-1}$ and $\lambda$ is a real number except zero.

$\begin{aligned} & \therefore p(x)=a(x-i \lambda)(x+i \lambda) \\ & \Rightarrow p(x)=a\left(x^2+\lambda^2\right) \end{aligned}$

Now, $\quad p(p(x))=0$

$\Rightarrow \quad a\left((p(x))^2+\lambda^2\right)=0$

$\Rightarrow a\left[a^2\left(x^2+\lambda^2\right)^2+\lambda^2\right]=0$

$\Rightarrow a^2\left(x^2+\lambda^2\right)^2+\lambda^2=0$

$\Rightarrow \quad\left(x^2+\lambda^2\right)^2=-\frac{\lambda^2}{a^2}$

$\Rightarrow \quad x^2+\lambda^2= \pm i \frac{\lambda}{a}$

$\Rightarrow \quad x^2= \pm i \frac{\lambda}{a}-\lambda^2$

$\Rightarrow \quad x= \pm \sqrt{ \pm i \frac{\lambda}{a}-\lambda^2}$

Hence, $p(p(x))=0$ has four roots but all the roots are neither purely real nor purely imaginary.

Hint:

(i) In a quadratic equation imaginary roots are always in conjugate pair i.e, if one root is $p+i q$, then other root must be $p-i q$

(ii) A quadratic equation with real coefficients has purely imaginary roots, then consider $\pm i \lambda$ are the roots of the given quadratic, where $\lambda \in \mathrm{R}-\{0\}$.

Let a + 1 = t⁶. Thus, when a $\to$ 0, t $\to$ 1.

$\therefore$ $({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$

$ \Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$,

as t $\to$ 1

$2{x^2} + 3x + 1 = 0$

$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$

$ \Rightarrow (2x + 1)(x + 1) = 0$

Thus, x = $-$1, $-$1/2

or, $\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$

and $\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$

We have,

${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$ ...... (1)

${3^{\ln x}} = {2^{\ln y}}$ ....... (2)

$ \Rightarrow (\log x)(\log 3) = (\log y)\log 2$

$ \Rightarrow \log y = {{(\log x)(\log 3)} \over {\log 2}}$ ........ (3)

Taking log both sides of Eq. (1), we get

$(\log 2)\{ \log 2 + \log x\} = \log 3\{ \log 3 + \log y\} $

${(\log 2)^2} + (\log 2)(\log x) = {(\log 3)^2} + {{{{(\log 3)}^2}(\log x)} \over {\log 2}}$ from Eq. (3)

$ \Rightarrow {(\log 2)^2} - {(\log 3)^2} = {{{{(\log 3)}^2} - {{(\log 2)}^2}} \over {\log 2}}(\log x)$

$ \Rightarrow - \log 2 = \log x$

$ \Rightarrow x = {1 \over 2} \Rightarrow {x_0} = {1 \over 2}$.

We have, ${a_n} = {\alpha ^n} - {\beta ^n}$

${\alpha ^2} - 6\alpha - 2 = 0$

Multiplying with $\alpha$⁸ on both sides, we get

${\alpha ^{10}} - 6{\alpha ^9} - 2{\alpha ^8} = 0$ ..... (1)

Similarly, ${\beta ^{10}} - 6{\beta ^9} - 2{\beta ^8} = 0$ ..... (2)

From Eqs. (1) and (2), we get

${\alpha ^{10}} - {\beta ^{10}} - 6({\alpha ^9} - {\beta ^9}) = 2({\alpha ^8} - {\beta ^8})$

$ \Rightarrow {a_{10}} - 6{a_9} = 2{a_8} \Rightarrow {{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$.

The given equations are

${x^2} + bx - 1 = 0$

${x^2} + x + b = 0$ ....... (1)

Common root is $(b - 1)x - 1 - b = 0$.

$ \Rightarrow x = {{b + 1} \over {b - 1}}$

This value of x satisfies Eq. (1), we get

${{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}} + {{b + 1} \over {b - 1}} + b = 0$

$ \Rightarrow b = i\sqrt 3 ,\, - i\sqrt 3 ,\,0$

Sum of roots = ${{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}$ and product = 1

Given, $\alpha$ + $\beta$ = $-$ p and $\alpha$³ + $\beta$³ = q

$ \Rightarrow (\alpha + \beta )({\alpha ^2} - \alpha \beta + {\beta ^2}) = q$

$\therefore$ ${\alpha ^2} + {\beta ^2} - \alpha \beta = {{ - q} \over p}$ ..... (i)

and ${(\alpha + \beta )^2} = {p^2}$

$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$ ..... (ii)

From Eq. (i) and (ii), we get

${\alpha ^2} + {\beta ^2} = {{{p^3} - 2q} \over {3p}}$

and $\alpha \beta = {{{p^3} + q} \over {3p}}$

$\therefore$ Required equation is

${x^2} - {{({p^2} - 2q)x} \over {({p^3} + q)}} + 1 = 0$

$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$

As a, b, c, p, q $\in$ R and the two given equations have exactly one common root.

$\Rightarrow$ Either both equations have real roots.

Or both equations have imaginary roots.

$\Rightarrow$ Either $\Delta_1\ge0$ and $\Delta_2 > 0$ or

$\Delta_1 < 0$ and $\Delta_2 < 0$

$\Rightarrow~p^2-q\ge0$ and $b^2-ac\ge0$

or $p^2-q\le0$ and $b^2-ac\le0$

$\Rightarrow~(p^2-q)(b^2-ac)\ge0$

$\therefore$ Statement 1 is true.

Also, we have $\alpha\beta=q$ and ${\alpha \over \beta } = {c \over a}$

$\therefore$ ${{\alpha \beta } \over {{\alpha \over \beta }}} = {q \over c} \times a \Rightarrow {\beta ^2} = {{qa} \over c}$

As $\beta \ne 1$ or $-1$

$ \Rightarrow {\beta ^2} \ne 1$

$ \Rightarrow {{qa} \over c} \ne 1$ or $c \ne qa$

Again as exactly one root $\alpha$ is common and $\beta\ne1$

$\therefore$ $\alpha + \beta \ne \alpha + {1 \over \beta } \Rightarrow {{ - 2b} \over a} \ne - 2P$

$ \Rightarrow b \ne ap$

$\therefore$ Statement 2 is correct.

But Statement 2 is not correct explanation of Statement 1.

Since $\alpha,\beta$ are the roots of

${x^2} - px + r = 0$

$\therefore$ $\alpha + \beta = p$ and $\alpha \beta = r$

We know that

${\alpha \over 2}$ and 2$\beta$ are the roots of the equation ${x^2} - qx + r = 0$

$\therefore$ ${\alpha \over 2} + 2\beta = q$ and ${\alpha \over 2} \times 2\beta = r$

Solving $\alpha + \beta = p$ and ${\alpha \over 2} + 2\beta = q$

We get $\alpha = {2 \over 3}(2p - q)$ and $\beta = {1 \over 3}(2q - p)$

$\alpha \beta = r$

$r = {2 \over 9}(2p - q)(2q - p)$

$x^{2}+2(a+b+c) x+3 \lambda(a b+b c+c a)=0$

For the given quadratic equation has real roots, we must have

$\{2(a+b+c)\}^{2}-4 \times 3 \lambda(a b+b c+c a) \geq 0$

i.e, Discriminant $\geq 0$

$\Rightarrow 4(a+b+c)^{2} \geq 12 \lambda(a b+b c+a c)$

$\Rightarrow \lambda \leq \frac{4}{12} \frac{(a+b+c)^{2}}{(a b+b c+a c)}$

$\Rightarrow \lambda \leq \frac{(a+b+c)^{2}}{3(a b+b c+a c)}=\frac{\left(a^{2}+b^{2}+c^{2}\right)}{3(a b+b c+a c)}+\frac{2}{3}$ ...... (i)

$\left(\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c\right)$

In $\triangle \mathrm{ABC}$

$|a-b| < c$

$\Rightarrow \quad(a-b)^{2} < c^{2}$

$\Rightarrow \quad a^{2}+b^{2}-2 a b < c^{2}$ .... (ii)

Similarly $b^{2}+c^{2}-2 b c < a^{2}$ ..... (iii)

$a^{2}+c^{2}-2 a c < b^{2}$ ..... (iv)

Adding (ii), (iii) and (iv)

$2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+a c) < $

$\left(a^{2}+b^{2}+c^{2}\right)$

$\Rightarrow a^{2}+b^{2}+c^{2} < 2(a b+b c+a c)$

$\Rightarrow \frac{a^{2}+b^{2}+c^{2}}{a b+b c+a c} < 2$

From (i), we have

$\lambda \leq \frac{a^{2}+b^{2}+c^{2}}{3(a b+b c+a c)}+\frac{2}{3}$

$\begin{array}{ll} \Rightarrow \lambda < \frac{2}{3}+\frac{2}{3} \quad \because \frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} < 2 \\ \Rightarrow \lambda < \frac{4}{3} \end{array}$