Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln

1.	Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15
Answer» We're given the inequality, $\longrightarrow 4(\ln x)^3-8(\ln x)^2-11\ln x+15\leq0$ Let us solve this inequality first. We can see the equality holds TRUE for $\ln x=1.$ So we can divide our inequality by $\ln x-1$ like, $\longrightarrow 4(\ln x)^3-4(\ln x)^2-4(\ln x)^2+4\ln x-15\ln x+15\leq0$ $\longrightarrow 4(\ln x)^2\left(\ln x-1\right)-4\ln x\left(\ln x-1\right)-15\left(\ln x-1\right)\leq0$ $\longrightarrow\left(4(\ln x)^2-4\ln x-15\right)\left(\ln x-1\right)\leq0$ $\longrightarrow\left(4(\ln x)^2-10\ln x+6\ln x-15\right)\left(\ln x-1\right)\leq0$ $\longrightarrow\left(2\ln x\left(2\ln x-5\right)+3(2\ln x-5)\right)\left(\ln x-1\right)\leq0$ $\longrightarrow\left(2\ln x+3\right)\left(2\ln x-5\right)\left(\ln x-1\right)\leq0$ Now the whole LHS is factorised here and we can apply wavy curve method, for $\ln x.$ $\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\thicklines\put(0,0){\vector(1,0){60}}\multiput(15,0)(15,0){3}{\circle{1}}\put(13,-4){$-\frac{3}{2}$}\put(29.5,-4){$1$}\put(43,-4){$\frac{5}{2}$}\put(60,-3){$\ln x$}\qbezier(45,0)(52.5,5)(60,10)\qbezier(30,0)(37.5,-5)(45,0)\qbezier(15,0)(22.5,5)(30,0)\qbezier(0,-10)(7.5,-5)(15,0)\end{picture}$ So the condition for $\ln x$ to satisfy our inequality is given by this wavy curve as, $\longrightarrow\ln x\in\left(-\infty,\ -\dfrac{3}{2}\right]\cup\left[1,\ \dfrac{5}{2}\right]$ Taking antilog (possible since $e^x$ is an increasing function), $\longrightarrow x\in\left(e^{-\infty},\ e^{-\frac{3}{2}}\right]\cup\left[e^1,\ e^{\frac{5}{2}}\right]$ $\longrightarrow x\in\left(0,\ e^{-\frac{3}{2}}\right]\cup\left[e,\ e^{\frac{5}{2}}\right]\quad\quad\dots(1)$ But in the QUESTION, the solution is given as, $\longrightarrow x\in\left(a,\ e^{-b}\right]\cup\left[e^c,\ e^d\right]\quad\quad\dots(2)$ COMPARING (1) and (2) we GET, $a=0$ $b=\dfrac{3}{2}$ $c=1$ $d=\dfrac{5}{2}$ So, $\longrightarrow I=a^2+b^2+c^2+d^2$ $\longrightarrow I=0^2+\left(\dfrac{3}{2}\right)^2+1^2+\left(\dfrac{5}{2}\right)^2$ $\longrightarrow I=\dfrac{38}{4}$ $\longrightarrow I=9.5$ $\longrightarrow\underline{\underline{[I]=9}}$ Hence (D)* is the ANSWER.

Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15

Answer»

We're given the inequality,

$\longrightarrow 4(\ln x)^3-8(\ln x)^2-11\ln x+15\leq0$

Let us solve this inequality first.

We can see the equality holds TRUE for $\ln x=1.$

So we can divide our inequality by $\ln x-1$ like,

$\longrightarrow 4(\ln x)^3-4(\ln x)^2-4(\ln x)^2+4\ln x-15\ln x+15\leq0$

$\longrightarrow 4(\ln x)^2\left(\ln x-1\right)-4\ln x\left(\ln x-1\right)-15\left(\ln x-1\right)\leq0$

$\longrightarrow\left(4(\ln x)^2-4\ln x-15\right)\left(\ln x-1\right)\leq0$

$\longrightarrow\left(4(\ln x)^2-10\ln x+6\ln x-15\right)\left(\ln x-1\right)\leq0$

$\longrightarrow\left(2\ln x\left(2\ln x-5\right)+3(2\ln x-5)\right)\left(\ln x-1\right)\leq0$

$\longrightarrow\left(2\ln x+3\right)\left(2\ln x-5\right)\left(\ln x-1\right)\leq0$

Now the whole LHS is factorised here and we can apply wavy curve method, for $\ln x.$

$\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\thicklines\put(0,0){\vector(1,0){60}}\multiput(15,0)(15,0){3}{\circle*{1}}\put(13,-4){$-\frac{3}{2}$}\put(29.5,-4){$1$}\put(43,-4){$\frac{5}{2}$}\put(60,-3){$\ln x$}\qbezier(45,0)(52.5,5)(60,10)\qbezier(30,0)(37.5,-5)(45,0)\qbezier(15,0)(22.5,5)(30,0)\qbezier(0,-10)(7.5,-5)(15,0)\end{picture}$

So the condition for $\ln x$ to satisfy our inequality is given by this wavy curve as,

$\longrightarrow\ln x\in\left(-\infty,\ -\dfrac{3}{2}\right]\cup\left[1,\ \dfrac{5}{2}\right]$

Taking antilog (possible since $e^x$ is an increasing function),

$\longrightarrow x\in\left(e^{-\infty},\ e^{-\frac{3}{2}}\right]\cup\left[e^1,\ e^{\frac{5}{2}}\right]$

$\longrightarrow x\in\left(0,\ e^{-\frac{3}{2}}\right]\cup\left[e,\ e^{\frac{5}{2}}\right]\quad\quad\dots(1)$

But in the QUESTION, the solution is given as,

$\longrightarrow x\in\left(a,\ e^{-b}\right]\cup\left[e^c,\ e^d\right]\quad\quad\dots(2)$

COMPARING (1) and (2) we GET,

$a=0$

$b=\dfrac{3}{2}$

$c=1$

$d=\dfrac{5}{2}$

So,

$\longrightarrow I=a^2+b^2+c^2+d^2$

$\longrightarrow I=0^2+\left(\dfrac{3}{2}\right)^2+1^2+\left(\dfrac{5}{2}\right)^2$

$\longrightarrow I=\dfrac{38}{4}$

$\longrightarrow I=9.5$

$\longrightarrow\underline{\underline{[I]=9}}$

Hence (D) is the ANSWER.

Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15

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