We're given the inequality,Let us solve this inequality first.We can see the equality holds TRUE for So we can divide our inequality by like,Now the whole LHS is factorised here and we can apply wavy

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Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15<=0 is(a, e^-b]U[e^c,e^d] and I = a^2+ b^2 + c^2 + d^2.If [.] represents greatest integer function, then [l] is equal to(A) 5(B) 7 (C) 8(D) 9

Math Secondary School in Math 8 months ago

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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.

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Posted on 23 Oct 2024, this text provides information on Math related to Secondary School in Math. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15<=0 is(a, e^-b]U[e^c,e^d] and I = a^2+ b^2 + c^2 + d^2.If [.] represents greatest integer function, then [l] is equal to(A) 5(B) 7 (C) 8(D) 9​

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Set of all real values of x satisfying the inequality 4(ln x)^3 -8(ln x)^2 - 11(en x) + 15<=0 is(a, e^-b]U[e^c,e^d] and I = a^2+ b^2 + c^2 + d^2.If [.] represents greatest integer function, then [l] is equal to(A) 5(B) 7 (C) 8(D) 9