Question

guys can anyone solve that problem immediately ​
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Answer 1

Here,

$\longrightarrow\sf{\dfrac{2x+1}{x^2-x-12}\leq\dfrac{1}{2}}$

$\longrightarrow\sf{\dfrac{2x+1}{x^2-x-12}-\dfrac{1}{2}\leq0}$

$\longrightarrow\sf{\dfrac{4x+2-x^2+x+12}{2(x^2-x-12)}\leq0}$

$\longrightarrow\sf{\dfrac{5x+14-x^2}{x^2-x-12}\leq0}$

$\longrightarrow\sf{\dfrac{x^2-5x-14}{x^2-x-12}\geq0}$

$\longrightarrow\sf{\dfrac{x^2-7x+2x-14}{x^2-4x+3x-12}\geq0}$

$\longrightarrow\sf{\dfrac{x(x-7)+2(x-7)}{x(x-4)+3(x-4)}\geq0}$

$\longrightarrow\sf{\dfrac{(x-7)(x+2)}{(x-4)(x+3)}\geq0}$

Case 1:- Let the numerator be non - negative and the denominator be positive.

$\longrightarrow\sf{(x-7)(x+2)\geq0\quad \land\quad (x-4)(x+3)\ \textgreater\ 0}$

$\Longrightarrow\sf{x\in\left[(-\infty,\ -2]\cup[7,\ \infty)\right]\cap[(-\infty,\ -3)\cup(4,\ \infty)]}$

$\longrightarrow\sf{x\in(-\infty,\ -3)\cup[7,\ \infty)\quad\quad\dots(1)}$

Case 2:- Let the numerator be non - positive and the denominator be negative.

$\longrightarrow\sf{(x-7)(x+2)\leq0\quad \land\quad (x-4)(x+3)\ \textless\ 0}$

$\Longrightarrow\sf{x\in[-2,\ 7]\cap(-3,\ 4)\quad\quad\dots(2)}$

Combining (1) and (2), we get,

$\longrightarrow\sf{x\in\big[(-\infty,\ -3)\cup[7,\ \infty)\big]\cup\big[[-2,\ 7]\cap(-3,\ 4)\big]}$

$\longrightarrow\underline{\underline{\sf{x\in(-\infty,\ -3)\cup[-2,\ 4)\cup[7,\ \infty)}}}$