MHT CET · Maths · Quadratic Equation
If \(\alpha\) and \(\beta\) are roots of the equation \(x^2+5|x|-6=0\) then the value of \(\left|\tan ^{-1} \alpha-\tan ^{-1} \beta\right|\) is
- A \(\frac{\pi}{2}\)
- B \(0\)
- C \(\pi\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(x^2+5|x|-6=0 \)
\( |x|^2+5|x|-6=0 \)
\( |x|^2+6|x|-|x|-6=0 \)
\( |x|(|x|+6)-1(|x|+6)=0 \)
\( (|x|-1)(|x|+6)=0\)
\(|x|=1\), nbsp \(;|x| \neq-6\) (Since modulus can not be giving negative values)
\(\therefore |x|=1\)
\(\therefore x= \pm 1\)
\(\alpha=1, \beta=-1\)
\(\therefore \left|\tan ^{-1} \alpha-\tan ^{-1} \beta\right|=\left|\tan ^{-1} 1-\tan ^{-1}(-1)\right|\)
\(=\left|\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right|\)
\(=\left|\frac{\pi}{2}\right|=\frac{\pi}{2}\)
\( |x|^2+5|x|-6=0 \)
\( |x|^2+6|x|-|x|-6=0 \)
\( |x|(|x|+6)-1(|x|+6)=0 \)
\( (|x|-1)(|x|+6)=0\)
\(|x|=1\), nbsp \(;|x| \neq-6\) (Since modulus can not be giving negative values)
\(\therefore |x|=1\)
\(\therefore x= \pm 1\)
\(\alpha=1, \beta=-1\)
\(\therefore \left|\tan ^{-1} \alpha-\tan ^{-1} \beta\right|=\left|\tan ^{-1} 1-\tan ^{-1}(-1)\right|\)
\(=\left|\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right|\)
\(=\left|\frac{\pi}{2}\right|=\frac{\pi}{2}\)
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