MHT CET 2017 Math Solved Paper

(a) It is given that, α and β are the roots of the equation
x2 + 5|x| - 6 = 0
Here, |x|2 + 6|x | - |x | - 6 = 0
⇒ |x|(|x| + 6) - 1(|x| + 6) = 0
⇒ (|x| + 6)(|x| - 1) = 0
|x| = - 6,1
As, modulus is always positive.
Therefore,|x| = 1 ⇒ x = ±1
Consider, α = land β = -1
Hence,
|tan−1α−tan−1β| = |tan−11−tan−1(−1)|
= |

π

4

−(−

π

4

)|=|

π

2

|