Manipal Entrance Test 2014 Math Paper

we know that funtion |x| is not differentiable at x=0
∴ |x2−3x+2 |=|(x−1)(x−2)|
Hence, it is not differentiable at x=1 and 2
Now, f(x)= (x2−1 ) |x2−3x+2 |+cos|x| is not differentiable at x=2
For 1<x<2,f(x)=−(x2−1 ) (x2−3x+2 )+cos‌x
For 2<x<3,f(x)= (x2−1 ) (x2−3x+2 )+cos‌x
Lf′(x)=− (x2−1 )(2x−3)−2x (x2−3x+2 )−sin‌x
L′(2)=−3‌sin‌2
Rf′(x)= (x2−1 )(2x−3)+2x (x2−3x+2 )−sin‌x
Rf′(2)=(4−1)(4−3)+0−sin‌2 =3−sin‌2
Hence, L′(2)≠Rf′(2)
So, f(x) is not differentable at x=2