KEAM 2013 Math Paper

Given curve is
y=5+x−x2.....(i)
On differentiating w.r.t. x, we get
‌

dy

dx

=1−2x
Slope of normal =‌

−1

(dy∕dx)

=‌

−1

1−2x

=‌

1

2x−1

......(ii)
since, normal makes equal intercepts.
∴θ=135∘
From Eq. (ii),
‌

1

2x−1

=tan‌135∘=−1
⇒‌‌−2x+1=1⇒x=0
Then, from Eq. (i), y=5
So, the required point is (0,5)