Sol. Graphs of the two curves y = |x 2 –1| and y = |x 2 –3| are shown in fig.

Clearly, two curves intersect at P and Q. Point P and Q are the intersection point of y = x 2 –1 and y = –(x 2 –3). Solving these two equations simultaneously, we get x = ± and y = ±1

Thus, the coordinates of P and Q are ( , 1) and (– , 1) respectively.

Angle of intersection at ( , 1):

The equations of the two curves are

y = x 2 –1 ... (i)

and y = –(x 2 –3) ... (ii)

⇒ = 2x [for curve (i)]

and, = –2x [for curve (ii)]

Let m 1 and m 2 be the slopes of the tangents at point P to curve (i) and (ii) respectively. Then,

m 1 = = 2 and m 2 = = –2

Let θ be the acute angle of intersection. Then,

tan θ =

⇒ tan θ = =

∴ θ = tan –1

Angle of intersection at Q (– , 1) :

In this case, we have

m 1 = Slope of the tangent to (i) curve at

Q =

= –2

and,

m 2 = slope of the tangent to (ii) curve at

Q =

= 2

So, the acute angle θ 1 between the tangents is given by

tan θ 1 = =

=

⇒ θ 1 = tan –1