NIMCET 2026 — Mathematics PYQ

To find the angle of intersection between two curves, we first determine their points of intersection and then calculate the angle between their tangents at those points.

Step 1: Find points of intersection

Equate the two functions:

$$(x-2{)}^2=-4+6x-x^2$$

$$x^2-4x+4=-4+6x-x^2$$

$$2x^2-10x+8=0$$

$$x^2-5x+4=0$$

$$(x-1)(x-4)=0$$

The curves intersect at $x=1$ and $x=4$.

Step 2: Find slopes of tangents at intersection points

Let $f(x)=(x-2{)}^2⟹f^{\prime }(x)=2(x-2)$

Let $g(x)=-4+6x-x^2⟹g^{\prime }(x)=6-2x$

At $x=1$:

$m_1=f^{\prime }(1)=2(1-2)=-2$

$m_2=g^{\prime }(1)=6-2(1)=4$

At $x=4$:

$m_1=f^{\prime }(4)=2(4-2)=4$

$m_2=g^{\prime }(4)=6-2(4)=-2$

Step 3: Calculate the acute angle

The angle $\theta$ between two curves is given by $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$:

$$\tan \theta =\left|\frac{-2-4}{1+(-2)(4)}\right|=\left|\frac{-6}{1-8}\right|=\left|\frac{-6}{-7}\right|=\frac{6}{7}$$

$$\theta ={\tan }^{-1}\frac{6}{7}$$

Correct Option: (b) ${\tan }^{-1}\frac{6}{7}$