QBManager

Application of Derivatives

85796 If $θ$ is the acute angle between the curves $x^{2} + y^{2} = 4$ and $y^{2} = 3 x$ then $\tan θ$

1

\frac{5}{\sqrt{3}}

2

\frac{\sqrt{3}}{4}

3

\frac{4}{\sqrt{3}}

4

\frac{\sqrt{3}}{5}

Explanation:

(A) : Given curve,
$x^{2} + y^{2} = 4$
And $y^{2} = 3 x$
Point of contact is obtained by solving above two equations-
From equation (i) and (ii)-
$x^{2} + 3 x - 4 = 0$
$\Rightarrow (x + 4) (x - 1) = 0 x = - 4, 1$
Neglecting $x = - 4$ , for $x = 1$
$y = \sqrt{3}$
$∴$ Point of contact is $(1, \sqrt{3})$
For curve $x^{2} + y^{2} = 1 = \frac{d y}{d x} = (\frac{- x}{y})$
$∴ m = {(\frac{dy}{dx})}_{(1, \sqrt{3})} = \frac{- 1}{\sqrt{3}}$
for $y^{2} = 3 x \Rightarrow {(\frac{d y}{d x})}_{(1, \sqrt{3})} = \frac{\sqrt{3}}{2}$
$∴$ Angle between the curve $=$ angle between their normal
$∴ \tan θ = \frac{\sqrt{3} + \frac{2}{\sqrt{3}}}{1 + \sqrt{3} (\frac{- 2}{\sqrt{3}})}$
$= \frac{- 5}{\sqrt{3}}$
$∵$ Angle is acute angle, then-
$\tan θ = \frac{5}{\sqrt{3}}$

Shift-I

Application of Derivatives

85797 The angle between the curves $2 x^{2} + y^{2} = 20$ and $4 y^{2} - x^{2} = 8$ at a point where they intersect in the $4^{th}$ quadrant $i$

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

(B): Given,
$2 x^{2} + y^{2} = 20$
$4 y^{2} - x^{2} = 8$
After solving both equation we get-
$(x, y) \to (2 \sqrt{2}, - 2)$
$C_{1} = 2 x^{2} + y^{2} = 20$
$4 x + 2 y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{- 2 x}{y}$
$\Rightarrow m_{1} = 2 \sqrt{2}$
$C_{2} \equiv 4 y^{2} - x^{2} = 8$
$8 y \frac{dy}{dx} - 2 x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{4 y}$
$\frac{dy}{dx} at (2 \sqrt{2}, - 2) = \frac{- 1}{2 \sqrt{2}} \Rightarrow | m_{2} = \frac{- 1}{2 \sqrt{2}}$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 \sqrt{2} + \frac{1}{2 \sqrt{2}}}{1 - (2 \sqrt{2}) \times \frac{1}{2 \sqrt{2}}} | = \frac{8 + 9}{\frac{2 \sqrt{2}}{0}} = \infty$
$θ = \frac{π}{2}$

TS EAMCET-2021-04.08.2021

Application of Derivatives

85798 The angle between the curves $y = \sin x$ and $y =$ $\cos x, 0 < x < \frac{π}{2}$

1

\tan^{- 1} (\sqrt{2})

2

\tan^{- 1} (2 \sqrt{2})

3

\tan^{- 1} (3 \sqrt{2})

4

\tan^{- 1} (3 \sqrt{3})

Explanation:

(B) : Given, $y = \sin x$ and $y = \cos x$
$\sin x = \cos x$
$\frac{\sin x}{\cos x} = 1$
$\tan x = 1$
$x = 45^{\circ} or \frac{π}{4}$
Differentiate the given Equation with respect to $x$
$y = \sin x$
$\frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \cos \frac{π}{4}$
$m_{1} = \frac{1}{\sqrt{2}}$
And,
$y = \cos x$
$\frac{d y}{d x} = - \sin x \Rightarrow \frac{d y}{d x} = - \sin \frac{π}{4}$
$m_{2} = - \frac{1}{\sqrt{2}}$
We know that
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
Substituting the value of $m_{1}$ and $m_{2}$
$\tan θ = | \frac{- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{1 + (\frac{1}{\sqrt{2}}) (- \frac{1}{\sqrt{2}})} | \Rightarrow \tan θ = | \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} |$
$\tan θ = | \frac{\frac{2}{\sqrt{2}}}{\frac{2 - 1}{2}} | \Rightarrow \tan θ = 2 \sqrt{2}$
$∴ θ = \tan^{- 1} 2 \sqrt{2}$

MHT CET-2022

Application of Derivatives

85799 If the angle between the curves $y = 2^{x}$ and $y =$ $3^{x}$ is $α$ , then the value of $\tan α$ , is equal to

1

\frac{\log (\frac{3}{2})}{1 + (\log 2) (\log 3)}

2

\frac{6}{7}

3

\frac{1}{7}

4

\frac{\log (6)}{1 + (\log 2) (\log 3)}

5 0

Explanation:

(A) : Given curves are $y = 2^{x}$ and $y = 3^{x}$
The point of intersection is $3^{x} = 2^{x} x = 0$
On differentiating w.r.t. $x$ , we get
${\begin{cases} \frac{d y}{d x} = 2^{x} \log 2 = m_{1} \\ And, \frac{d y}{d x} = 3^{x} \log 3 = m_{2} \end{cases}$
Therefore, $\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$
$= \frac{3^{x} \log 3 - 2^{x} \log 2}{1 + 3^{x} \times 2^{x} \log 3 \times \log 2}$
At, $x = 0$
$\tan a = \frac{3^{0} \log 3 - 2^{0} \log 2}{1 + 3^{0} \times 2^{0} \log 2 \log 3}$
$= \frac{\log \frac{3}{2}}{1 + \log 2 \log 3}$

Kerala CEE-2012

Application of Derivatives

85796 If $θ$ is the acute angle between the curves $x^{2} + y^{2} = 4$ and $y^{2} = 3 x$ then $\tan θ$

1

\frac{5}{\sqrt{3}}

2

\frac{\sqrt{3}}{4}

3

\frac{4}{\sqrt{3}}

4

\frac{\sqrt{3}}{5}

Explanation:

(A) : Given curve,
$x^{2} + y^{2} = 4$
And $y^{2} = 3 x$
Point of contact is obtained by solving above two equations-
From equation (i) and (ii)-
$x^{2} + 3 x - 4 = 0$
$\Rightarrow (x + 4) (x - 1) = 0 x = - 4, 1$
Neglecting $x = - 4$ , for $x = 1$
$y = \sqrt{3}$
$∴$ Point of contact is $(1, \sqrt{3})$
For curve $x^{2} + y^{2} = 1 = \frac{d y}{d x} = (\frac{- x}{y})$
$∴ m = {(\frac{dy}{dx})}_{(1, \sqrt{3})} = \frac{- 1}{\sqrt{3}}$
for $y^{2} = 3 x \Rightarrow {(\frac{d y}{d x})}_{(1, \sqrt{3})} = \frac{\sqrt{3}}{2}$
$∴$ Angle between the curve $=$ angle between their normal
$∴ \tan θ = \frac{\sqrt{3} + \frac{2}{\sqrt{3}}}{1 + \sqrt{3} (\frac{- 2}{\sqrt{3}})}$
$= \frac{- 5}{\sqrt{3}}$
$∵$ Angle is acute angle, then-
$\tan θ = \frac{5}{\sqrt{3}}$

Shift-I

Application of Derivatives

85797 The angle between the curves $2 x^{2} + y^{2} = 20$ and $4 y^{2} - x^{2} = 8$ at a point where they intersect in the $4^{th}$ quadrant $i$

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

(B): Given,
$2 x^{2} + y^{2} = 20$
$4 y^{2} - x^{2} = 8$
After solving both equation we get-
$(x, y) \to (2 \sqrt{2}, - 2)$
$C_{1} = 2 x^{2} + y^{2} = 20$
$4 x + 2 y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{- 2 x}{y}$
$\Rightarrow m_{1} = 2 \sqrt{2}$
$C_{2} \equiv 4 y^{2} - x^{2} = 8$
$8 y \frac{dy}{dx} - 2 x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{4 y}$
$\frac{dy}{dx} at (2 \sqrt{2}, - 2) = \frac{- 1}{2 \sqrt{2}} \Rightarrow | m_{2} = \frac{- 1}{2 \sqrt{2}}$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 \sqrt{2} + \frac{1}{2 \sqrt{2}}}{1 - (2 \sqrt{2}) \times \frac{1}{2 \sqrt{2}}} | = \frac{8 + 9}{\frac{2 \sqrt{2}}{0}} = \infty$
$θ = \frac{π}{2}$

TS EAMCET-2021-04.08.2021

Application of Derivatives

85798 The angle between the curves $y = \sin x$ and $y =$ $\cos x, 0 < x < \frac{π}{2}$

1

\tan^{- 1} (\sqrt{2})

2

\tan^{- 1} (2 \sqrt{2})

3

\tan^{- 1} (3 \sqrt{2})

4

\tan^{- 1} (3 \sqrt{3})

Explanation:

(B) : Given, $y = \sin x$ and $y = \cos x$
$\sin x = \cos x$
$\frac{\sin x}{\cos x} = 1$
$\tan x = 1$
$x = 45^{\circ} or \frac{π}{4}$
Differentiate the given Equation with respect to $x$
$y = \sin x$
$\frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \cos \frac{π}{4}$
$m_{1} = \frac{1}{\sqrt{2}}$
And,
$y = \cos x$
$\frac{d y}{d x} = - \sin x \Rightarrow \frac{d y}{d x} = - \sin \frac{π}{4}$
$m_{2} = - \frac{1}{\sqrt{2}}$
We know that
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
Substituting the value of $m_{1}$ and $m_{2}$
$\tan θ = | \frac{- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{1 + (\frac{1}{\sqrt{2}}) (- \frac{1}{\sqrt{2}})} | \Rightarrow \tan θ = | \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} |$
$\tan θ = | \frac{\frac{2}{\sqrt{2}}}{\frac{2 - 1}{2}} | \Rightarrow \tan θ = 2 \sqrt{2}$
$∴ θ = \tan^{- 1} 2 \sqrt{2}$

MHT CET-2022

Application of Derivatives

85799 If the angle between the curves $y = 2^{x}$ and $y =$ $3^{x}$ is $α$ , then the value of $\tan α$ , is equal to

1

\frac{\log (\frac{3}{2})}{1 + (\log 2) (\log 3)}

2

\frac{6}{7}

3

\frac{1}{7}

4

\frac{\log (6)}{1 + (\log 2) (\log 3)}

5 0

Explanation:

(A) : Given curves are $y = 2^{x}$ and $y = 3^{x}$
The point of intersection is $3^{x} = 2^{x} x = 0$
On differentiating w.r.t. $x$ , we get
${\begin{cases} \frac{d y}{d x} = 2^{x} \log 2 = m_{1} \\ And, \frac{d y}{d x} = 3^{x} \log 3 = m_{2} \end{cases}$
Therefore, $\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$
$= \frac{3^{x} \log 3 - 2^{x} \log 2}{1 + 3^{x} \times 2^{x} \log 3 \times \log 2}$
At, $x = 0$
$\tan a = \frac{3^{0} \log 3 - 2^{0} \log 2}{1 + 3^{0} \times 2^{0} \log 2 \log 3}$
$= \frac{\log \frac{3}{2}}{1 + \log 2 \log 3}$

Kerala CEE-2012

Application of Derivatives

85796 If $θ$ is the acute angle between the curves $x^{2} + y^{2} = 4$ and $y^{2} = 3 x$ then $\tan θ$

1

\frac{5}{\sqrt{3}}

2

\frac{\sqrt{3}}{4}

3

\frac{4}{\sqrt{3}}

4

\frac{\sqrt{3}}{5}

Explanation:

(A) : Given curve,
$x^{2} + y^{2} = 4$
And $y^{2} = 3 x$
Point of contact is obtained by solving above two equations-
From equation (i) and (ii)-
$x^{2} + 3 x - 4 = 0$
$\Rightarrow (x + 4) (x - 1) = 0 x = - 4, 1$
Neglecting $x = - 4$ , for $x = 1$
$y = \sqrt{3}$
$∴$ Point of contact is $(1, \sqrt{3})$
For curve $x^{2} + y^{2} = 1 = \frac{d y}{d x} = (\frac{- x}{y})$
$∴ m = {(\frac{dy}{dx})}_{(1, \sqrt{3})} = \frac{- 1}{\sqrt{3}}$
for $y^{2} = 3 x \Rightarrow {(\frac{d y}{d x})}_{(1, \sqrt{3})} = \frac{\sqrt{3}}{2}$
$∴$ Angle between the curve $=$ angle between their normal
$∴ \tan θ = \frac{\sqrt{3} + \frac{2}{\sqrt{3}}}{1 + \sqrt{3} (\frac{- 2}{\sqrt{3}})}$
$= \frac{- 5}{\sqrt{3}}$
$∵$ Angle is acute angle, then-
$\tan θ = \frac{5}{\sqrt{3}}$

Shift-I

Application of Derivatives

85797 The angle between the curves $2 x^{2} + y^{2} = 20$ and $4 y^{2} - x^{2} = 8$ at a point where they intersect in the $4^{th}$ quadrant $i$

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

(B): Given,
$2 x^{2} + y^{2} = 20$
$4 y^{2} - x^{2} = 8$
After solving both equation we get-
$(x, y) \to (2 \sqrt{2}, - 2)$
$C_{1} = 2 x^{2} + y^{2} = 20$
$4 x + 2 y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{- 2 x}{y}$
$\Rightarrow m_{1} = 2 \sqrt{2}$
$C_{2} \equiv 4 y^{2} - x^{2} = 8$
$8 y \frac{dy}{dx} - 2 x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{4 y}$
$\frac{dy}{dx} at (2 \sqrt{2}, - 2) = \frac{- 1}{2 \sqrt{2}} \Rightarrow | m_{2} = \frac{- 1}{2 \sqrt{2}}$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 \sqrt{2} + \frac{1}{2 \sqrt{2}}}{1 - (2 \sqrt{2}) \times \frac{1}{2 \sqrt{2}}} | = \frac{8 + 9}{\frac{2 \sqrt{2}}{0}} = \infty$
$θ = \frac{π}{2}$

TS EAMCET-2021-04.08.2021

Application of Derivatives

85798 The angle between the curves $y = \sin x$ and $y =$ $\cos x, 0 < x < \frac{π}{2}$

1

\tan^{- 1} (\sqrt{2})

2

\tan^{- 1} (2 \sqrt{2})

3

\tan^{- 1} (3 \sqrt{2})

4

\tan^{- 1} (3 \sqrt{3})

Explanation:

(B) : Given, $y = \sin x$ and $y = \cos x$
$\sin x = \cos x$
$\frac{\sin x}{\cos x} = 1$
$\tan x = 1$
$x = 45^{\circ} or \frac{π}{4}$
Differentiate the given Equation with respect to $x$
$y = \sin x$
$\frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \cos \frac{π}{4}$
$m_{1} = \frac{1}{\sqrt{2}}$
And,
$y = \cos x$
$\frac{d y}{d x} = - \sin x \Rightarrow \frac{d y}{d x} = - \sin \frac{π}{4}$
$m_{2} = - \frac{1}{\sqrt{2}}$
We know that
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
Substituting the value of $m_{1}$ and $m_{2}$
$\tan θ = | \frac{- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{1 + (\frac{1}{\sqrt{2}}) (- \frac{1}{\sqrt{2}})} | \Rightarrow \tan θ = | \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} |$
$\tan θ = | \frac{\frac{2}{\sqrt{2}}}{\frac{2 - 1}{2}} | \Rightarrow \tan θ = 2 \sqrt{2}$
$∴ θ = \tan^{- 1} 2 \sqrt{2}$

MHT CET-2022

Application of Derivatives

85799 If the angle between the curves $y = 2^{x}$ and $y =$ $3^{x}$ is $α$ , then the value of $\tan α$ , is equal to

1

\frac{\log (\frac{3}{2})}{1 + (\log 2) (\log 3)}

2

\frac{6}{7}

3

\frac{1}{7}

4

\frac{\log (6)}{1 + (\log 2) (\log 3)}

5 0

Explanation:

(A) : Given curves are $y = 2^{x}$ and $y = 3^{x}$
The point of intersection is $3^{x} = 2^{x} x = 0$
On differentiating w.r.t. $x$ , we get
${\begin{cases} \frac{d y}{d x} = 2^{x} \log 2 = m_{1} \\ And, \frac{d y}{d x} = 3^{x} \log 3 = m_{2} \end{cases}$
Therefore, $\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$
$= \frac{3^{x} \log 3 - 2^{x} \log 2}{1 + 3^{x} \times 2^{x} \log 3 \times \log 2}$
At, $x = 0$
$\tan a = \frac{3^{0} \log 3 - 2^{0} \log 2}{1 + 3^{0} \times 2^{0} \log 2 \log 3}$
$= \frac{\log \frac{3}{2}}{1 + \log 2 \log 3}$

Kerala CEE-2012

Application of Derivatives

85796 If $θ$ is the acute angle between the curves $x^{2} + y^{2} = 4$ and $y^{2} = 3 x$ then $\tan θ$

1

\frac{5}{\sqrt{3}}

2

\frac{\sqrt{3}}{4}

3

\frac{4}{\sqrt{3}}

4

\frac{\sqrt{3}}{5}

Explanation:

(A) : Given curve,
$x^{2} + y^{2} = 4$
And $y^{2} = 3 x$
Point of contact is obtained by solving above two equations-
From equation (i) and (ii)-
$x^{2} + 3 x - 4 = 0$
$\Rightarrow (x + 4) (x - 1) = 0 x = - 4, 1$
Neglecting $x = - 4$ , for $x = 1$
$y = \sqrt{3}$
$∴$ Point of contact is $(1, \sqrt{3})$
For curve $x^{2} + y^{2} = 1 = \frac{d y}{d x} = (\frac{- x}{y})$
$∴ m = {(\frac{dy}{dx})}_{(1, \sqrt{3})} = \frac{- 1}{\sqrt{3}}$
for $y^{2} = 3 x \Rightarrow {(\frac{d y}{d x})}_{(1, \sqrt{3})} = \frac{\sqrt{3}}{2}$
$∴$ Angle between the curve $=$ angle between their normal
$∴ \tan θ = \frac{\sqrt{3} + \frac{2}{\sqrt{3}}}{1 + \sqrt{3} (\frac{- 2}{\sqrt{3}})}$
$= \frac{- 5}{\sqrt{3}}$
$∵$ Angle is acute angle, then-
$\tan θ = \frac{5}{\sqrt{3}}$

Shift-I

Application of Derivatives

85797 The angle between the curves $2 x^{2} + y^{2} = 20$ and $4 y^{2} - x^{2} = 8$ at a point where they intersect in the $4^{th}$ quadrant $i$

1 0

2

\frac{π}{2}

3

\frac{π}{4}

4

\frac{π}{6}

Explanation:

(B): Given,
$2 x^{2} + y^{2} = 20$
$4 y^{2} - x^{2} = 8$
After solving both equation we get-
$(x, y) \to (2 \sqrt{2}, - 2)$
$C_{1} = 2 x^{2} + y^{2} = 20$
$4 x + 2 y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{- 2 x}{y}$
$\Rightarrow m_{1} = 2 \sqrt{2}$
$C_{2} \equiv 4 y^{2} - x^{2} = 8$
$8 y \frac{dy}{dx} - 2 x = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{4 y}$
$\frac{dy}{dx} at (2 \sqrt{2}, - 2) = \frac{- 1}{2 \sqrt{2}} \Rightarrow | m_{2} = \frac{- 1}{2 \sqrt{2}}$
$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 \sqrt{2} + \frac{1}{2 \sqrt{2}}}{1 - (2 \sqrt{2}) \times \frac{1}{2 \sqrt{2}}} | = \frac{8 + 9}{\frac{2 \sqrt{2}}{0}} = \infty$
$θ = \frac{π}{2}$

TS EAMCET-2021-04.08.2021

Application of Derivatives

85798 The angle between the curves $y = \sin x$ and $y =$ $\cos x, 0 < x < \frac{π}{2}$

1

\tan^{- 1} (\sqrt{2})

2

\tan^{- 1} (2 \sqrt{2})

3

\tan^{- 1} (3 \sqrt{2})

4

\tan^{- 1} (3 \sqrt{3})

Explanation:

(B) : Given, $y = \sin x$ and $y = \cos x$
$\sin x = \cos x$
$\frac{\sin x}{\cos x} = 1$
$\tan x = 1$
$x = 45^{\circ} or \frac{π}{4}$
Differentiate the given Equation with respect to $x$
$y = \sin x$
$\frac{dy}{dx} = \cos x \Rightarrow \frac{dy}{dx} = \cos \frac{π}{4}$
$m_{1} = \frac{1}{\sqrt{2}}$
And,
$y = \cos x$
$\frac{d y}{d x} = - \sin x \Rightarrow \frac{d y}{d x} = - \sin \frac{π}{4}$
$m_{2} = - \frac{1}{\sqrt{2}}$
We know that
$\tan θ = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} |$
Substituting the value of $m_{1}$ and $m_{2}$
$\tan θ = | \frac{- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{1 + (\frac{1}{\sqrt{2}}) (- \frac{1}{\sqrt{2}})} | \Rightarrow \tan θ = | \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} |$
$\tan θ = | \frac{\frac{2}{\sqrt{2}}}{\frac{2 - 1}{2}} | \Rightarrow \tan θ = 2 \sqrt{2}$
$∴ θ = \tan^{- 1} 2 \sqrt{2}$

MHT CET-2022

Application of Derivatives

85799 If the angle between the curves $y = 2^{x}$ and $y =$ $3^{x}$ is $α$ , then the value of $\tan α$ , is equal to

1

\frac{\log (\frac{3}{2})}{1 + (\log 2) (\log 3)}

2

\frac{6}{7}

3

\frac{1}{7}

4

\frac{\log (6)}{1 + (\log 2) (\log 3)}

5 0

Explanation:

(A) : Given curves are $y = 2^{x}$ and $y = 3^{x}$
The point of intersection is $3^{x} = 2^{x} x = 0$
On differentiating w.r.t. $x$ , we get
${\begin{cases} \frac{d y}{d x} = 2^{x} \log 2 = m_{1} \\ And, \frac{d y}{d x} = 3^{x} \log 3 = m_{2} \end{cases}$
Therefore, $\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$
$= \frac{3^{x} \log 3 - 2^{x} \log 2}{1 + 3^{x} \times 2^{x} \log 3 \times \log 2}$
At, $x = 0$
$\tan a = \frac{3^{0} \log 3 - 2^{0} \log 2}{1 + 3^{0} \times 2^{0} \log 2 \log 3}$
$= \frac{\log \frac{3}{2}}{1 + \log 2 \log 3}$

Kerala CEE-2012