Let θ be the acute angle between the curves y = x x ln x and y =(2 x -2/ ln 2) at their point of intersection on the line y=0. The value of tan θ is equal to

Question

Let θ be the acute angle between the curves y = x x ln x and y =(2 x -2/ ln 2) at their point of intersection on the line y=0. The value of tan θ is equal to (A) (1/2) (B) 2 (C) (1/3) (D) 3

Tardigrade Admin · Accepted Answer

y = x x ln x ; y =(2 x -2/ ln 2) At x =1, m 1=( dy / dx )= x x ⋅ (1/ x )+ x x ( ln x +1) ln x =1 At x =1, m 2=( dy / dx )=2 x ( ln 2/ ln 2)=2 ∴ tan θ=|(2-1/1+2)|=(1/3)

Q. Let $θ$ be the acute angle between the curves $y = x^{x} ln x$ and $y = \frac{2 ^{x} - 2}{l n 2}$ at their point of intersection on the line $y = 0$ . The value of $tan θ$ is equal to

$\frac{1}{2}$

2

$\frac{1}{3}$

3

$y = x^{x} ln x; y = \frac{2 ^{x} - 2}{l n 2}$
At $x = 1, m_{1} = \frac{d y}{d x} = x^{x} \cdot \frac{1}{x} + x^{x} (ln x + 1) ln x = 1$
At $x = 1, m_{2} = \frac{d y}{d x} = 2^{x} \frac{l n 2}{l n 2} = 2$
$∴ tan θ = ∣ ∣ \frac{2 - 1}{1 + 2} ∣ ∣ = \frac{1}{3}$

Q. Let θ be the acute angle between the curves y=xxlnx and y=ln22x−2​ at their point of intersection on the line y=0. The value of tanθ is equal to

21​

2

31​

3

y=xxlnx;y=ln22x−2​ At x=1,m1​=dxdy​=xx⋅x1​+xx(lnx+1)lnx=1 At x=1,m2​=dxdy​=2xln2ln2​=2 ∴tanθ=∣∣​1+22−1​∣∣​=31​

Q. Let $θ$ be the acute angle between the curves $y = x^{x} ln x$ and $y = \frac{2 ^{x} - 2}{l n 2}$ at their point of intersection on the line $y = 0$ . The value of $tan θ$ is equal to

$\frac{1}{2}$

$\frac{1}{3}$

$y = x^{x} ln x; y = \frac{2 ^{x} - 2}{l n 2}$
At $x = 1, m_{1} = \frac{d y}{d x} = x^{x} \cdot \frac{1}{x} + x^{x} (ln x + 1) ln x = 1$
At $x = 1, m_{2} = \frac{d y}{d x} = 2^{x} \frac{l n 2}{l n 2} = 2$
$∴ tan θ = ∣ ∣ \frac{2 - 1}{1 + 2} ∣ ∣ = \frac{1}{3}$