CUET Mathematics 2024 - If ^-1(23^-x+1)= ^-1(33^x+1), then which one of the following is true ? | PYQs + Solutions

CUET Mathematics 2024

Geometry

Trigonometry

Medium

If $\tan ^{-1}\left(\frac{2}{3^{-x}+1}\right)=\cot ^{-1}\left(\frac{3}{3^{x}+1}\right)$ , then which one of the following is true ?

There is no real value of x satisfying the above equation.

There is one positive and one negative real value of

x

satisfying the above equation.

There are two real positive values of x satisfying the above equation.

There are two real negative values of

x

satisfying the above equation.

✅ Correct Option: 2

First, let's recall a key relationship between inverse tangent and inverse cotangent:

\cot^{-1}(z) = \tan^{-1}\left(\dfrac{1}{z}\right)

This relationship exists because cotangent is the reciprocal of tangent. Geometrically, if an angle has a tangent value of z, then the same angle has a cotangent value of 1/z.

Using this relationship, let's rewrite the right side of our equation:

\cot^{-1}\left(\dfrac{3}{3^x+1}\right) = \tan^{-1}\left(\dfrac{3^x+1}{3}\right)

Our equation becomes:

\tan^{-1}\left(\dfrac{2}{3^{-x}+1}\right) = \tan^{-1}\left(\dfrac{3^x+1}{3}\right)

For two inverse tangent expressions to be equal, their arguments must be equal (since inverse tangent is a one-to-one function in its principal domain):

\dfrac{2}{3^{-x}+1} = \dfrac{3^x+1}{3}

Cross-multiplying to solve:

2 \cdot 3 = (3^{-x}+1)(3^x+1)

6 = 3^{-x} \cdot 3^x + 3^{-x} + 3^x + 1

6 = 1 + 3^{-x} + 3^x + 1

4 = 3^{-x} + 3^x

Remember that

3^{-x} = \dfrac{1}{3^x}

, so we can rewrite:

4 = \dfrac{1}{3^x} + 3^x

\newline

4 \cdot 3^x = 1 + (3^x)^2

\newline

(3^x)^2 - 4(3^x) + 1 = 0

Let's substitute

y = 3^x

to make this easier to solve:

y^2 - 4y + 1 = 0

Using the quadratic formula:

y = \dfrac{4 \pm \sqrt{16-4}}{2} = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}

So we have two solutions:

\newline

y = 3^x = 2 + \sqrt{3}

\newline

y = 3^x = 2 - \sqrt{3}

Now let's find the x-values by taking logarithm base 3:

For the first solution:

\newline

3^x = 2 + \sqrt{3}

\newline

x = \log_3(2 + \sqrt{3})

Since

2 + \sqrt{3} \approx 3.732 > 1

, this value of x is positive.

For the second solution:

\newline

3^x = 2 - \sqrt{3}

\newline

x = \log_3(2 - \sqrt{3})

Since

2 - \sqrt{3} \approx 0.268 < 1

, this value of x is negative.

A quick conceptual check: When we have an equation with exponential terms like

3^x

and

3^{-x}

, it's common to get solutions in pairs - one positive and one negative. This happens because of the symmetrical nature of the equation. We can see this symmetry in the final form

4 = 3^{-x} + 3^x

where replacing x with -x gives us the same equation.

Therefore, there is one positive real value

x = \log_3(2 + \sqrt{3})

and one negative real value

x = \log_3(2 - \sqrt{3})

satisfying the equation.

The correct option is: There is one positive and one negative real value of x satisfying the above equation.

CUET Mathematics 2024 PYQs

If tan⁡−1(23−x+1)=cot⁡−1(33x+1)\tan ^{-1}\left(\frac{2}{3^{-x}+1}\right)=\cot ^{-1}\left(\frac{3}{3^{x}+1}\right)tan−1(3−x+12​)=cot−1(3x+13​), then which one of the following is true ?

Keyboard Shortcuts

If $\tan ^{-1}\left(\frac{2}{3^{-x}+1}\right)=\cot ^{-1}\left(\frac{3}{3^{x}+1}\right)$ , then which one of the following is true ?