If k= ( 4+ 12 ^-1 ( 23)) + ( 12 ^-1 ( 23)), then the number of solutions of the equation ^-1(kx-1)= ^-1x- ^-1x is:

Given: k= ( 4+ 12 ^-1 ( 23))+ ( 12 ^-1 ( 23)) and the equation ^-1(kx-1)= ^-1x- ^-1x Find: The number of solutions. Step 1: Evaluate k. Let = ^-1 ( 23) Then = 53 Using the identity 2= 1- we get 2= 1- 23 53= 1 5 Now, ( 4+ 2)= 1+ ( /2)1- ( /2)= 1+ 1 51- 1 5= 3+ 52 Let = ^-1 ( 23) Then = 53 and 2= 1- = 3- 52 Hence, k= 3+ 52+ 3- 52=3 Step 2: Simplify the equation. Substitute k=3: ^-1(3x-1)= ^-1x- ^-1x Using ^-1x- ^-1x= ^-1x- ( 2- ^-1x)=2 ^-1x- 2 so the equation becomes ^-1(3x-1)=2 ^-1x- 2 Step 3: Determine the domain. For ^-1(3x-1) to be defined, -1 3x-1 1 0 x 23 Also, the right-hand side must lie in the range of ^-1, that is [- 2, 2 ]. Since 0 ^-1x 2 we obtain the effective domain 0 x 23 Step 4: Solve the equation. Let t= ^-1x Then (2t- 2)=3x-1 Since (2t- 2)=- 2t and 2t=1-2 ^2 t=1-2x^2 we get -(1-2x^2)=3x-1 So, 2x^2=3x which gives x(2x-3)=0 Hence, x=0 or x= 32 Step 5: Check admissible solutions. Only x=0 lies in [0, 23 ]. Therefore, exactly one solution is admissible. Step 6: Verify. At x=0, ^-1(-1)=- 2 and ^-1(0)- ^-1(0)=0- 2=- 2 Hence, the equation is satisfied. Therefore, the number of solutions is 1.

If k= ( 4+ 12 ^-1 ( 23)) + ( 12 ^-1 ( 23)), then the number of so… — Solution | JEE Mathematics

NVAMediumJEE 2026Inverse Trigonometric Functions

If $k=\tan\!\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\!\left(\frac{2}{3}\right)\right) +\tan\!\left(\frac{1}{2}\sin^{-1}\!\left(\frac{2}{3}\right)\right),$ then the number of solutions of the equation $\sin^{-1}(kx-1)=\sin^{-1}x-\cos^{-1}x$ is:

Answer

Correct answer:1

Step-by-step solution

Standard Method

Given:

k=\tan\!\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\!\left(\frac{2}{3}\right)\right)+\tan\!\left(\frac{1}{2}\sin^{-1}\!\left(\frac{2}{3}\right)\right)

and the equation

\sin^{-1}(kx-1)=\sin^{-1}x-\cos^{-1}x

Find: The number of solutions.

Step 1: Evaluate $k$ .

Let

\alpha=\cos^{-1}\!\left(\frac{2}{3}\right)

Then

\sin\alpha=\frac{\sqrt5}{3}

Using the identity

\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}

we get

\tan\frac{\alpha}{2}=\frac{1-\frac{2}{3}}{\frac{\sqrt5}{3}}=\frac{1}{\sqrt5}

Now,

\tan\!\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\frac{1+\tan(\alpha/2)}{1-\tan(\alpha/2)}=\frac{1+\frac{1}{\sqrt5}}{1-\frac{1}{\sqrt5}}=\frac{3+\sqrt5}{2}

Let

\beta=\sin^{-1}\!\left(\frac{2}{3}\right)

Then

\cos\beta=\frac{\sqrt5}{3}

and

\tan\frac{\beta}{2}=\frac{1-\cos\beta}{\sin\beta}=\frac{3-\sqrt5}{2}

Hence,

k=\frac{3+\sqrt5}{2}+\frac{3-\sqrt5}{2}=3

Step 2: Simplify the equation.

Substitute $k=3$ :

\sin^{-1}(3x-1)=\sin^{-1}x-\cos^{-1}x

Using

\sin^{-1}x-\cos^{-1}x=\sin^{-1}x-\left(\frac{\pi}{2}-\sin^{-1}x\right)=2\sin^{-1}x-\frac{\pi}{2}

so the equation becomes

\sin^{-1}(3x-1)=2\sin^{-1}x-\frac{\pi}{2}

Step 3: Determine the domain.

For $\sin^{-1}(3x-1)$ to be defined,

-1\le 3x-1\le 1 \Rightarrow 0\le x\le \frac{2}{3}

Also, the right-hand side must lie in the range of $\sin^{-1}$ , that is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ . Since

0\le \sin^{-1}x\le \frac{\pi}{2}

we obtain the effective domain

0\le x\le \frac{2}{3}

Step 4: Solve the equation.

Let

t=\sin^{-1}x

Then

\sin\left(2t-\frac{\pi}{2}\right)=3x-1

Since

\sin\left(2t-\frac{\pi}{2}\right)=-\cos 2t

and

\cos 2t=1-2\sin^2 t=1-2x^2

we get

-(1-2x^2)=3x-1

So,

2x^2=3x

which gives

x(2x-3)=0

Hence,

x=0 \quad \text{or} \quad x=\frac{3}{2}

Step 5: Check admissible solutions.

Only $x=0$ lies in $\left[0,\frac{2}{3}\right]$ . Therefore, exactly one solution is admissible.

Step 6: Verify.

At $x=0$ ,

\sin^{-1}(-1)=-\frac{\pi}{2}

and

\sin^{-1}(0)-\cos^{-1}(0)=0-\frac{\pi}{2}=-\frac{\pi}{2}

Hence, the equation is satisfied.

Therefore, the number of solutions is $1$ .

Common mistakes

Ignoring the range restriction of $\sin^{-1}$ after transforming the equation. This is wrong because solving the algebraic equation alone can produce extraneous roots. Always check that both sides lie in the principal range and then test the obtained values in the original equation.
Using the identity between $\sin^{-1}x$ and $\cos^{-1}x$ incorrectly. The correct relation is $\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$ , so $\sin^{-1}x-\cos^{-1}x=2\sin^{-1}x-\frac{\pi}{2}$ . A sign error here changes the entire equation.
Finding $k$ incorrectly from the half-angle expressions. This is wrong because both tangent terms must be evaluated using the correct half-angle identity before substitution. Compute each term carefully and confirm that $k=3$ before solving for $x$ .

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