Question

If \(sin^{-1}(tan\frac{π}{4})-sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{π}{6}\) then \(x\) is a root of the equation

• A. \(x^2-x-6 = 0\)
• B. \(x^2-x-12 = 0\)
• C. \(x^2+x-12 = 0\)
• D. \(x^2+x-6 = 0\)

Hint:

Evaluate sin^-1(tan(pi/4)) = pi/2, then solve for the second term.

Answer 1

The Correct Option is B

Step 1: Simplify the First Term:
\(\tan\dfrac\pi4=1\), so \(\sin^{-1}(1)=\dfrac\pi2\).

Step 2: Solve for the Second Term:
\[ \frac\pi2-\sin^{-1}\sqrt{\frac3x}=\frac\pi6\Rightarrow\sin^{-1}\sqrt{\frac3x}=\frac\pi2-\frac\pi6=\frac\pi3 \]

Step 3: Find x:
\[ \sqrt{\frac3x}=\sin\frac\pi3=\frac{\sqrt3}2\Rightarrow\frac3x=\frac34\Rightarrow x=4 \]

Step 4: Test the Equations:
\(x^2-x-6\) at 4: \(16-4-6=6\neq0\).
\(x^2-x-12\) at 4: \(16-4-12=0\). True.
\(x^2+x-12\) at 4: \(16+4-12=8\neq0\).
\(x^2+x-6\) at 4: \(16+4-6=14\neq0\).
Only (B) has 4 as a root.

Final Answer:
\(x=4\) is a root of \(x^2-x-12=0\), option (B). \[ \boxed{\text{(B) } x^2-x-12=0} \]