Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

(I) ${\cos }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{1-x}+{\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}$.

Let ${\cos }^{-1}\sqrt{x}=\alpha$. Then $\sqrt{x}=\cos \alpha$.

So, ${\cos }^{-1}\sqrt{1-x}={\cos }^{-1}\sqrt{1-{\cos }^2\alpha }={\cos }^{-1}(\sin \alpha )=\frac{\pi }{2}-\alpha$.

The equation becomes $\alpha +\frac{\pi }{2}-\alpha +{\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}$.

$\frac{\pi }{2}+{\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}$.

${\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}-\frac{\pi }{2}=\frac{4\pi -3\pi }{6}=\frac{\pi }{6}$.

$\sqrt{1-y}=\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$.

$1-y={\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}$.

$y=1-\frac{3}{4}=\frac{1}{4}$.

Since ${\cos }^{-1}\sqrt{x}$ and ${\cos }^{-1}\sqrt{1-x}$ are defined, $0\le x\le 1$.

The maximum value of ${\cos }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{1-x}$ is $\frac{\pi }{2}$.

The domain of ${\cos }^{-1}z$ is $[-1,1]$.

For ${\cos }^{-1}\sqrt{x}$, $0\le \sqrt{x}\le 1⟹0\le x\le 1$.

For ${\cos }^{-1}\sqrt{1-x}$, $0\le \sqrt{1-x}\le 1⟹0\le 1-x\le 1⟹0\le x\le 1$.

For ${\cos }^{-1}\sqrt{1-y}$, $0\le \sqrt{1-y}\le 1⟹0\le 1-y\le 1⟹0\le y\le 1$.

From ${\cos }^{-1}\sqrt{1-y}=\frac{\pi }{6}$, we have $y=\frac{1}{4}$, which is in the domain.

The given equation is ${\cos }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{1-x}+{\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}$.

We know that ${\cos }^{-1}A+{\sin }^{-1}A=\frac{\pi }{2}$.

So, ${\cos }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{1-x}={\cos }^{-1}\sqrt{x}+{\sin }^{-1}\sqrt{x}=\frac{\pi }{2}$.

Thus, $\frac{\pi }{2}+{\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}$.

${\cos }^{-1}\sqrt{1-y}=\frac{2\pi }{3}-\frac{\pi }{2}=\frac{\pi }{6}$.

$\sqrt{1-y}=\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$.

$1-y=\frac{3}{4}⟹y=\frac{1}{4}$.

The value of $4(x+y)$ can be $4(x+\frac{1}{4})=4x+1$.

Since $0\le x\le 1$, $1\le 4x+1\le 5$.

The options for (I) are (P) -1, (Q) 1, (R) 3, (T) 5, (U) 6.

So $4(x+y)$ can be 1, 3, 5.

(II) ${\lim }_{z\to 0}\left[\left\{\max \left({\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2,\min \left(x^2+4x+7\right)\right)\right\}\cdot \frac{{\sin }^{-1}z}{z}\right]$

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ for $x\in [-1,1]$.

So, ${\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2={\left(\frac{\pi }{2}\right)}^2=\frac{{\pi }^2}{4}$.

$\frac{{\pi }^2}{4}\approx \frac{(3.14{)}^2}{4}=\frac{9.8596}{4}\approx 2.46$.

Now consider $\min (x^2+4x+7)$.

$x^2+4x+7=(x^2+4x+4)+3=(x+2{)}^2+3$.

The minimum value of $(x+2{)}^2+3$ is 3 (when $x=-2$).

So, $\max \left(\frac{{\pi }^2}{4},3\right)=\max (2.46,3)=3$.

The expression inside the limit becomes $\left[3\cdot \frac{{\sin }^{-1}z}{z}\right]$.

We know that ${\lim }_{z\to 0}\frac{{\sin }^{-1}z}{z}=1$.

So, the limit is ${\lim }_{z\to 0}[3\cdot 1]=[3]=3$.

The value for (II) is (R) 3.

(III) If $\alpha$ is a solution of equation $|\sin x-\cos x|=\sin 2x-1$ in $[0,2\pi ]$, then $\frac{1}{\sin \alpha }$ can be equal to

Let $t=\sin x-\cos x$.

Then $t^2=(\sin x-\cos x{)}^2={\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=1-\sin 2x$.

So, $\sin 2x=1-t^2$.

The equation becomes $|t|=(1-t^2)-1=-t^2$.

$|t|=-t^2$.

Since $|t|\ge 0$ and $-t^2\le 0$, this equation can only hold if both sides are 0.

So, $|t|=0⟹t=0$.

If $t=0$, then $\sin x-\cos x=0⟹\sin x=\cos x$.

This implies $\tan x=1$.

In the interval $[0,2\pi ]$, the solutions are $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$.

Let $\alpha =\frac{\pi }{4}$. Then $\sin \alpha =\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$.

So, $\frac{1}{\sin \alpha }=\sqrt{2}$.

Let $\alpha =\frac{5\pi }{4}$. Then $\sin \alpha =\sin \left(\frac{5\pi }{4}\right)=-\frac{1}{\sqrt{2}}$.

So, $\frac{1}{\sin \alpha }=-\sqrt{2}$.

The options for (III) are (P) -1, (Q) 1, (R) 3, (T) 5, (U) 6.

None of the options match $\sqrt{2}$ or $-\sqrt{2}$.

Let's recheck the question. The image for (III) is not rendered. Assuming it asks for $1/\sin \alpha$.

If the question was asking for $\sin \alpha$, then $\frac{1}{\sqrt{2}}\approx 0.707$ and $-\frac{1}{\sqrt{2}}\approx -0.707$.

If the question was asking for $\sin 2\alpha$, then $\sin 2(\frac{\pi }{4})=\sin (\frac{\pi }{2})=1$.

And $\sin 2(\frac{5\pi }{4})=\sin (\frac{5\pi }{2})=\sin (\frac{\pi }{2}+2\pi )=1$.

So $\sin 2\alpha =1$. This matches option (Q).

Given the options, it is highly probable that the question intended to ask for $\sin 2\alpha$.

So, (III) $\to$ (Q).

(IV) The integers in range of $f(x)=\frac{x^2+x+1}{x^2+x-1}$

Let $y=\frac{x^2+x+1}{x^2+x-1}$.

$y(x^2+x-1)=x^2+x+1$.

$y{x}^2+yx-y=x^2+x+1$.

$(y-1)x^2+(y-1)x-(y+1)=0$.

If $y=1$, then $-(1+1)=0⟹-2=0$, which is false. So $y\ne 1$.

For $x$ to be real, the discriminant $D$ must be $\ge 0$.

$D=(y-1{)}^2-4(y-1)(-(y+1))\ge 0$.

$(y-1{)}^2+4(y-1)(y+1)\ge 0$.

$(y-1)[(y-1)+4(y+1)]\ge 0$.

$(y-1)[y-1+4y+4]\ge 0$.

$(y-1)(5y+3)\ge 0$.

This inequality holds when $y\le -\frac{3}{5}$ or $y\ge 1$.

Since $y\ne 1$, the range is $y\in \left(-\infty ,-\frac{3}{5}\right]\cup (1,\infty )$.

The integers in this range are $\dots ,-2,-1$ and $2,3,4,\dots$.

The options for (IV) are (P) -1, (Q) 1, (R) 3, (T) 5, (U) 6.

So, the integers can be -1, 3, 5, 6.

Therefore, the only correct combination is (II) $\to$ (R).

The final answer is 2