Let f(x) = \sin ^ - 1x and g(x) = x^2 - x - 2 \over 2x^2 - x - 6 . If g(2) = \mathop \lim \limits_x \to 2 g(x) , then the domain of the function fog is :

( - \infty , - 2] \cup [ - 4 \over 3,\infty )

( - \infty , - 2] \cup [ - 1,\infty )

( - \infty , - 2] \cup [ - 3 \over 2,\infty )

( - \infty , - 1] \cup [2,\infty )

Limits Continuity and Differentiability PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q35 with Solution

View written solutionFree

Correct answer: A

Given functions

We have $f(x)=\sin^{-1}x$ and $g(x)=\frac{x^2-x-2}{2x^2-x-6}$ with the additional condition $g(2)=\lim_{x\to 2}g(x).$

We need the domain of $f\circ g$ , i.e. of $f(g(x))=\sin^{-1}(g(x)).$

For $\sin^{-1}(y)$ to be defined, we need $-1\le y\le 1.$ So we must find all $x$ such that $-1\le g(x)\le 1.$

Simplify $g(x)$

Factor numerator and denominator: $x^2-x-2=(x-2)(x+1),$ $2x^2-x-6=(x-2)(2x+3).$

Thus, for $x\ne 2$ , $g(x)=\frac{(x-2)(x+1)}{(x-2)(2x+3)}=\frac{x+1}{2x+3}.$

Since it is given that $g(2)=\lim_{x\to 2}g(x),$ we compute: $\lim_{x\to 2}g(x)=\lim_{x\to 2}\frac{x+1}{2x+3}=\frac{3}{7}.$ So $g(2)$ is defined as $\frac37$ .

Hence effectively, $g(x)=\frac{x+1}{2x+3}, \quad x\ne -\frac32,$ and also $x=2$ is included with value $\frac37$ .

So the only excluded point from the domain of $g$ is $x=-\frac32.$

Condition for $f\circ g$ to exist

We need $-1\le \frac{x+1}{2x+3}\le 1, \qquad x\ne -\frac32.$

We solve the two inequalities separately.

Solve $\displaystyle \frac{x+1}{2x+3}\le 1$

Bring all terms to one side: $\frac{x+1}{2x+3}-1\le 0$ $\frac{x+1-(2x+3)}{2x+3}\le 0$ $\frac{-x-2}{2x+3}\le 0$ $\frac{x+2}{2x+3}\ge 0.$

Critical points are $x=-2,\quad x=-\frac32.$

Now check intervals:

For $x< -2$ : both numerator and denominator are negative, so ratio $\ge 0$ .
For $-2<x< -\frac32$ : numerator positive, denominator negative, so ratio $<0$ .
For $x> -\frac32$ : both positive, so ratio $\ge 0$ .

Thus, $\frac{x+1}{2x+3}\le 1 \iff x\in (-\infty,-2]\cup\left(-\frac32,\infty\right).$

Solve $\displaystyle \frac{x+1}{2x+3}\ge -1$

Bring all terms to one side: $\frac{x+1}{2x+3}+1\ge 0$ $\frac{x+1+(2x+3)}{2x+3}\ge 0$ $\frac{3x+4}{2x+3}\ge 0.$

Critical points are $x=-\frac43,\quad x=-\frac32.$

Check intervals:

For $x< -\frac32$ : numerator and denominator both negative, so ratio $\ge 0$ .
For $-\frac32<x< -\frac43$ : numerator negative, denominator positive, so ratio $<0$ .
For $x\ge -\frac43$ : numerator and denominator positive, so ratio $\ge 0$ .

Hence, $\frac{x+1}{2x+3}\ge -1 \iff x\in (-\infty,-\tfrac32)\cup\left[-\frac43,\infty\right).$

Take intersection

We need both conditions simultaneously:

\cap \left[(-\infty,-\frac32)\cup\left[-\frac43,\infty\right)\right].$$ Compute intersection: - From the left side, we get $$(-\infty,-2]$$ - From the right side, for values greater than $-\frac32$, the common part is $$\left[-\frac43,\infty\right).$$ So the domain is $$(-\infty,-2]\cup\left[-\frac43,\infty\right).$$ Note that $x=2$ is included, since by definition $g(2)=\frac37$, and $$\frac37\in[-1,1].$$ --- 7. **Final answer** Therefore, the domain of $f\circ g$ is $$\boxed{(-\infty,-2]\cup\left[-\frac43,\infty\right)}.$$ This corresponds to **Option A**.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: A

Given functions

We have $f(x)=\sin^{-1}x$ and $g(x)=\frac{x^2-x-2}{2x^2-x-6}$ with the additional condition $g(2)=\lim_{x\to 2}g(x).$

We need the domain of $f\circ g$ , i.e. of $f(g(x))=\sin^{-1}(g(x)).$

For $\sin^{-1}(y)$ to be defined, we need $-1\le y\le 1.$ So we must find all $x$ such that $-1\le g(x)\le 1.$

Simplify $g(x)$

Factor numerator and denominator: $x^2-x-2=(x-2)(x+1),$ $2x^2-x-6=(x-2)(2x+3).$

Thus, for $x\ne 2$ , $g(x)=\frac{(x-2)(x+1)}{(x-2)(2x+3)}=\frac{x+1}{2x+3}.$

Since it is given that $g(2)=\lim_{x\to 2}g(x),$ we compute: $\lim_{x\to 2}g(x)=\lim_{x\to 2}\frac{x+1}{2x+3}=\frac{3}{7}.$ So $g(2)$ is defined as $\frac37$ .

Hence effectively, $g(x)=\frac{x+1}{2x+3}, \quad x\ne -\frac32,$ and also $x=2$ is included with value $\frac37$ .

So the only excluded point from the domain of $g$ is $x=-\frac32.$

Condition for $f\circ g$ to exist

We need $-1\le \frac{x+1}{2x+3}\le 1, \qquad x\ne -\frac32.$

We solve the two inequalities separately.

Solve $\displaystyle \frac{x+1}{2x+3}\le 1$

Bring all terms to one side: $\frac{x+1}{2x+3}-1\le 0$ $\frac{x+1-(2x+3)}{2x+3}\le 0$ $\frac{-x-2}{2x+3}\le 0$ $\frac{x+2}{2x+3}\ge 0.$

Critical points are $x=-2,\quad x=-\frac32.$

Now check intervals:

For $x< -2$ : both numerator and denominator are negative, so ratio $\ge 0$ .
For $-2<x< -\frac32$ : numerator positive, denominator negative, so ratio $<0$ .
For $x> -\frac32$ : both positive, so ratio $\ge 0$ .

Thus, $\frac{x+1}{2x+3}\le 1 \iff x\in (-\infty,-2]\cup\left(-\frac32,\infty\right).$

Solve $\displaystyle \frac{x+1}{2x+3}\ge -1$

Bring all terms to one side: $\frac{x+1}{2x+3}+1\ge 0$ $\frac{x+1+(2x+3)}{2x+3}\ge 0$ $\frac{3x+4}{2x+3}\ge 0.$

Critical points are $x=-\frac43,\quad x=-\frac32.$

Check intervals:

For $x< -\frac32$ : numerator and denominator both negative, so ratio $\ge 0$ .
For $-\frac32<x< -\frac43$ : numerator negative, denominator positive, so ratio $<0$ .
For $x\ge -\frac43$ : numerator and denominator positive, so ratio $\ge 0$ .

Hence, $\frac{x+1}{2x+3}\ge -1 \iff x\in (-\infty,-\tfrac32)\cup\left[-\frac43,\infty\right).$

Take intersection

We need both conditions simultaneously:

\cap \left[(-\infty,-\frac32)\cup\left[-\frac43,\infty\right)\right].$$ Compute intersection: - From the left side, we get $$(-\infty,-2]$$ - From the right side, for values greater than $-\frac32$, the common part is $$\left[-\frac43,\infty\right).$$ So the domain is $$(-\infty,-2]\cup\left[-\frac43,\infty\right).$$ Note that $x=2$ is included, since by definition $g(2)=\frac37$, and $$\frac37\in[-1,1].$$ --- 7. **Final answer** Therefore, the domain of $f\circ g$ is $$\boxed{(-\infty,-2]\cup\left[-\frac43,\infty\right)}.$$ This corresponds to **Option A**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability