The domain of the function f(x) = \cos ^ - 1( x^2 - 5x + 6 \over x^2 - 9 ) \over \log _e(x^2 - 3x + 2) is :

[ - 1 \over 2,1 ) \cup (2,\infty ) - \ 3,3 + \sqrt 5 \over 2,3 - \sqrt 5 \over 2 \

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

Inverse Trigonometric Functions PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q34 with Solution

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Correct answer: D

We need the domain of $f(x)=\frac{\cos^{-1}\!\left(\dfrac{x^2-5x+6}{x^2-9}\right)}{\ln(x^2-3x+2)}.$

For the function to be defined, all of the following must hold:

the argument of $\cos^{-1}$ must lie in $[-1,1]$ ,
the denominator $\ln(x^2-3x+2)$ must be defined, i.e. $x^2-3x+2>0$ ,
and also $\ln(x^2-3x+2)\neq 0$ since it is in the denominator.

Condition from the logarithm: $x^2-3x+2=(x-1)(x-2).$ So $x^2-3x+2>0 \iff (x-1)(x-2)>0 \iff x<1 \text{ or } x>2.$ Thus, $x\in (-\infty,1)\cup(2,\infty).$

Now, denominator should not be zero: $\ln(x^2-3x+2)\neq 0 \iff x^2-3x+2\neq 1.$ So, $x^2-3x+1\neq 0.$ Its roots are $x=\frac{3\pm\sqrt5}{2}.$ Hence these two points must be excluded.

So far, $x\in \left(( -\infty,1)\cup(2,\infty)\right)\setminus\left\{\frac{3-\sqrt5}{2},\frac{3+\sqrt5}{2}\right\}.$

Condition from the inverse cosine: $-1\le \frac{x^2-5x+6}{x^2-9}\le 1,$ with also $x^2-9\neq 0 \iff x\neq \pm 3.$

Factor the numerator and denominator: $x^2-5x+6=(x-2)(x-3),\qquad x^2-9=(x-3)(x+3).$ So for $x\neq 3$ , $\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3},$ provided $x\neq -3,3$ .

Thus we need $-1\le \frac{x-2}{x+3}\le 1, \qquad x\neq -3,3.$

Solve $\frac{x-2}{x+3}\le 1.$ This gives $\frac{x-2}{x+3}-1\le 0$ $\frac{x-2-(x+3)}{x+3}\le 0$ $\frac{-5}{x+3}\le 0.$ Since numerator is negative constant, this holds when $x+3>0 \iff x>-3.$

Solve $\frac{x-2}{x+3}\ge -1.$ This gives $\frac{x-2}{x+3}+1\ge 0$ $\frac{x-2+x+3}{x+3}\ge 0$ $\frac{2x+1}{x+3}\ge 0.$ Critical points are $x=-3$ and $x=-\frac12$ . Sign analysis gives $\frac{2x+1}{x+3}\ge 0 \iff x\in (-\infty,-3)\cup\left[-\frac12,\infty\right).$

Intersect the two cosine conditions:

from Step 4: $x>-3$
from Step 5: $x\in (-\infty,-3)\cup\left[-\frac12,\infty\right)$

Therefore, $-1\le \frac{x-2}{x+3}\le 1 \iff x\in \left[-\frac12,\infty\right),$ with $x\neq 3$ and also $x\neq -3$ already outside this set.

So the $\cos^{-1}$ condition gives $x\in \left[-\frac12,\infty\right)\setminus\{3\}.$

Now intersect with the logarithm condition:

=\left[-\frac12,1\right)\cup(2,\infty).$$ Now exclude the points where denominator becomes zero: $$x\neq \frac{3-\sqrt5}{2},\quad x\neq \frac{3+\sqrt5}{2}.$$ Also exclude $x=3$ from the $\cos^{-1}$ expression. Hence the domain is $$\left[-\frac12,1\right)\cup(2,\infty)\; -\;\left\{3,\frac{3+\sqrt5}{2},\frac{3-\sqrt5}{2}\right\}.$$ --- 8. Comparing with the options, this is exactly **Option D**.

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View written solutionFree

Correct answer: D

We need the domain of $f(x)=\frac{\cos^{-1}\!\left(\dfrac{x^2-5x+6}{x^2-9}\right)}{\ln(x^2-3x+2)}.$

For the function to be defined, all of the following must hold:

the argument of $\cos^{-1}$ must lie in $[-1,1]$ ,
the denominator $\ln(x^2-3x+2)$ must be defined, i.e. $x^2-3x+2>0$ ,
and also $\ln(x^2-3x+2)\neq 0$ since it is in the denominator.

Condition from the logarithm: $x^2-3x+2=(x-1)(x-2).$ So $x^2-3x+2>0 \iff (x-1)(x-2)>0 \iff x<1 \text{ or } x>2.$ Thus, $x\in (-\infty,1)\cup(2,\infty).$

Now, denominator should not be zero: $\ln(x^2-3x+2)\neq 0 \iff x^2-3x+2\neq 1.$ So, $x^2-3x+1\neq 0.$ Its roots are $x=\frac{3\pm\sqrt5}{2}.$ Hence these two points must be excluded.

So far, $x\in \left(( -\infty,1)\cup(2,\infty)\right)\setminus\left\{\frac{3-\sqrt5}{2},\frac{3+\sqrt5}{2}\right\}.$

Condition from the inverse cosine: $-1\le \frac{x^2-5x+6}{x^2-9}\le 1,$ with also $x^2-9\neq 0 \iff x\neq \pm 3.$

Factor the numerator and denominator: $x^2-5x+6=(x-2)(x-3),\qquad x^2-9=(x-3)(x+3).$ So for $x\neq 3$ , $\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3},$ provided $x\neq -3,3$ .

Thus we need $-1\le \frac{x-2}{x+3}\le 1, \qquad x\neq -3,3.$

Solve $\frac{x-2}{x+3}\le 1.$ This gives $\frac{x-2}{x+3}-1\le 0$ $\frac{x-2-(x+3)}{x+3}\le 0$ $\frac{-5}{x+3}\le 0.$ Since numerator is negative constant, this holds when $x+3>0 \iff x>-3.$

Solve $\frac{x-2}{x+3}\ge -1.$ This gives $\frac{x-2}{x+3}+1\ge 0$ $\frac{x-2+x+3}{x+3}\ge 0$ $\frac{2x+1}{x+3}\ge 0.$ Critical points are $x=-3$ and $x=-\frac12$ . Sign analysis gives $\frac{2x+1}{x+3}\ge 0 \iff x\in (-\infty,-3)\cup\left[-\frac12,\infty\right).$

Intersect the two cosine conditions:

from Step 4: $x>-3$
from Step 5: $x\in (-\infty,-3)\cup\left[-\frac12,\infty\right)$

Therefore, $-1\le \frac{x-2}{x+3}\le 1 \iff x\in \left[-\frac12,\infty\right),$ with $x\neq 3$ and also $x\neq -3$ already outside this set.

So the $\cos^{-1}$ condition gives $x\in \left[-\frac12,\infty\right)\setminus\{3\}.$

Now intersect with the logarithm condition:

=\left[-\frac12,1\right)\cup(2,\infty).$$ Now exclude the points where denominator becomes zero: $$x\neq \frac{3-\sqrt5}{2},\quad x\neq \frac{3+\sqrt5}{2}.$$ Also exclude $x=3$ from the $\cos^{-1}$ expression. Hence the domain is $$\left[-\frac12,1\right)\cup(2,\infty)\; -\;\left\{3,\frac{3+\sqrt5}{2},\frac{3-\sqrt5}{2}\right\}.$$ --- 8. Comparing with the options, this is exactly **Option D**.

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