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Inverse Trigonometric Functions question

2021 · 31 Aug · Shift 2 · Q23
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Inverse Trigonometric Functions question

2021 · 31 Aug · Shift 2 · Q23

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The domain of the function f(x)=sin⁡−1(3x2+x−1(x−1)2)+cos⁡−1(x−1x+1)f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)f(x)=sin−1((x−1)23x2+x−1​)+cos−1(x+1x−1​) is :
  1. A
    [0,14]\left[ {0,{1 \over 4}} \right][0,41​]
  2. B
    [−2,0]∪[14,12][ - 2,0] \cup \left[ {{1 \over 4},{1 \over 2}} \right][−2,0]∪[41​,21​]
  3. C
    [14,12]∪{0}\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\}[41​,21​]∪{0}
  4. D
    [0,12]\left[ {0,{1 \over 2}} \right][0,21​]
View written solutionFree

Correct answer: C

We need the domain of

\sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right)+\cos^{-1}\left(\frac{x-1}{x+1}\right).$$ For the function to be defined, both inverse-trigonometric terms must be defined. --- ## 1. Condition for the first term For $$\sin^{-1}(u)$$ to exist, we need $$-1\le u\le 1.$$ So, $$-1\le \frac{3x^2+x-1}{(x-1)^2}\le 1,$$ with also $$(x-1)^2\ne 0 \Rightarrow x\ne 1.$$ Let $$A=\frac{3x^2+x-1}{(x-1)^2}.$$ Since $(x-1)^2>0$ for $x\ne 1$, we can multiply inequalities safely. ### (i) Condition $A\le 1$ $$\frac{3x^2+x-1}{(x-1)^2}\le 1$$ $$3x^2+x-1\le (x-1)^2=x^2-2x+1$$ $$2x^2+3x-2\le 0$$ $$(2x-1)(x+2)\le 0.$$ Hence, $$-2\le x\le \frac12.$$ ### (ii) Condition $A\ge -1$ $$\frac{3x^2+x-1}{(x-1)^2}\ge -1$$ $$3x^2+x-1\ge -(x-1)^2=-(x^2-2x+1)$$ $$3x^2+x-1\ge -x^2+2x-1$$ $$4x^2-x\ge 0$$ $$x(4x-1)\ge 0.$$ Hence, $$x\in (-\infty,0]\cup\left[\frac14,\infty\right).$$ ### Combining (i) and (ii) We need both: $$x\in \left[-2,\frac12\right]\cap \left[(-\infty,0]\cup\left[\frac14,\infty\right)\right].$$ Therefore, from the first term, $$x\in [-2,0]\cup\left[\frac14,\frac12\right].$$ --- ## 2. Condition for the second term For $$\cos^{-1}(v)$$ to exist, we need $$-1\le v\le 1,$$ with denominator nonzero: $x+1\ne 0$. So, $$-1\le \frac{x-1}{x+1}\le 1, \qquad x\ne -1.$$ Let $$B=\frac{x-1}{x+1}.$$ We solve $$\left|\frac{x-1}{x+1}\right|\le 1.$$ Squaring both sides: $$\frac{(x-1)^2}{(x+1)^2}\le 1$$ $$ (x-1)^2\le (x+1)^2,

provided x≠−1x\ne -1x=−1.

Now, x2−2x+1≤x2+2x+1x^2-2x+1\le x^2+2x+1x2−2x+1≤x2+2x+1 −2x≤2x-2x\le 2x−2x≤2x x≥0.x\ge 0.x≥0.

So the second term requires x≥0.x\ge 0.x≥0. (And this automatically avoids x=−1x=-1x=−1.)

3. Final domain

Intersect the conditions from both terms:

First term: [−2,0]∪[14,12][-2,0]\cup\left[\frac14,\frac12\right][−2,0]∪[41​,21​]

Second term: [0,∞)[0,\infty)[0,∞)

Hence,

=\{0\}\cup\left[\frac14,\frac12\right].$$ Thus the domain is $$\boxed{\left[\frac14,\frac12\right]\cup\{0\}}.$$ --- ## 4. Checking options - **A:** $\left[0,\frac14\right]$ — incorrect - **B:** $[-2,0]\cup\left[\frac14,\frac12\right]$ — misses restriction from second term - **C:** $\left[\frac14,\frac12\right]\cup\{0\}$ — correct - **D:** $\left[0,\frac12\right]$ — incorrect So the correct option is $$\boxed{\text{C}}.$$
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