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Inverse Trigonometric Functions question

2021 · 22 Jul · Shift 2 · Q35
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  5. /2021 · 22 Jul · Shift 2 · Q35

Inverse Trigonometric Functions question

2021 · 22 Jul · Shift 2 · Q35

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If the domain of the function f(x)=cos⁡−1x2−x+1sin⁡−1(2x−12)f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}f(x)=sin−1(22x−1​)​cos−1x2−x+1​​ is the interval (α\alphaα, β\betaβ], then α\alphaα+β\betaβ is equal to :
  1. A
    32{3 \over 2}23​
  2. B
    2
  3. C
    12{1 \over 2}21​
  4. D
    1
View written solutionFree

Correct answer: A

  1. Given function
f(x)=cos⁡−1 ⁣(x2−x+1)sin⁡−1 ⁣(2x−12) f(x)=\frac{\cos^{-1}\!\left(\sqrt{x^2-x+1}\right)}{\sqrt{\sin^{-1}\!\left(\frac{2x-1}{2}\right)}}f(x)=sin−1(22x−1​)​cos−1(x2−x+1​)​

We need the domain of this function.

  1. Condition from cos⁡−1(x2−x+1)\cos^{-1}\left(\sqrt{x^2-x+1}\right)cos−1(x2−x+1​)

For cos⁡−1(t)\cos^{-1}(t)cos−1(t) to be defined,

−1≤t≤1-1 \le t \le 1−1≤t≤1

Here,

t=x2−x+1t=\sqrt{x^2-x+1}t=x2−x+1​

Since a square root is always non-negative,

x2−x+1≥0\sqrt{x^2-x+1} \ge 0x2−x+1​≥0

So only the upper bound matters:

x2−x+1≤1\sqrt{x^2-x+1} \le 1x2−x+1​≤1

Squaring both sides,

x2−x+1≤1x^2-x+1 \le 1x2−x+1≤1 x2−x≤0x^2-x \le 0x2−x≤0 x(x−1)≤0x(x-1) \le 0x(x−1)≤0

Hence,

0≤x≤10 \le x \le 10≤x≤1
  1. Condition from sin⁡−1(2x−12)\sqrt{\sin^{-1}\left(\frac{2x-1}{2}\right)}sin−1(22x−1​)​

Since this is in the denominator, we need:

  • sin⁡−1(2x−12)\sin^{-1}\left(\frac{2x-1}{2}\right)sin−1(22x−1​) to be defined,
  • its value must be strictly positive because it is inside a square root and also in denominator.

(i) Argument of sin⁡−1\sin^{-1}sin−1 must lie in [−1,1][-1,1][−1,1]

−1≤2x−12≤1-1 \le \frac{2x-1}{2} \le 1−1≤22x−1​≤1

Multiply by 2:

−2≤2x−1≤2-2 \le 2x-1 \le 2−2≤2x−1≤2

Add 1:

−1≤2x≤3-1 \le 2x \le 3−1≤2x≤3 −12≤x≤32-\frac12 \le x \le \frac32−21​≤x≤23​

(ii) Quantity inside outer square root must be positive

We need

sin⁡−1(2x−12)>0\sin^{-1}\left(\frac{2x-1}{2}\right) > 0sin−1(22x−1​)>0

Now sin⁡−1(y)>0\sin^{-1}(y)>0sin−1(y)>0 iff y>0y>0y>0 (since principal range of sin⁡−1\sin^{-1}sin−1 is [−π/2,π/2][-\pi/2,\pi/2][−π/2,π/2] and it is increasing).

So,

2x−12>0\frac{2x-1}{2} > 022x−1​>0 2x−1>02x-1>02x−1>0 x>12x>\frac12x>21​

Thus denominator condition gives

12<x≤32\frac12 < x \le \frac3221​<x≤23​
  1. Intersect all conditions

From numerator:

0≤x≤10 \le x \le 10≤x≤1

From denominator:

12<x≤32\frac12 < x \le \frac3221​<x≤23​

Intersection:

(12,1]\left(\frac12,1\right](21​,1]

So,

α=12,β=1\alpha=\frac12, \qquad \beta=1α=21​,β=1

Therefore,

α+β=12+1=32\alpha+\beta=\frac12+1=\frac32α+β=21​+1=23​
  1. Option check
  • A: 32\frac3223​ ✅
  • B: 222 ❌
  • C: 12\frac1221​ ❌
  • D: 111 ❌

So the correct option is A.

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