Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Inverse Trigonometric Functions question

2022 · 29 Jul · Shift 2 · Q36
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Inverse Trigonometric Functions
  5. /2022 · 29 Jul · Shift 2 · Q36

Inverse Trigonometric Functions question

2022 · 29 Jul · Shift 2 · Q36

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The domain of the function f(x)=sin⁡−1(x2−3x+2x2+2x+7)f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)f(x)=sin−1(x2+2x+7x2−3x+2​) is :
  1. A
    [1,∞)[1, \infty)[1,∞)
  2. B
    [−1,2][-1,2][−1,2]
  3. C
    [−1,∞)[-1, \infty)[−1,∞)
  4. D
    (−∞,2](-\infty, 2](−∞,2]
View written solutionFree

Correct answer: C

  1. For the function f(x)=sin⁡−1(x2−3x+2x2+2x+7),f(x)=\sin^{-1}\left(\frac{x^2-3x+2}{x^2+2x+7}\right),f(x)=sin−1(x2+2x+7x2−3x+2​), the argument of sin⁡−1\sin^{-1}sin−1 must satisfy −1≤x2−3x+2x2+2x+7≤1.-1 \le \frac{x^2-3x+2}{x^2+2x+7} \le 1.−1≤x2+2x+7x2−3x+2​≤1.

  2. First check the denominator: x2+2x+7=(x+1)2+6>0for all x.x^2+2x+7=(x+1)^2+6>0 \quad \text{for all } x.x2+2x+7=(x+1)2+6>0for all x. So the expression is defined for every real xxx, and we only need to enforce the range condition.

  3. Solve the right inequality: x2−3x+2x2+2x+7≤1.\frac{x^2-3x+2}{x^2+2x+7} \le 1.x2+2x+7x2−3x+2​≤1. Since the denominator is always positive, multiply directly: x2−3x+2≤x2+2x+7.x^2-3x+2 \le x^2+2x+7.x2−3x+2≤x2+2x+7. −3x+2≤2x+7-3x+2 \le 2x+7−3x+2≤2x+7 −5x≤5-5x \le 5−5x≤5 x≥−1.x \ge -1.x≥−1.

  4. Solve the left inequality: x2−3x+2x2+2x+7≥−1.\frac{x^2-3x+2}{x^2+2x+7} \ge -1.x2+2x+7x2−3x+2​≥−1. Again multiplying by the positive denominator: x2−3x+2≥−(x2+2x+7).x^2-3x+2 \ge -(x^2+2x+7).x2−3x+2≥−(x2+2x+7). x2−3x+2≥−x2−2x−7x^2-3x+2 \ge -x^2-2x-7x2−3x+2≥−x2−2x−7 2x2−x+9≥0.2x^2-x+9 \ge 0.2x2−x+9≥0.

  5. Check whether 2x2−x+9≥02x^2-x+9 \ge 02x2−x+9≥0 is always true. Its discriminant is Δ=(−1)2−4(2)(9)=1−72=−71<0.\Delta = (-1)^2-4(2)(9)=1-72=-71<0.Δ=(−1)2−4(2)(9)=1−72=−71<0. Since the coefficient of x2x^2x2 is positive and the discriminant is negative, 2x2−x+9>0for all x.2x^2-x+9>0 \quad \text{for all } x.2x2−x+9>0for all x. So this inequality holds for every real number.

  6. Therefore the only restriction comes from x≥−1.x \ge -1.x≥−1.

Hence the domain is [−1,∞).[-1,\infty).[−1,∞).

  1. Comparing with the options:
  • A: [1,∞)[1,\infty)[1,∞) ❌
  • B: [−1,2][-1,2][−1,2] ❌
  • C: [−1,∞)[-1,\infty)[−1,∞) ✅
  • D: (−∞,2](-\infty,2](−∞,2] ❌

So the correct option is C.

PreviousNext

More from Inverse Trigonometric Functions

  • The domain of the function cos−1(π2sin−1(4x2−11​)​) is :2022 · MCQ
  • 50tan(3tan−1(21​)+2cos−1(5​1​))+42​tan(21​tan−1(22​)) is equal to ​…2022 · Numerical
  • Let m and M respectively be the minimum and the maximum values of f(x)=sin−12x+sin2x+cos−12x+cos2x,x∈[0,8π​]. Then m + M is equal to :2022 · MCQ
  • Let α=tan(165π​sin(2cos−1(5​1​))) and β=cos(sin−1(54​)+sec−1(35​))…2022 · MCQ
  • cos−1(cos(−5))+sin−1(sin(6))−tan−1(tan(12)) is equal to : (The inverse trigonometric functions take the principal values)2021 · MCQ
  • Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin−1(53x​)+sin−1(54x​)=sin−1x…2021 · MCQ
  • The sum of possible values of x for tan − 1(x + 1) + cot − 1 (x−11​) = tan − 1 (318​) is :2021 · MCQ
  • If cot − 1(α) = cot − 1 2 + cot − 1 8 + cot − 1 18 + cot − 1 32 + ...... upto 100 terms, then α is :2021 · MCQ