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 · 24 Jun · Shift 1 · Q28
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 · 24 Jun · Shift 1 · Q28

Inverse Trigonometric Functions question

2022 · 24 Jun · Shift 1 · Q28

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The set of all values of k for which (tan⁡−1x)3+(cot⁡−1x)3=kπ3, x∈R{({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R(tan−1x)3+(cot−1x)3=kπ3,x∈R, is the interval :
  1. A
    [132,78)\left[ {{1 \over {32}},{7 \over 8}} \right)[321​,87​)
  2. B
    (124,1316)\left( {{1 \over {24}},{{13} \over {16}}} \right)(241​,1613​)
  3. C
    [148,1316]\left[ {{1 \over {48}},{{13} \over {16}}} \right][481​,1613​]
  4. D
    [132,98)\left[ {{1 \over {32}},{9 \over 8}} \right)[321​,89​)
View written solutionFree

Correct answer: A

  1. Use the relation between inverse trigonometric functions

For real xxx, an−1x+cot⁡−1x=π2 an^{-1}x + \cot^{-1}x = \frac{\pi}{2}an−1x+cot−1x=2π​ when we take principal values with tan⁡−1x∈(−π2,π2),cot⁡−1x∈(0,π).\tan^{-1}x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right), \qquad \cot^{-1}x \in (0,\pi).tan−1x∈(−2π​,2π​),cot−1x∈(0,π).

Let a=tan⁡−1x.a=\tan^{-1}x.a=tan−1x. Then cot⁡−1x=π2−a.\cot^{-1}x=\frac{\pi}{2}-a.cot−1x=2π​−a.

So the given expression becomes a3+(π2−a)3=kπ3.a^3+\left(\frac{\pi}{2}-a\right)^3=k\pi^3.a3+(2π​−a)3=kπ3.

  1. Determine the range of aaa

Since x∈Rx\in \mathbb Rx∈R, a=tan⁡−1x∈(−π2,π2).a=\tan^{-1}x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right).a=tan−1x∈(−2π​,2π​).

Thus we need the range of f(a)=a3+(π2−a)3,a∈(−π2,π2).f(a)=a^3+\left(\frac{\pi}{2}-a\right)^3, \qquad a\in \left(-\frac{\pi}{2},\frac{\pi}{2}\right).f(a)=a3+(2π​−a)3,a∈(−2π​,2π​).

  1. Simplify the function

Using (u−v)3=u3−3u2v+3uv2−v3,(u-v)^3=u^3-3u^2v+3uv^2-v^3,(u−v)3=u3−3u2v+3uv2−v3, we get (π2−a)3=π38−3π24a+3π2a2−a3.\left(\frac{\pi}{2}-a\right)^3=\frac{\pi^3}{8}-\frac{3\pi^2}{4}a+\frac{3\pi}{2}a^2-a^3.(2π​−a)3=8π3​−43π2​a+23π​a2−a3.

Hence

=\frac{\pi^3}{8}-\frac{3\pi^2}{4}a+\frac{3\pi}{2}a^2.$$ So $$k=\frac{f(a)}{\pi^3}.
  1. Find the minimum value

Differentiate: f′(a)=−3π24+3πa=3π(a−π4).f'(a)=-\frac{3\pi^2}{4}+3\pi a=3\pi\left(a-\frac{\pi}{4}\right).f′(a)=−43π2​+3πa=3π(a−4π​).

Thus critical point is at a=π4.a=\frac{\pi}{4}.a=4π​.

Also, f′′(a)=3π>0,f''(a)=3\pi>0,f′′(a)=3π>0, so this gives the minimum.

Now,

=2\cdot \frac{\pi^3}{64}=\frac\pi^3{32}.$$ Therefore, $$k_{\min}=\frac{1}{32}.$$ This value is attained when $$\tan^{-1}x=\frac{\pi}{4}\Rightarrow x=1.$$ --- 5. **Find the largest possible value** Since the interval for $a$ is open, endpoints are not included: $$a\to -\frac{\pi}{2}^{+} \quad \text{or} \quad a\to \frac{\pi}{2}^{-}.$$ Evaluate endpoint behavior: - As $a\to -\frac{\pi}{2}^{+}$, $$f(a)\to \left(-\frac{\pi}{2}\right)^3+\pi^3=-\frac{\pi^3}{8}+\pi^3=\frac{7\pi^3}{8}.$$ - As $a\to \frac{\pi}{2}^{-}$, $$f(a)\to \left(\frac{\pi}{2}\right)^3+0=\frac{\pi^3}{8}.$$ So the supremum is $$\frac{7\pi^3}{8},$$ but it is **not attained** because $a=-\frac{\pi}{2}$ is not in the domain. Hence, $$k\in \left[\frac{1}{32},\frac{7}{8}\right).$$ --- 6. **Match with the options** This is exactly **Option A**: $$\boxed{\left[\frac{1}{32},\frac{7}{8}\right)}.$$ --- 7. **Comparison with stored correct answer** Stored correct answer: **A** Our derived answer: **A** So they agree.
PreviousNext

More from Inverse Trigonometric Functions

  • The domain of the function f(x)=loge​(x2−3x+2)cos−1(x2−9x2−5x+6​)​ is :2022 · MCQ
  • Let x∗y=x2+y3 and (x∗1)∗1=x∗(1∗1). Then a value of 2sin−1(x4+x2+2x4+x2−2​) is :2022 · MCQ
  • Let x=sin(2tan−1α) and y=sin(21​tan−134​). If S={a∈R:y2=1−x}, then α∈S∑​16α3 is equal to ​…2022 · Numerical
  • The value of tan−1(sin(4π​)cos(415π​)−1​) is equal to :2022 · MCQ
  • tan(2tan−151​+sec−125​​+2tan−181​) is equal to :2022 · MCQ
  • If 0<x<2​1​ and αsin−1x​=βcos−1x​, then the value of sin(α+β2πα​) is :2022 · MCQ
  • If the inverse trigonometric functions take principal values then cos−1(103​cos(tan−1(34​))+52​sin(tan−1(34​)))…2022 · MCQ
  • For k∈R, let the solutions of the equation cos(sin−1(xcot(tan−1(cos(sin−1x)))))=k,0<∣x∣<2​1​ be α and β,…2022 · Numerical