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Functions question

2015 · Shift 1 · Q40
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Functions question

2015 · Shift 1 · Q40

JEE AdvancedMathematicsFunctionsMultiple correct+4 / −2
Let f(x)=sin⁡(π6sin⁡(π2sin⁡x))f(x) = \sin \left( {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right)f(x)=sin(6π​sin(2π​sinx)) for all x∈Rx \in Rx∈R and g(x) =π2sin⁡x{{\pi \over 2}\sin x}2π​sinx for all x ∈\in∈ R. Let (f∘g)(x)(f \circ g)(x)(f∘g)(x) denote f(g(x)) and (g∘f)(x)(g \circ f)(x)(g∘f)(x) denote g(f(x)). Then which of the following is/are true?
  1. A
    Range of f is [−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right][−21​,21​].
  2. B
    Range of f ∘\circ∘ g is [−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right][−21​,21​].
  3. C
    lim⁡x→0f(x)g(x)=π6\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}} = {\pi \over 6}x→0lim​g(x)f(x)​=6π​.
  4. D
    There is an x ∈\in∈ R such that (g ∘\circ∘ f)(x) = 1.
View written solutionFree

Correct answer: A, B, C

  1. Given functions

f(x)=sin⁡(π6sin⁡(π2sin⁡x)),g(x)=π2sin⁡x.f(x)=\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right),\qquad g(x)=\frac{\pi}{2}\sin x.f(x)=sin(6π​sin(2π​sinx)),g(x)=2π​sinx.

We must check each option.

  1. Option A: Range of fff

Let t=sin⁡(π2sin⁡x).t=\sin\left(\frac{\pi}{2}\sin x\right).t=sin(2π​sinx). Since sin⁡x∈[−1,1]\sin x\in[-1,1]sinx∈[−1,1], we get π2sin⁡x∈[−π2,π2].\frac{\pi}{2}\sin x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].2π​sinx∈[−2π​,2π​]. Now on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][−2π​,2π​], the sine function takes all values from −1-1−1 to 111. Hence t∈[−1,1].t\in[-1,1].t∈[−1,1].

Therefore π6t∈[−π6,π6].\frac{\pi}{6}t\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right].6π​t∈[−6π​,6π​]. Again, sine is increasing on [−π6,π6]\left[-\frac{\pi}{6},\frac{\pi}{6}\right][−6π​,6π​], so f(x)=sin⁡(π6t)∈[sin⁡(−π6),sin⁡(π6)]=[−12,12].f(x)=\sin\left(\frac{\pi}{6}t\right)\in\left[\sin\left(-\frac{\pi}{6}\right),\sin\left(\frac{\pi}{6}\right)\right]=\left[-\frac12,\frac12\right].f(x)=sin(6π​t)∈[sin(−6π​),sin(6π​)]=[−21​,21​].

Also, the endpoints are attained:

  • if sin⁡x=1\sin x=1sinx=1, then f(x)=sin⁡(π6sin⁡π2)=sin⁡(π6)=12;f(x)=\sin\left(\frac{\pi}{6}\sin\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{6}\right)=\frac12;f(x)=sin(6π​sin2π​)=sin(6π​)=21​;
  • if sin⁡x=−1\sin x=-1sinx=−1, then f(x)=sin⁡(−π6)=−12.f(x)=\sin\left(-\frac{\pi}{6}\right)=-\frac12.f(x)=sin(−6π​)=−21​.

So the range of fff is exactly [−12,12].\left[-\frac12,\frac12\right].[−21​,21​].

Thus, A is true.

  1. Option B: Range of f∘gf\circ gf∘g

We have (f∘g)(x)=f(g(x))=sin⁡(π6sin⁡(π2sin⁡(π2sin⁡x))).(f\circ g)(x)=f(g(x))=\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\sin x\right)\right)\right).(f∘g)(x)=f(g(x))=sin(6π​sin(2π​sin(2π​sinx))).

Let u=sin⁡(π2sin⁡x).u=\sin\left(\frac{\pi}{2}\sin x\right).u=sin(2π​sinx). As above, since sin⁡x∈[−1,1]\sin x\in[-1,1]sinx∈[−1,1], we get u∈[−1,1].u\in[-1,1].u∈[−1,1]. Hence π2u∈[−π2,π2],\frac{\pi}{2}u\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],2π​u∈[−2π​,2π​], and so sin⁡(π2u)∈[−1,1].\sin\left(\frac{\pi}{2}u\right)\in[-1,1].sin(2π​u)∈[−1,1]. Thus the inner sine in f(g(x))f(g(x))f(g(x)) again runs through all values in [−1,1][-1,1][−1,1].

Therefore, exactly as in Option A, (f∘g)(x)∈[−12,12].(f\circ g)(x)\in\left[-\frac12,\frac12\right].(f∘g)(x)∈[−21​,21​].

Now check endpoints:

  • if sin⁡x=1\sin x=1sinx=1, then g(x)=π2g(x)=\frac{\pi}{2}g(x)=2π​, so f(g(x))=f(π2)=sin⁡(π6sin⁡(π2⋅1))=sin⁡(π6)=12;f(g(x))=f\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\cdot 1\right)\right)=\sin\left(\frac{\pi}{6}\right)=\frac12;f(g(x))=f(2π​)=sin(6π​sin(2π​⋅1))=sin(6π​)=21​;
  • if sin⁡x=−1\sin x=-1sinx=−1, similarly f(g(x))=−12f(g(x))=-\frac12f(g(x))=−21​.

So the range of f∘gf\circ gf∘g is [−12,12].\left[-\frac12,\frac12\right].[−21​,21​].

Thus, B is true.

  1. Option C: Evaluate lim⁡x→0f(x)g(x)\displaystyle \lim_{x\to 0}\frac{f(x)}{g(x)}x→0lim​g(x)f(x)​

We use small-angle behavior.

First, g(x)=π2sin⁡x.g(x)=\frac{\pi}{2}\sin x.g(x)=2π​sinx. As x→0x\to 0x→0, g(x)∼π2x.g(x)\sim \frac{\pi}{2}x.g(x)∼2π​x.

Now, f(x)=sin⁡(π6sin⁡(π2sin⁡x)).f(x)=\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right).f(x)=sin(6π​sin(2π​sinx)). As x→0x\to 0x→0:

  • sin⁡x∼x\sin x\sim xsinx∼x,
  • so π2sin⁡x∼π2x\frac{\pi}{2}\sin x\sim \frac{\pi}{2}x2π​sinx∼2π​x,
  • hence sin⁡(π2sin⁡x)∼π2x,\sin\left(\frac{\pi}{2}\sin x\right)\sim \frac{\pi}{2}x,sin(2π​sinx)∼2π​x,
  • therefore π6sin⁡(π2sin⁡x)∼π6⋅π2x=π212x,\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\sim \frac{\pi}{6}\cdot \frac{\pi}{2}x=\frac{\pi^2}{12}x,6π​sin(2π​sinx)∼6π​⋅2π​x=12π2​x,
  • and thus f(x)∼π212x.f(x)\sim \frac{\pi^2}{12}x.f(x)∼12π2​x.

So lim⁡x→0f(x)g(x)=lim⁡x→0π212xπ2x=π212⋅2π=π6.\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{\frac{\pi^2}{12}x}{\frac{\pi}{2}x}=\frac{\pi^2}{12}\cdot\frac{2}{\pi}=\frac{\pi}{6}.limx→0​g(x)f(x)​=limx→0​2π​x12π2​x​=12π2​⋅π2​=6π​.

Thus, C is true.

  1. Option D: Is there x∈Rx\in\mathbb Rx∈R such that (g∘f)(x)=1(g\circ f)(x)=1(g∘f)(x)=1?

We have (g∘f)(x)=g(f(x))=π2sin⁡(f(x)).(g\circ f)(x)=g(f(x))=\frac{\pi}{2}\sin(f(x)).(g∘f)(x)=g(f(x))=2π​sin(f(x)). From Option A, we know f(x)∈[−12,12].f(x)\in\left[-\frac12,\frac12\right].f(x)∈[−21​,21​]. Since sine is increasing on this interval, sin⁡(f(x))∈[−sin⁡12,sin⁡12].\sin(f(x))\in\left[-\sin\frac12,\sin\frac12\right].sin(f(x))∈[−sin21​,sin21​]. Hence (g∘f)(x)∈[−π2sin⁡12,π2sin⁡12].(g\circ f)(x)\in\left[-\frac{\pi}{2}\sin\frac12,\frac{\pi}{2}\sin\frac12\right].(g∘f)(x)∈[−2π​sin21​,2π​sin21​]. Now π2sin⁡12≈1.5708×0.4794≈0.753<1.\frac{\pi}{2}\sin\frac12\approx 1.5708\times 0.4794\approx 0.753<1.2π​sin21​≈1.5708×0.4794≈0.753<1. So (g∘f)(x)(g\circ f)(x)(g∘f)(x) can never equal 111.

Thus, D is false.

  1. Final conclusion

The true statements are: A, B, C.\boxed{A,\ B,\ C}.A, B, C​.

  1. Comparison with stored correct answer

Stored correct answer: A, B, C

Our derived answer: A, B, C

So the answers agree.

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