JEE Challenger
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MathMedAdv 2015 • Paper 1

Range and Limit of Nested Sine Composite Functions

Suppose f(x)=sin⁡(π6sin⁡(π2sin⁡x))f(x) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right) for all x∈Rx \in \mathbb{R} and g(x)=π2sin⁡xg(x) = \frac{\pi}{2}\sin x for all x∈Rx \in \mathbb{R}. Let (f∘g)(x)(f \circ g)(x) represent f(g(x))f(g(x)) and (g∘f)(x)(g \circ f)(x) represent g(f(x))g(f(x)). Which of the following statements is (are) correct?

Options

A

Range of ff is [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]

Correct
B

Range of f∘gf \circ g is [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]

Correct
C

lim⁡x→0f(x)g(x)=π6\lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}

Correct
D

There is an x∈Rx \in \mathbb{R} such that (g∘f)(x)=1(g \circ f)(x) = 1

Topics & Concepts

Relations and FunctionsFunctionsDomain and RangeLimits

Step-by-Step Solution

To determine which of the given statements are correct, let us analyze each statement individually:

1. Range of f(x)f(x) (Option A)

The function is given by: f(x)=sin⁡(π6sin⁡(π2sin⁡x)),x∈Rf(x) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right), \quad x \in \mathbb{R}

  • Since x∈Rx \in \mathbb{R}, we know that: sin⁡x∈[−1,1]\sin x \in [-1, 1]
  • Multiplying by π2\frac{\pi}{2}: π2sin⁡x∈[−π2,π2]\frac{\pi}{2}\sin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
  • Applying the sine function, which is continuous and monotonically increasing on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]: sin⁡(π2sin⁡x)∈[sin⁡(−π2),sin⁡(π2)]=[−1,1]\sin\left(\frac{\pi}{2}\sin x\right) \in \left[\sin\left(-\frac{\pi}{2}\right), \sin\left(\frac{\pi}{2}\right)\right] = [-1, 1]
  • Multiplying by π6\frac{\pi}{6}: π6sin⁡(π2sin⁡x)∈[−π6,π6]\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right) \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]
  • Finally, taking the sine again: f(x)=sin⁡(π6sin⁡(π2sin⁡x))∈[sin⁡(−π6),sin⁡(π6)]=[−12,12]f(x) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right) \in \left[\sin\left(-\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right)\right] = \left[-\frac{1}{2}, \frac{1}{2}\right]

Therefore, the range of ff is [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right].
Hence, Option A is correct.

2. Range of (f∘g)(x)(f \circ g)(x) (Option B)

The composite function is (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)), where g(x)=π2sin⁡xg(x) = \frac{\pi}{2}\sin x.

  • For x∈Rx \in \mathbb{R}, the range of g(x)g(x) is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
  • Let u=g(x)∈[−π2,π2]u = g(x) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Then: (f∘g)(x)=f(u)=sin⁡(π6sin⁡(π2sin⁡u))(f \circ g)(x) = f(u) = \sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin u\right)\right)
  • As uu varies over [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], sin⁡u\sin u attains all values in [−1,1][-1, 1].
  • Following the exact same steps as for f(x)f(x): sin⁡u∈[−1,1]  ⟹  π2sin⁡u∈[−π2,π2]  ⟹  sin⁡(π2sin⁡u)∈[−1,1]\sin u \in [-1, 1] \implies \frac{\pi}{2}\sin u \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \implies \sin\left(\frac{\pi}{2}\sin u\right) \in [-1, 1]   ⟹  π6sin⁡(π2sin⁡u)∈[−π6,π6]  ⟹  f(u)∈[−12,12]\implies \frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin u\right) \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \implies f(u) \in \left[-\frac{1}{2}, \frac{1}{2}\right]

Thus, the range of f∘gf \circ g is [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right].
Hence, Option B is correct.

3. Evaluation of lim⁡x→0f(x)g(x)\lim_{x \to 0} \frac{f(x)}{g(x)} (Option C)

We have: lim⁡x→0f(x)g(x)=lim⁡x→0sin⁡(π6sin⁡(π2sin⁡x))π2sin⁡x\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{\sin\left(\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\right)}{\frac{\pi}{2}\sin x}

Notice that g(x)=π2sin⁡x→0g(x) = \frac{\pi}{2}\sin x \to 0 as x→0x \to 0. We can rewrite the expression as: f(x)g(x)=sin⁡(π6sin⁡(g(x)))π6sin⁡(g(x))⋅π6sin⁡(g(x))g(x)\frac{f(x)}{g(x)} = \frac{\sin\left(\frac{\pi}{6}\sin(g(x))\right)}{\frac{\pi}{6}\sin(g(x))} \cdot \frac{\frac{\pi}{6}\sin(g(x))}{g(x)}

Taking the limit as x→0x \to 0 (which implies g(x)→0g(x) \to 0 and sin⁡(g(x))→0\sin(g(x)) \to 0): lim⁡x→0sin⁡(π6sin⁡(g(x)))π6sin⁡(g(x))=1\lim_{x \to 0} \frac{\sin\left(\frac{\pi}{6}\sin(g(x))\right)}{\frac{\pi}{6}\sin(g(x))} = 1 lim⁡x→0sin⁡(g(x))g(x)=1\lim_{x \to 0} \frac{\sin(g(x))}{g(x)} = 1

Therefore: lim⁡x→0f(x)g(x)=1⋅π6⋅1=π6\lim_{x \to 0} \frac{f(x)}{g(x)} = 1 \cdot \frac{\pi}{6} \cdot 1 = \frac{\pi}{6}

Hence, Option C is correct.

4. Existence of xx such that (g∘f)(x)=1(g \circ f)(x) = 1 (Option D)

We have: (g∘f)(x)=g(f(x))=π2sin⁡(f(x))(g \circ f)(x) = g(f(x)) = \frac{\pi}{2}\sin(f(x))

Since the range of f(x)f(x) is [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right], we have: sin⁡(f(x))∈[sin⁡(−12),sin⁡(12)]\sin(f(x)) \in \left[\sin\left(-\frac{1}{2}\right), \sin\left(\frac{1}{2}\right)\right]

The maximum possible value of (g∘f)(x)(g \circ f)(x) is: (g∘f)max⁡=π2sin⁡(12)(g \circ f)_{\max} = \frac{\pi}{2}\sin\left(\frac{1}{2}\right)

Since sin⁡θ<θ\sin \theta < \theta for all θ>0\theta > 0: sin⁡(12)<12\sin\left(\frac{1}{2}\right) < \frac{1}{2}

Thus: (g∘f)max⁡<π2⋅12=π4≈0.785<1(g \circ f)_{\max} < \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4} \approx 0.785 < 1

Since the maximum value of (g∘f)(x)(g \circ f)(x) is strictly less than 11, there is no x∈Rx \in \mathbb{R} such that (g∘f)(x)=1(g \circ f)(x) = 1.
Hence, Option D is incorrect.

Conclusion:

The correct statements are A, B, and C.

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