Question:
If the co-domain of the function
\[
f(x)=
\begin{cases}
3\sin x-4\cos x, & x\in \left(\tan^{-1}\frac{4}{3}-\frac{\pi}{2},\,\tan^{-1}\frac{4}{3}+\frac{\pi}{2}\right)\\[2mm]
\log(\sin x), & x\in \left(\frac{5\pi}{6},\pi\right)
\end{cases}
\]
is \((-\infty,5)\), then \(f\) is
If the co-domain of the function
\[
f(x)=
\begin{cases}
3\sin x-4\cos x, & x\in \left(\tan^{-1}\frac{4}{3}-\frac{\pi}{2},\,\tan^{-1}\frac{4}{3}+\frac{\pi}{2}\right)\\[2mm]
\log(\sin x), & x\in \left(\frac{5\pi}{6},\pi\right)
\end{cases}
\]
is \((-\infty,5)\), then \(f\) is
Show Hint
For piecewise functions, first find the range of each branch separately and then take their union. Compare the resulting range with the given co-domain to test surjectivity.
Updated On: Jul 29, 2026
- an injection but not a surjection
- a surjection but not an injection
- a bijection
- neither an injection nor a surjection
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
A function is surjective if its range is equal to its co-domain. A function is injective if distinct elements of the domain have distinct images.
Step 1: Find the range of \(3\sin x-4\cos x\). Write \[ 3\sin x-4\cos x=5\sin(x-\alpha), \] where \[ \cos\alpha=\frac{3}{5},\qquad \sin\alpha=\frac{4}{5}. \] Thus, \[ \alpha=\tan^{-1}\frac{4}{3}. \] Hence, \[ 3\sin x-4\cos x=5\sin\left(x-\tan^{-1}\frac{4}{3}\right). \] Given \[ x\in\left(\tan^{-1}\frac{4}{3}-\frac{\pi}{2},\,\tan^{-1}\frac{4}{3}+\frac{\pi}{2}\right), \] so \[ x-\tan^{-1}\frac{4}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \] Since \(\sin t\) is strictly increasing on \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \sin t\in(-1,1). \] Therefore, \[ 3\sin x-4\cos x\in(-5,5). \]
Step 2: Find the range of \(\log(\sin x)\). For \[ x\in\left(\frac{5\pi}{6},\pi\right), \] we have \[ \sin x\in\left(0,\frac{1}{2}\right). \] Hence, \[ \log(\sin x)\in(-\infty,\log\tfrac12). \] Since \[ \log\tfrac12\lt 0, \] this interval is contained in \((-\infty,5)\).
Step 3: Find the overall range of \(f\). Combining both parts, \[ \text{Range}(f)=(-5,5)\cup(-\infty,\log\tfrac12). \] Since \[ \log\tfrac12\gt -5, \] the union becomes \[ (-\infty,5). \] Thus, \[ \text{Range}(f)=(-\infty,5). \] Since the co-domain is also \((-\infty,5)\), \(f\) is surjective.
Step 4: Check injectivity. The interval \[ (-\infty,\log\tfrac12) \] lies inside \[ (-5,5). \] Hence, values in this interval are attained by both branches of the function. Therefore, different domain elements can have the same image. So, \(f\) is not injective.
Conclusion: \(f\) is a surjection but not an injection. \[ \boxed{\text{\(f\) is a surjection but not an injection}} \]
Step 1: Find the range of \(3\sin x-4\cos x\). Write \[ 3\sin x-4\cos x=5\sin(x-\alpha), \] where \[ \cos\alpha=\frac{3}{5},\qquad \sin\alpha=\frac{4}{5}. \] Thus, \[ \alpha=\tan^{-1}\frac{4}{3}. \] Hence, \[ 3\sin x-4\cos x=5\sin\left(x-\tan^{-1}\frac{4}{3}\right). \] Given \[ x\in\left(\tan^{-1}\frac{4}{3}-\frac{\pi}{2},\,\tan^{-1}\frac{4}{3}+\frac{\pi}{2}\right), \] so \[ x-\tan^{-1}\frac{4}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \] Since \(\sin t\) is strictly increasing on \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[ \sin t\in(-1,1). \] Therefore, \[ 3\sin x-4\cos x\in(-5,5). \]
Step 2: Find the range of \(\log(\sin x)\). For \[ x\in\left(\frac{5\pi}{6},\pi\right), \] we have \[ \sin x\in\left(0,\frac{1}{2}\right). \] Hence, \[ \log(\sin x)\in(-\infty,\log\tfrac12). \] Since \[ \log\tfrac12\lt 0, \] this interval is contained in \((-\infty,5)\).
Step 3: Find the overall range of \(f\). Combining both parts, \[ \text{Range}(f)=(-5,5)\cup(-\infty,\log\tfrac12). \] Since \[ \log\tfrac12\gt -5, \] the union becomes \[ (-\infty,5). \] Thus, \[ \text{Range}(f)=(-\infty,5). \] Since the co-domain is also \((-\infty,5)\), \(f\) is surjective.
Step 4: Check injectivity. The interval \[ (-\infty,\log\tfrac12) \] lies inside \[ (-5,5). \] Hence, values in this interval are attained by both branches of the function. Therefore, different domain elements can have the same image. So, \(f\) is not injective.
Conclusion: \(f\) is a surjection but not an injection. \[ \boxed{\text{\(f\) is a surjection but not an injection}} \]
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