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Class 12 Maths · Relations And Functions · Mcq Single Correct #83

Let \(f:\mathbb R\to[0,\pi/2)\), \(f(x)=\tan^{-1}(x^2+x+a)\). The set of real \(a\) for which \(f\) is onto is

✅ Step-by-Step Solution

Final Answer: Option (B): \(\{1/4\}\)
  1. \(x^2+x+a=(x+1/2)^2+a-1/4\), whose range is \([a-1/4,\infty)\). Applying principal arctangent gives \([\arctan(a-1/4),\pi/2)\). This equals \([0,\pi/2)\) exactly when \(a=1/4\). The inequality \(a\ge1/4\) alone only makes the outputs nonnegative; it does not guarantee onto.

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