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Class 12 Relations And Functions · Mcq Single Correct1 Markstypes_of_functions

Class 12 Maths · Relations And Functions · Mcq Single Correct #94

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

✅ Step-by-Step Solution

Final Answer: Option (D): \(\{1/8\}\)
  1. The minimum of \(x^2+x+2a\) is \(2a-1/4\). Its arctangent has range \([\arctan(2a-1/4),\pi/2)\). Equality with \([0,\pi/2)\) requires \(2a-1/4=0\), hence \(a=1/8\).

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