Primitive pair over finite fields with specified quadratic coefficient

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Let $ {\mathbb{F}}_{q^m} $ be the $ m $-th degree extension of the finite field $ {\mathbb{F}}_{q} $, for some prime power $ q $ and a positive integer $ m $. For $ n \in \mathbb{N} $, consider the rational function $ f(x) = \frac{f_{1}(x)}{f_{2}(x)} \in {\mathbb{F}}_{q^m}(x) $, where $ f_{1} $ and $ f_{2} $ are co-prime polynomials over $ {\mathbb{F}}_{q^m} $ with $ f_1 $ square-free, and deg($ f_{1} $)+deg($ f_{2} $$ ) = n $. In this article, we have established a condition which is sufficient to ensure the existence of a primitive pair $ (\alpha,f(\alpha))\in $ $ {\mathbb{F}}_{q^m}^*\times {\mathbb{F}}_{q^m}^* $ such that $ \mathrm{Qc}_{ {\mathbb{F}_{q^m}}/ {\mathbb{F}_{q}}}(\alpha) = a $, where $ a\in {\mathbb{F}}_{q}^* $ and $ \mathrm{Qc}_{ {\mathbb{F}_{q^m}}/ {\mathbb{F}_{q}}}(\alpha) = \underset{0\leq i<j\leq m-1}{\sum}\Bigg(\underset{k\neq i,j}{\underset{0\leq k\leq m-1}{\prod_{}^{}}}\alpha^{q^k}\Bigg) $. Moreover, when $ n = 2 $, $ m\geq 9 $ and $ q $ is an odd prime power, we show that such a pair will definitely exist, apart from $ 21 $ possible choices.