WBJEE 2024 Math Paper

Let's analyze the function $f(x)=e^x+e^{-x}$ where $f: \mathbb{R} \to \mathbb{R}$.First, we check if the function is one-one (injective).An injective function means that if $f(a)=f(b)$, then $a=b$. Let's assume $f(a)=f(b)$.So, we have:$e^a+e^{-a}=e^b+e^{-b}$Let's rewrite the equation as:$e^a+\;\frac{1}{e^a}=e^b+\;\frac{1}{e^b}$ Given that the exponential function $e^x$ is strictly increasing, and $e^{-x}$ is strictly decreasing, the sum is minimized when $x=0$. This involves checking various values, but combining these two observations, we can see that the function is symmetric and not strictly monotonic. Therefore, it is not one-one.Next, we check if the function is onto (surjective).A surjective function means that for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x)=y$. Here, considering $f(x)=e^x+e^{-x}$, let's find the range.Define a new function $g(x)=e^x+e^{-x}$. We use the fact that: $g(0)=2$When $x \to \infty$, then $e^x \to \infty$ and $e^{-x} \to 0$, so $g(x) \to \infty$.When $x \to -\infty$, then $e^x \to 0$ and $e^{-x} \to \infty$, so $g(x) \to \infty$. But at no point does the function $e^x+e^{-x}$ produce negative values. Therefore, the range of $f(x) \geq 2$. Hence, the function is not surjective.Given that the function is neither injective nor surjective, it cannot be bijective.Thus, the correct answer is:Option D: not bijective