Problem 101

Analysis

By utilizing the monotonicity and the odd-even property of the function given in the problem, there is only one value of $x$ that makes $f(2x^{2}+1)=f(x-\lambda)$. That is, there is only one value of $x$ that satisfies $2x^{2}+1=x-\lambda$. By setting the discriminant to zero, we can solve for the value of $\lambda$. This problem tests the concept of the zeros of a function, the monotonicity of a function, and the odd-even property of a function. It is a medium-level problem as long as the basics are solid, making it relatively easy to solve.

Solution

Since the function $y=f(x^{2})+f(k-x)$ has only one zero point, there is only one value of $x$ that makes $f(2x^{2}+1)+f(\lambda-x)=0$.

Since $f(x)$ is an odd function, there is only one value of $x$ that makes $f(2x^{2}+1)=f(x-\lambda)$.

Furthermore, since $f(x)$ is a monotonic function on $\mathbb{R}$, there is only one value of $x$ that satisfies $2x^{2}+1=x-\lambda$.

That is, the equation $2x^{2}-x+\lambda+1=0$ has exactly one solution.

Therefore, $\Delta =1-8(\lambda+1)=0$, solving for $\lambda$ gives $\lambda=-\dfrac{7}{8}$.

Hence, the correct choice is $\boxed{C}$.