1. The force varies with position as

$F(x)=\alpha+\beta x^2$

Given:

• $\alpha=1\,\text{N}$
• displacement from $x=0$ to $x=1\,\text{m}$
• work done $W=5\,\text{J}$

2. Work done by a variable force is

$W=\int_{0}^{1} F(x)\,dx$

So,

$5=\int_0^1 (\alpha+\beta x^2)dx$

Substitute $\alpha=1$:

$5=\int_0^1 (1+\beta x^2)dx$

3. Integrate term by term:

$5=\left[x+\beta\frac{x^3}{3}\right]_0^1$

$5=\left(1+\frac{\beta}{3}\right)-0$

$5=1+\frac{\beta}{3}$

4. Solve for $\beta$:

$\frac{\beta}{3}=4$

$\beta=12\,\text{N/m}^2$

5. Checking options:

• A: $15\,\text{N/m}^2$
• B: $10\,\text{N/m}^2$
• C: $12\,\text{N/m}^2$
• D: $8\,\text{N/m}^2$

Hence, the correct option is C.