JEE Mathematics 2026 Question with Solution

Given:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4}\,dx$$

where $[x]$ denotes the greatest integer less than or equal to $x$.

Find: The value of the integral and the correct option.

For the greatest integer function, split the interval at integer points inside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Since

$$-\frac{\pi}{2}\approx -1.57 \quad \text{and} \quad \frac{\pi}{2}\approx 1.57,$$

the interval is divided as

$$\left[-\frac{\pi}{2},-1\right),\ \left[-1,0\right),\ \left[0,1\right),\ \left[1,\frac{\pi}{2}\right].$$

Hence,

• on $\left[-\frac{\pi}{2},-1\right)$, $[x]=-2$, so the integrand is $\frac{1}{2}$,
• on $\left[-1,0\right)$, $[x]=-1$, so the integrand is $\frac{1}{3}$,
• on $\left[0,1\right)$, $[x]=0$, so the integrand is $\frac{1}{4}$,
• on $\left[1,\frac{\pi}{2}\right]$, $[x]=1$, so the integrand is $\frac{1}{5}$.

Therefore,

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4}\,dx = \int_{-\frac{\pi}{2}}^{-1} \frac{1}{2}\,dx + \int_{-1}^{0} \frac{1}{3}\,dx + \int_{0}^{1} \frac{1}{4}\,dx + \int_{1}^{\frac{\pi}{2}} \frac{1}{5}\,dx$$

Now evaluate each part:

$$\int_{-\frac{\pi}{2}}^{-1} \frac{1}{2}\,dx = \frac{1}{2}\left(-1+\frac{\pi}{2}\right)$$ $$\int_{-1}^{0} \frac{1}{3}\,dx = \frac{1}{3}$$ $$\int_{0}^{1} \frac{1}{4}\,dx = \frac{1}{4}$$ $$\int_{1}^{\frac{\pi}{2}} \frac{1}{5}\,dx = \frac{1}{5}\left(\frac{\pi}{2}-1\right)$$

Adding,

$$\frac{1}{2}\left(\frac{\pi}{2}-1\right)+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$$ $$= \left(\frac{\pi}{2}-1\right)\left(\frac{1}{2}+\frac{1}{5}\right)+\frac{1}{3}+\frac{1}{4}$$ $$= \frac{7}{10}\left(\frac{\pi}{2}-1\right)+\frac{7}{12}$$ $$= \frac{7\pi}{20}-\frac{7}{10}+\frac{7}{12}$$ $$= \frac{21\pi-7}{60}$$ $$= \frac{7}{60}(3\pi-1)$$

Therefore, the value of the integral is $\dfrac{7}{60}(3\pi-1)$ and the correct option is C.

The solution's intermediate integrands contain a mismatch, but the final result and option selection clearly give C.](streamdown:incomplete-link)

Given: An integral involving the greatest integer function.

Find: The exact value by evaluating the integrand on each integer subinterval.

The key idea is that $[x]$ remains constant between consecutive integers. So first identify the integer breakpoints inside the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, namely $-1,0,1$.

Then write the four constant values:

$$\begin{aligned} [x] &= -2, && x \in \left[-\frac{\pi}{2},-1\right) \\ [x] &= -1, && x \in \left[-1,0\right) \\ [x] &= 0, && x \in \left[0,1\right) \\ [x] &= 1, && x \in \left[1,\frac{\pi}{2}\right] \end{aligned}$$

Hence,

$$\frac{1}{[x]+4}= \begin{cases} \frac{1}{2}, & x \in \left[-\frac{\pi}{2},-1\right) \\ \frac{1}{3}, & x \in \left[-1,0\right) \\ \frac{1}{4}, & x \in \left[0,1\right) \\ \frac{1}{5}, & x \in \left[1,\frac{\pi}{2}\right] \end{cases}$$

Now integrate constant functions over their interval lengths:

$$\begin{aligned} I &= \frac{1}{2}\left(-1+\frac{\pi}{2}\right)+\frac{1}{3}(1)+\frac{1}{4}(1)+\frac{1}{5}\left(\frac{\pi}{2}-1\right) \\ &= \frac{1}{2}\left(\frac{\pi}{2}-1\right)+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi}{2}-1\right) \\ &= \frac{7}{10}\left(\frac{\pi}{2}-1\right)+\frac{7}{12} \\ &= \frac{7\pi}{20}-\frac{7}{10}+\frac{7}{12} \\ &= \frac{21\pi-42+35}{60} \\ &= \frac{21\pi-7}{60} \\ &= \frac{7}{60}(3\pi-1) \end{aligned}$$

So the correct option is C.](streamdown:incomplete-link)