1. Forces along the incline

Take upward along the plane as positive.

The forces along the plane are:

• Applied force: $+P$
• Component of weight down the plane: $-mg\sin\theta$
• Friction force: $f$ (its direction depends on tendency of motion)

Since the block is held stationary, net force along plane is zero: $P+f-mg\sin\theta=0$ So, $f=mg\sin\theta-P$

This is the friction required for equilibrium.

2. Range in which static friction can adjust

For static equilibrium, friction must satisfy $|f|\le \mu N$ Here, $N=mg\cos\theta$ So, $|f|\le \mu mg\cos\theta$

Using $f=mg\sin\theta-P$, $-\mu mg\cos\theta \le mg\sin\theta-P \le \mu mg\cos\theta$ This gives $mg(\sin\theta-\mu\cos\theta) \le P \le mg(\sin\theta+\mu\cos\theta)$

This is exactly the given range from $P_1=mg(\sin\theta-\mu\cos\theta)$ to $P_2=mg(\sin\theta+\mu\cos\theta)$

3. Nature of the graph $f$ vs $P$

Since $f=mg\sin\theta-P$ this is a straight line with slope $-1$.

Now check endpoints:

• At $P=P_1=mg(\sin\theta-\mu\cos\theta)$, $f=mg\sin\theta-mg(\sin\theta-\mu\cos\theta)=+\mu mg\cos\theta$

• At $P=P_2=mg(\sin\theta+\mu\cos\theta)$, $f=mg\sin\theta-mg(\sin\theta+\mu\cos\theta)=-\mu mg\cos\theta$

• At $P=mg\sin\theta$, $f=0$

So the graph is a straight line decreasing from $+\mu mg\cos\theta$ at $P_1$ to $-\mu mg\cos\theta$ at $P_2$, crossing zero at $P=mg\sin\theta$.

4. Physical interpretation of sign of friction

• For smaller $P$, the block tends to slide down, so friction acts up the plane: $f>0$.
• For larger $P$, the block tends to move up, so friction acts down the plane: $f<0$.

Thus the graph must cross from positive to negative linearly.

5. Conclusion

The correct graph is the straight line with negative slope joining $(P_1,+\mu mg\cos\theta)$ to $(P_2,-\mu mg\cos\theta)$.

Hence, the correct option is A.