NIMCET 2007 — Mathematics PYQ

To find the minimum value of the function $f(x)=\max (2x+1,3-4x)$, we need to look at how the two linear functions behave and where they intersect.

Step 1: Find the point of intersection

The function $f(x)$ takes the upper boundary (the maximum) of the two straight lines:

1. $y_1=2x+1$ (an increasing line with a positive slope)

2. $y_2=3-4x$ (a decreasing line with a negative slope)

The minimum value of a function defined as $\max (y_1,y_2)$ occurs exactly at their point of intersection. Let's find this intersection point by setting them equal to each other:

$$2x+1=3-4x$$

Step 2: Solve for $x$

Add $4x$ to both sides:

$$2x+4x+1=3$$

$$6x+1=3$$

Subtract $1$ from both sides:

$$6x=2$$

Divide by $6$:

$$x=\frac{2}{6}=\frac{1}{3}$$

Step 3: Calculate the minimum value of $f(x)$

Now, substitute $x=\frac{1}{3}$ back into either of the original equations to find the corresponding $y$-value (which represents the minimum value of $f(x)$).

Using $y_1=2x+1$:

$$f\left(\frac{1}{3}\right)=2\left(\frac{1}{3}\right)+1$$

$$f\left(\frac{1}{3}\right)=\frac{2}{3}+1=\frac{5}{3}$$

Conclusion

The minimum value of $f(x)$ is $5/3$.

Correct Option: (b)