Match List-I with List-II

List-I	List-II
(A) Point of minima of $f(x) = \vert x+1\vert$	(I) 1
(B) Minimum value of $f(x) = \vert x\vert$	(II) -1
(C) Maximum value of $f(x) = 1 - x^2$	(III) 2
(D) Minimum value of $f(x) = 2 + \sin^2 x$	(IV) 0

Choose the correct answer from the options given below:

Let me break down each function so you can understand exactly what's happening!

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(A) Point of minima of $f(x) = |x+1|$

We need to find the x-value WHERE the function reaches its lowest point.

The absolute value function $|x+1|$ looks like a "V" shape.

When does $|x+1| = 0$? When $x+1 = 0$, so $x = -1$

At $x = -1$: $f(-1) = |(-1)+1| = |0| = 0$

For any other x value, $f(x) > 0$

The lowest point happens at $x = -1$

(A) → (II) $= -1$

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(B) Minimum value of $f(x) = |x|$

We need the LOWEST Y-VALUE (not the x-position).

The graph of $|x|$ is a "V" with its tip at the origin.

When $x = 0$: $f(0) = |0| = 0$

For any other x (positive or negative), $|x|$ is always positive.

The minimum VALUE is $0$

(B) → (IV) $= 0$

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(C) Maximum value of $f(x) = 1 - x^2$

Find the HIGHEST Y-VALUE this function can reach.

This is an upside-down parabola (∩ shape).

$x^2$ is always $\geq 0$

So $-x^2$ is always $\leq 0$

Therefore, $1 - x^2$ is always $\leq 1$

When is it exactly $1$? When $x^2 = 0$, meaning $x = 0$

$f(0) = 1 - 0^2 = 1$

The maximum VALUE is $1$

(C) → (I) $= 1$

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(D) Minimum value of $f(x) = 2 + \sin^2 x$

Find the LOWEST Y-VALUE.

Remember: $\sin^2 x$ always stays between $0$ and $1$

Minimum of $\sin^2 x = 0$ (when $\sin x = 0$)

Maximum of $\sin^2 x = 1$ (when $\sin x = \pm 1$)

So: $f(x) = 2 + \sin^2 x$

Minimum $= 2 + 0 = 2$

Maximum $= 2 + 1 = 3$

The minimum VALUE is $2$

(D) → (III) $= 2$

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Final Matching:

(A) → (II) $-1$

(B) → (IV) $0$

(C) → (I) $1$

(D) → (III) $2$

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Quick Tips to Remember:

Point of minima means the x-coordinate where minimum happens

Minimum/Maximum value means the actual y-value (height)

$|\text{anything}|$ is always $\geq 0$

$\sin^2 x$ ranges from $0$ to $1$

$x^2$ is always $\geq 0$