Modulus Function PYQ | JEE Main 2025 | Fastest Shortcut

 

❓ Question

If

$$f(x)=\sqrt{x^2+2x+1}-\sqrt{x^2-2x+1},$$

find

$$f\!\left(-\frac54\right)+f\!\left(\frac34\right)+f\!\left(\frac74\right).$$

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✍️ Solution

First, simplify the function.

$$x^2+2x+1=(x+1)^2$$
$$x^2-2x+1=(x-1)^2$$

Hence,

$$f(x)=\sqrt{(x+1)^2}-\sqrt{(x-1)^2} =|x+1|-|x-1|.$$

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1. Calculate $f\left(-\frac54\right)$

$$f\left(-\frac54\right) =\left|-\frac54+1\right| -\left|-\frac54-1\right|$$
$$=\left|-\frac14\right| -\left|-\frac94\right| =\frac14-\frac94 =-2.$$

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2. Calculate $f\left(\frac34\right)$

$$f\left(\frac34\right) =\left|\frac34+1\right| -\left|\frac34-1\right|$$
$$=\frac74-\frac14 =\frac64 =\frac32.$$

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3. Calculate $f\left(\frac74\right)$

$$f\left(\frac74\right) =\left|\frac74+1\right| -\left|\frac74-1\right|$$
$$=\frac{11}{4}-\frac34 =\frac84 =2.$$

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Sum

$$-2+\frac32+2 =\frac32.$$

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✅ Final Answer

$$\boxed{\frac32}$$