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Functions question

2025 · 2 Apr · Shift 2 · Q36

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Functions
2025 · 2 Apr · Shift 2 · Q36

Functions question

2025 · 2 Apr · Shift 2 · Q36

JEE MainMathematicsFunctionsMCQ+4 / −1

If the domain of the function

f(x)=\frac{1}{\sqrt{10+3 x-x^2}}+\frac{1}{\sqrt{x+|x|}}

is

(a, b)

, then

(1+a)^2+b^2

is equal to :

A
29
B
30
C
25
D
26

View written solutionFree

Correct answer: D

We need the domain of $f(x)=\frac{1}{\sqrt{10+3x-x^2}}+\frac{1}{\sqrt{x+|x|}}.$

Since both terms are in denominators with square roots, their radicands must be strictly positive.

For the first term, $\frac{1}{\sqrt{10+3x-x^2}}$ we need $10+3x-x^2>0.$

Rewrite: $-x^2+3x+10>0$ $x^2-3x-10<0$ $(x-5)(x+2)<0.$

So, $-2<x<5.$

For the second term, $\frac{1}{\sqrt{x+|x|}},$ we need $x+|x|>0.$

Now consider cases:

If $x\ge 0$ , then $|x|=x$ , so $x+|x|=x+x=2x>0 \implies x>0.$
If $x<0$ , then $|x|=-x$ , so $x+|x|=x-x=0,$ which is not allowed because denominator becomes $\sqrt{0}=0$ .

Hence the second term is defined only for $x>0.$

Intersect both conditions: $-2<x<5$ and $x>0.$

Therefore the domain is $(0,5).$ So, $a=0,\quad b=5.$

Compute the required value: $(1+a)^2+b^2=(1+0)^2+5^2=1+25=26.$

Comparing with options, the correct option is $\boxed{26}$ which is Option D.

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