If the domain of the function is (a, b), then is equal to: [2025]

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Solution

Q.
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: [2025]

1 29

2 26

3 30

4 25

Ans.
(2)

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

For f(x) to be defined, we must have $10 + 3 x - x^{2} > 0$

$\Rightarrow x^{2} - 3 x - 10 < 0 \Rightarrow (x - 5) (x + 2) < 0 \Rightarrow - 2 < x < 5$

Also, x + |x| > 0

Now, if x > 0, then x + |x| > 0

If x < 0, then |x| = –x $\Rightarrow$ x + |x| = x – x = 0

$∴$ The domain of $\frac{1}{\sqrt{x + | x |}}$ is x > 0

$∴$ Domain of f(x) is 0 < x < 5 i.e., (0, 5)

$∴$ a = 0 and b = 5

$\Rightarrow {(1 + a)}^{2} + b^{2} = 1^{2} + 5^{2} = 26$ .

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