Find the number of integers in the domain of the function f(x)=√x2-|x|+(1/√9-x2) .

Question

Find the number of integers in the domain of the function f(x)=√x2-|x|+(1/√9-x2) . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(x)=√x2-|x|+(1/√9-x2) Let | x |= t f(x)=√t2-t+(1/√9-t2) t(t-1) ≥ 0 text and 9-t2>0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/5d7b0ba9ca309fc9da8f42510c8e7ffd-.png" /> Hence number of integer in the domain of function is 5

Q. Find the number of integers in the domain of the function $f (x) = x^{2} - ∣ x ∣ + \frac{1}{9 - x ^{2}} .$

$f (x) = x^{2} - ∣ x ∣ + \frac{1}{9 - x ^{2}}$
Let $∣ x ∣ = t$
$f (x) = t^{2} - t + \frac{1}{9 - t ^{2}}$
$t (t - 1) \geq 0 and 9 - t^{2} > 0$

Hence number of integer in the domain of function is $5$

Q. Find the number of integers in the domain of the function f(x)=x2−∣x∣​+9−x2​1​.

f(x)=x2−∣x∣​+9−x2​1​ Let ∣x∣=t f(x)=t2−t​+9−t2​1​ t(t−1)≥0 and 9−t2>0 Hence number of integer in the domain of function is 5

Q. Find the number of integers in the domain of the function $f (x) = x^{2} - ∣ x ∣ + \frac{1}{9 - x ^{2}} .$

$f (x) = x^{2} - ∣ x ∣ + \frac{1}{9 - x ^{2}}$
Let $∣ x ∣ = t$
$f (x) = t^{2} - t + \frac{1}{9 - t ^{2}}$
$t (t - 1) \geq 0 and 9 - t^{2} > 0$

Hence number of integer in the domain of function is $5$