Let f: R → [0, (π/2)) defined by f(x) = tan-1(x2 + x + 2a),then the set of values of âa' for which f is onto, is

Question

Let f: R → [0, (π/2)) defined by f(x) = tan-1(x2 + x + 2a),then the set of values of âa' for which f is onto, is (A) (-(1/4), ∞) (B) [-(1/4), ∞) (C) [-(1/8), ∞) (D) [(1/8), ∞)

Tardigrade Admin · Accepted Answer

Since for given function co-domain is 0 le x < (π/2), For onto function, we have Co-domain = Range = 0 le x < (π/2) This will be possible if x2 + x + 2a ge 0 Fact: Quadratic function f(x) ge 0 i.e. Ax2 + Bx + C ge 0 then D le 0 if A > 0 ∴ x2 + x + 2a ge 0 ⇒ 12 - 4(2a) le 0 ⇒ a ge 1/8 ∴ a ∈ [(1/8), ∞)

Q. Let $f : R \to [0, \frac{π}{2})$ defined by $f (x) = t a n^{- 1} (x^{2} + x + 2 a)$ ,then the set of values of ‘ $a$ ' for which $f$ is onto, is

$(- \frac{1}{4}, \infty)$

$[- \frac{1}{4}, \infty)$

$[- \frac{1}{8}, \infty)$

$[\frac{1}{8}, \infty)$

Q. Let f:R→[0,2π​) defined by f(x)=tan−1(x2+x+2a),then the set of values of ‘a' for which f is onto, is

(−41​,∞)

[−41​,∞)

[−81​,∞)

[81​,∞)

Since for given function co-domain is 0≤x<2π​, For onto function, we have Co-domain = Range = 0≤x<2π​ This will be possible if x2+x+2a≥0 Fact: Quadratic function f(x)≥0 i.e. Ax2+Bx+C≥0 then D≤0 if A>0 ∴x2+x+2a≥0 ⇒12−4(2a)≤0 ⇒a≥1/8 ∴a∈[81​,∞)

Q. Let $f : R \to [0, \frac{π}{2})$ defined by $f (x) = t a n^{- 1} (x^{2} + x + 2 a)$ ,then the set of values of ‘ $a$ ' for which $f$ is onto, is

$(- \frac{1}{4}, \infty)$

$[- \frac{1}{4}, \infty)$

$[- \frac{1}{8}, \infty)$

$[\frac{1}{8}, \infty)$