If tan θ and cot θ are roots of x2+2 a x+b=0, then least value of |a| is

Question

If tan θ and cot θ are roots of x2+2 a x+b=0, then least value of |a| is (A) (1/2) (B) 1 (C) 2 (D) cannot be found.

Tardigrade Admin · Accepted Answer

tan θ+ cot θ=2 a and tan θ cot θ=b Now, 2 a= tan θ+ cot θ ≥ 2 if tan θ>0 and 2 a= tan θ+ cot θ ≤-2 if tan θ<0. ⇒ 2|a| ≥ 2 ⇒|a| ≥ 1 Thus, least value of |a| is 1 .

Q. If $tan θ$ and $cot θ$ are roots of $x^{2} + 2 a x + b = 0$ , then least value of $∣ a ∣$ is

$\frac{1}{2}$

1

2

cannot be found.

$tan θ + cot θ = 2 a$ and $tan θ cot θ = b$
Now, $2 a = tan θ + cot θ \geq 2$ if $tan θ > 0$
and $2 a = tan θ + cot θ \leq - 2$ if $tan θ < 0$ .
$\Rightarrow 2∣ a ∣ \geq 2 \Rightarrow ∣ a ∣ \geq 1$
Thus, least value of $∣ a ∣$ is 1 .

Q. If tanθ and cotθ are roots of x2+2ax+b=0, then least value of ∣a∣ is

21​

1

2

cannot be found.

tanθ+cotθ=2a and tanθcotθ=b Now, 2a=tanθ+cotθ≥2 if tanθ>0 and 2a=tanθ+cotθ≤−2 if tanθ<0. ⇒2∣a∣≥2⇒∣a∣≥1 Thus, least value of ∣a∣ is 1 .

Q. If $tan θ$ and $cot θ$ are roots of $x^{2} + 2 a x + b = 0$ , then least value of $∣ a ∣$ is

$\frac{1}{2}$

$tan θ + cot θ = 2 a$ and $tan θ cot θ = b$
Now, $2 a = tan θ + cot θ \geq 2$ if $tan θ > 0$
and $2 a = tan θ + cot θ \leq - 2$ if $tan θ < 0$ .
$\Rightarrow 2∣ a ∣ \geq 2 \Rightarrow ∣ a ∣ \geq 1$
Thus, least value of $∣ a ∣$ is 1 .