The largest integral value of p for which graph of the function f(x)=2 p2-3 p tan x+ tan 2 x+1 does not lie below x-axis for all x ∈((-π/2), (π/2)) is

Question

The largest integral value of p for which graph of the function f(x)=2 p2-3 p tan x+ tan 2 x+1 does not lie below x-axis for all x ∈((-π/2), (π/2)) is (A) 0 (B) 2 (C) 3 (D) 8

Tardigrade Admin · Accepted Answer

text Put tan x=t y=t2-3 p t+2 p2+1, t ∈ R D ≤ 0 9 p2-4(2 p2+1) ≤ 0 ⇒ p2-4 ≤ 0 ⇒ p ∈[-2,2] Hence largest integral value is 2

Q. The largest integral value of $p$ for which graph of the function $f (x) = 2 p^{2} - 3 p tan x + tan^{2} x + 1$ does not lie below $x$ -axis for all $x \in (\frac{- π}{2}, \frac{π}{2})$ is

0

2

3

8

$Put tan x = t$
$y = t^{2} - 3 pt + 2 p^{2} + 1, t \in R$
$D \leq 0$
$9 p^{2} - 4 (2 p^{2} + 1) \leq 0 \Rightarrow p^{2} - 4 \leq 0 \Rightarrow p \in [- 2, 2]$
Hence largest integral value is 2

Q. The largest integral value of p for which graph of the function f(x)=2p2−3ptanx+tan2x+1 does not lie below x-axis for all x∈(2−π​,2π​) is

0

2

3

8

Put tanx=t y=t2−3pt+2p2+1,t∈R D≤0 9p2−4(2p2+1)≤0⇒p2−4≤0⇒p∈[−2,2] Hence largest integral value is 2

Q. The largest integral value of $p$ for which graph of the function $f (x) = 2 p^{2} - 3 p tan x + tan^{2} x + 1$ does not lie below $x$ -axis for all $x \in (\frac{- π}{2}, \frac{π}{2})$ is

$Put tan x = t$
$y = t^{2} - 3 pt + 2 p^{2} + 1, t \in R$
$D \leq 0$
$9 p^{2} - 4 (2 p^{2} + 1) \leq 0 \Rightarrow p^{2} - 4 \leq 0 \Rightarrow p \in [- 2, 2]$
Hence largest integral value is 2