If f (θ)= min .(|2 x -7|+| x -4|+| x -2- sin θ|), where x , θ ∈ R, then maximum value of f (θ) is

Question

If f (θ)= min .(|2 x -7|+| x -4|+| x -2- sin θ|), where x , θ ∈ R, then maximum value of f (θ) is (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

f (θ)= min .(|2 x -7|+| x -4|+| x -2- sin θ|) Let g(x)=|2 x-7|+|x-4|+|x-2- sin θ| .g(x)| min =g((7/2))=(1/2)+(3/2)- sin θ=2- sin θ=f(θ) .∴ f (θ)| max =3

Q. If $f (θ) = min . (∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣)$ , where $x, θ \in R$ , then maximum value of $f (θ)$ is

2

3

4

5

$f (θ) = min . (∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣)$
Let $g (x) = ∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣$
$g (x) ∣_{m i n} = g (\frac{7}{2}) = \frac{1}{2} + \frac{3}{2} - sin θ = 2 - sin θ = f (θ)$
$∴ f (θ) ∣_{m a x} = 3$

Q. If f(θ)=min.(∣2x−7∣+∣x−4∣+∣x−2−sinθ∣), where x,θ∈R, then maximum value of f(θ) is

2

3

4

5

f(θ)=min.(∣2x−7∣+∣x−4∣+∣x−2−sinθ∣) Let g(x)=∣2x−7∣+∣x−4∣+∣x−2−sinθ∣ g(x)∣min​=g(27​)=21​+23​−sinθ=2−sinθ=f(θ) ∴f(θ)∣max​=3

Q. If $f (θ) = min . (∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣)$ , where $x, θ \in R$ , then maximum value of $f (θ)$ is

$f (θ) = min . (∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣)$
Let $g (x) = ∣2 x - 7∣ + ∣ x - 4∣ + ∣ x - 2 - sin θ ∣$
$g (x) ∣_{m i n} = g (\frac{7}{2}) = \frac{1}{2} + \frac{3}{2} - sin θ = 2 - sin θ = f (θ)$
$∴ f (θ) ∣_{m a x} = 3$