If the maximum and the minimum values of 1-sin-4-2 cos-4- for all real values of - are - and - respectively- then - is

The correct option is C 6sin(π4+θ)= cos( π2 (π4+θ))=cos(π4 θ) nbsp; ∴ 1+sin(π4+θ)+2cos(π4 θ) =1+3cos(π4 θ) 3≤ 3cos(π4 θ)≤ 3 ⇒ 2≤ 1+3cos(π4 θ)≤4 ∴λ=4 and μ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Range of Trigonometric Expressions

If the maximu...

Question

If the maximum and the minimum values of $1 + sin (\frac{π}{4} + θ) + 2 cos (\frac{π}{4} - θ)$ for all real values of $θ$ are $λ$ and $μ$ respectively, then $λ - μ$ is

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $6$
$sin (\frac{π}{4} + θ) = cos (\frac{π}{2} - (\frac{π}{4} + θ)) = cos (\frac{π}{4} - θ)$
$∴ 1 + sin (\frac{π}{4} + θ) + 2 cos (\frac{π}{4} - θ)$
$= 1 + 3 cos (\frac{π}{4} - θ)$

$- 3 \leq 3 cos (\frac{π}{4} - θ) \leq 3$
$\Rightarrow - 2 \leq 1 + 3 cos (\frac{π}{4} - θ) \leq 4$
$∴ λ = 4$ and $μ = - 2$
$\Rightarrow λ - μ = 6$

Suggest Corrections

Similar questions

Q. If tangent of any of family of hyperbola

x y = 4 s i n^{2} θ, θ \in (0, 2 π) - {π}

is not a normal of family of circles

x^{2} + y^{2} - 2 x - 2 y + μ = 0

, where

μ

is any real parameter, then set of all values of

θ

is:

Q. Find the maximum value of

1 + sin (\frac{π}{4} + θ) + 2 sin (\frac{π}{4} - θ)

for all real values of

θ

Q. If the points

(0, 1. - 2), (3, λ, - 1)

and

(μ, - 3, - 4)

are collinear, then the value of

λ

and

μ

are given by

Q. The maximum value of

1 + sin (\frac{π}{4} + θ) + 2 cos (\frac{π}{4} - θ)

for real value of

θ

Q. If

c o s^{2} x = c o s^{4} x

then find the minimum and maximum value of

{sin}^{2} x + {cos}^{4} x,

for all real values of

θ

Join BYJU'S Learning Program

If the maximum and the minimum values of 1+sin(π4+θ)+2cos(π4−θ) for all real values of θ are λ and μ respectively, then λ−μ is

The correct option is C 6sin(π4+θ)=cos(π2−(π4+θ))=cos(π4−θ) ∴1+sin(π4+θ)+2cos(π4−θ) =1+3cos(π4−θ) −3≤3cos(π4−θ)≤3 ⇒−2≤1+3cos(π4−θ)≤4 ∴λ=4 and μ=−2 ⇒λ−μ=6

If the maximum and the minimum values of $1 + sin (\frac{π}{4} + θ) + 2 cos (\frac{π}{4} - θ)$ for all real values of $θ$ are $λ$ and $μ$ respectively, then $λ - μ$ is