Mathematics · 2019
JEE Main · 12 January 2019, Shift 1 · Q88
The maximum value of 3 cos θ+5 sin (θ-(π)/6) for any real value of θ is:
The maximum value of $\displaystyle 3 \cos \theta+5 \sin \left(\theta-\frac{\pi}{6}\right)$ for any real value of $\displaystyle \theta$ is :
Official answer
(2)
$\displaystyle \sqrt{19}$
More from Trigonometric Identities
- Let (π)/2<θ<π and cot θ=-1/(2 √2). Then the value of sin ((15 θ)/2)( cos 8 θ+ sin 8 θ)+ cos ((15 θ)/2)( cos 8 θ- sin 8 θ) is equal to2026
- If (cos^2 48^°- sin^2 12^°)/(sin^2 24^°- sin^2 6^°)=(α+β √5)/2, where α, β ∈ N, then α+β is equal to ____2026
- If L= sin^2((π)/16)- sin^2((π)/8) and M= cos^2((π)/16)- sin^2((π)/8), then:2020
- Let tan A, tan B, where A, B ∈(-(π)/2, (π)/2), be the roots of the quadratic equation x^2-2 x-5=0. Then 20 sin^2((A+B)/2) is equal to:2026
- If the line x=y=z intersects the line x sin A+y sin B+z sin C-18=0=x sin 2 A+y sin 2 B+z sin 2 C-9, where A, B, C are the angles of a triangle A B C,…2023
- For the functions f(θ)=α tan^2 θ+β cot^2 θ, and g(θ)=α sin^2 θ+β cos^2 θ, α>β>0, let min_0<θ<(π)/2 f(θ)= max_0<θ<π g(θ). If the first term of a G.P.…2026
- Let cos (α+β)=-1/10 and sin (α-β)=3/8, where 0<α<(π)/3 and 0<β<(π)/4. If tan 2 α=(3(1-r √5))/(√11(s+√5)), r, s ∈ N, then r+s is equal to ____.2026
- If tan A is equal to the integral solution of the inequality 4 x^2-24 x+35<0 and sin B is equal to the slope of the angle bisector of the second…2021
JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.