The function f(x)=esin x + cos ⁡ x ∀ x∈ [0 , 2 π ] attains local extrema at x=α and x=β , then α +β is equal to

Question

The function f(x)=esin x + cos ⁡ x ∀ x∈ [0 , 2 π ] attains local extrema at x=α and x=β , then α +β is equal to (A) π  (B) 2π  (C) (3 π /2) (D) (π /2)

Tardigrade Admin · Accepted Answer

f' (x) = esin x + cos x ⋅ (cos x - sin x) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-paa010nwpado.png" /> ∴ f(x) attains local extrema at x=(π /4),(5 π /4) i.e. α +β =(6 π /4)=(3 π /2)

Q. The function $f (x) = e^{s in x + cos x}$ $\forall x \in [0, 2 π]$ attains local extrema at $x = α$ and $x = β,$ then $α + β$ is equal to

$π$

$2 π$

$\frac{3 π}{2}$

$\frac{π}{2}$

$f^{^{'}} (x) = e^{s in x + cos x} \cdot (cos x - s in x)$

$∴ f (x)$ attains local extrema at $x = \frac{π}{4}, \frac{5 π}{4}$
i.e. $α + β = \frac{6 π}{4} = \frac{3 π}{2}$

Q. The function f(x)=esinx+cos⁡x ∀x∈[0,2π] attains local extrema at x=α and x=β, then α+β is equal to

π

2π

23π​

2π​

f′(x)=esinx+cosx⋅(cosx−sinx) ∴f(x) attains local extrema at x=4π​,45π​ i.e. α+β=46π​=23π​

Q. The function $f (x) = e^{s in x + cos x}$ $\forall x \in [0, 2 π]$ attains local extrema at $x = α$ and $x = β,$ then $α + β$ is equal to

$π$

$2 π$

$\frac{3 π}{2}$

$\frac{π}{2}$

$f^{^{'}} (x) = e^{s in x + cos x} \cdot (cos x - s in x)$

$∴ f (x)$ attains local extrema at $x = \frac{π}{4}, \frac{5 π}{4}$
i.e. $α + β = \frac{6 π}{4} = \frac{3 π}{2}$