If lim x → 0 [ 1 + x log (1 + b2) ](1/x) = 2 b sin2 θ, b 0 and θ ∈ ( - π , π), then the value of θ is

Question

If lim x → 0 [ 1 + x log (1 + b2) ](1/x) = 2 b sin2 θ, b > 0 and θ ∈ ( - π , π), then the value of θ is (A)  ± (π/4) (B)  ± (π/3) (C)  ± (π/6) (D)  ± (π/2)

Tardigrade Admin · Accepted Answer

Here, lim x → 0 1 + x log (1 + b2) 1/x = lim x → 0 x log (1 + b2) . (1/x) = elog (1 + b2) = (1 + b2) Given, lim x → 0 1 + x log (1 + b2 ) 1/x = 2b sin2 θ ⇒ (1 + b2) = 2b sin2 θ ∴ sin2 θ = (1 + b2/2b) By AM ge GM, ( b + (1/b)/ 2) ge (b. (1/b))1/2 ⇒ (b2 + 1/2b) ge 1 From Eqs. (ii) and (iii), sin2 θ = 1 ⇒ = ± (π/2), as θ ∈ (- π, π]

Q. If $l i m_{x \to 0} [1 + x l o g (1 + b^{2})]^{\frac{1}{x}} = 2 b s i n^{2} θ, b > 0 an d θ \in (- π, π),$ then the value of $θ$ is

$\pm \frac{π}{4}$

$\pm \frac{π}{3}$

$\pm \frac{π}{6}$

$\pm \frac{π}{2}$

Q. If limx→0​[1+xlog(1+b2)]x1​=2bsin2θ,b>0andθ∈(−π,π), then the value of θ is

±4π​

±3π​

±6π​

±2π​

Here, limx→0​{1+xlog(1+b2)}1/x = limx→0​{xlog(1+b2)}.x1​ = elog(1+b2)=(1+b2) Given, limx→0​{1+xlog(1+b2)}1/x=2bsin2θ ⇒(1+b2)=2bsin2θ ∴ sin2θ=2b1+b2​ By AM ≥ GM, 2b+b1​​≥(b.b1​)1/2 ⇒2bb2+1​≥1 From Eqs. (ii) and (iii), sin2θ=1⇒=±2π​,asθ∈(−π,π]

Q. If $l i m_{x \to 0} [1 + x l o g (1 + b^{2})]^{\frac{1}{x}} = 2 b s i n^{2} θ, b > 0 an d θ \in (- π, π),$ then the value of $θ$ is

$\pm \frac{π}{4}$

$\pm \frac{π}{3}$

$\pm \frac{π}{6}$

$\pm \frac{π}{2}$