Value of the expression 2sinx – cos2x is

Q: Value of the expression 2sinx – cos2x is

(A) greater than -3/2

(B) equal to – 3/2

(C) less than -3/2

(D) none of these

Sol. 2 sinx – cos2x = 2 (sin2x + sinx) – 1

$\large = 2 [(sinx + \frac{1}{2})^2 – \frac{1}{4}] – 1 $

$\large = 2(sinx + \frac{1}{2})^2 – \frac{3}{2} \ge -\frac{3}{2}$

⇒ 2 sinx – cos2x ≥ – 3/2

Hence (A), (B) are the correct answers.

If Tn = sin^nθ + cos^nθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

Q: If Tn = sinnθ + cosnθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

(A) $\large m \in [-1 , \frac{1}{3}]$

(B) $\large m \in [0 , \frac{1}{3}]$

(C) $\large m \in [-1 , 0]$

(D) $\large m \in [-1 , -\frac{1}{2}]$

Sol: $\large \frac{T_6 – T_4}{T_6} = \frac{-sin^2 2\theta}{4-3sin^2 2\theta} = m $

$\large sin^2 2\theta = \frac{4 m}{-1 + 3 m}$

$\large 0 \le \frac{4 m}{-1 + 3 m} \le 1 $

$\large m \in [-1 , 0]$

Hence (C), (D) are the correct answers

If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

Q: If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

(A) 2 tanβ

(B) 2 tanα

(C) tanα + tanβ

(D) none of these

Sol. $\large \frac{sin(2\alpha + \beta)}{sin\beta} = \frac{3}{1}$

$\large \frac{sin(2\alpha + \beta) + sin\beta}{sin(2\alpha + \beta) – sin\beta} = \frac{3+1}{3-1} = 2 $

$\large \frac{2 sin(\alpha + \beta) cos\alpha}{2sin\alpha cos(\alpha + \beta)} = 2$

⇒ tan(α + β) = 2 tanα

Hence (B) is the correct answer.

If tanθ = √n , for some non-square natural number n , then sec2θ is

Q: If tanθ = √n , for some non-square natural number n , then sec2θ is

(A) a rational number

(B) an irrational number

(C) a positive number

(D) none of these

Sol. $\large sec2\theta = \frac{1 + tan^2\theta}{1- tan^2 \theta}= \frac{1+n}{1-n} $

where n is a non-square natural number so 1 – n ≠ 0.

⇒ sec2θ is a rational number.

Hence (A) is the correct answer.

If in a ΔABC , ∠B = π/3, then the maximum value of sinA sinC is

Q: If in a ΔABC , ∠B = π/3, then the maximum value of sinA sinC is

(A) 1/2

(B) 3/4

(C) 2/3

(D) None of these

Sol: $\large A + C = \frac{2 \pi}{3} $

$\large C = \frac{2 \pi}{3} – A $

Now , $\large sin A sin C = sin A . sin(\frac{2 \pi}{3} – A) $

$\large = \frac{\sqrt{3}}{4} sin2A + \frac{1}{2}sin^2 A $

$\large = \frac{\sqrt{3}}{4} sin2A + \frac{1}{4}(1-cos2A) $

$\large = \frac{\sqrt{3}}{4} sin2A – \frac{1}{4}cos2A + \frac{1}{4}$

$\large – \frac{1}{2} \le sinA sinC – \frac{1}{4} \le \frac{1}{2} $

$\large – \frac{1}{4} \le sinA sinC \le \frac{3}{4} $

Hence (B) is the correct answer.