Question

1. Find the maximum and minimum value of f(x) = 2(1 + sin^2x)

2. Find the minimum and maximum value of

|sin2x| - 3​

Answer 1

Answer:

Using method of completing the squares:

f(x)=sin2(x)−sin(x)−2

=sin2(x)−2(12)sin(x)+14−14–2

=(sin(x)−12)2−94

Now,

−1≤sin(x)≤1

−1–12≤sin(x)−12≤1–12 (Subtracting 12 throughout)

Taking higher of the two bounds

0≤(sin(x)−12)2≤(−32)2

Therefore,

0–94≤(sin(x)−12)2−94≤94–94

$−94≤f(x)≤0$

fmax=0

$fmin=−9/4$

Answer 2

$\large\underline{\sf{Solution-1}}$

Given function is

$\rm \: f(x) = 2(1 + {sin}^{2}x)$

We know,

$\rm \: - 1 \leqslant sinx \leqslant 1$

So,

$\rm\implies \:0 \leqslant {sin}^{2}x \leqslant 1$

On adding 1 in each term we get

$\rm \: \:0 + 1 \leqslant {sin}^{2}x + 1 \leqslant 1 + 1$

$\rm \: \:1 \leqslant {sin}^{2}x + 1 \leqslant 2$

On multiply by 2, each term, we get

$\rm \: \:2 \leqslant 2({sin}^{2}x + 1) \leqslant 4$

$\rm\implies \:2 \leqslant f(x) \leqslant 4$

So, It implies

Minimum value of f(x) = 2

Maximum value of f(x) = 4

$\green{\large\underline{\sf{Solution-2}}}$

Given function is

$\rm \: f(x) = |sin2x| - 3$

We know,

$\rm \: - 1 \leqslant sin2x \leqslant 1$

So,

$\rm\implies \:0 \leqslant |sin2x| \leqslant 1$

On Subtracting 3 from each term, we get

$\rm \: 0 - 3 \leqslant |sin2x| - 3 \leqslant 1 - 3$

$\rm \: - 3 \leqslant |sin2x| - 3 \leqslant - 2$

$\rm\implies \: - 3 \leqslant f(x) \leqslant - 2$

So, it implies

Minimum value of f(x) = - 3

Maximum value of f(x) = - 2