Question

The least value of 2sin²Φ + 3cos²Φ
is

(a)1 (b)2 (c)3 (d)5


Solve it ...............​

Answer 1

Hi there,

sin²Ф = 1 - cos²Ф

=> 2(1 - cos²Ф) + 3 cos²Ф

i.e. 2 + cos²Ф.......(1)

cos 2Ф= cos²x - sin²x

=> cos 2Ф = 2 cos²Ф - 1

so, cos²Ф = (cos 2Ф + 1)/2

From (1) => 5/2 + (1/2)cos 2Ф

Now, cos 2Ф minimum is -1

minimum 5/2 - 1/2 = 2

Ans is B

:)

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Answer 2

αɳรωεɾ

2 (Option b ) is the correct option.

รσℓµƭเσɳ

Firstly we should know that :-

sin²x + cos²x = 1

Range of cos x is

$- 1 \leqslant \cos(x) \leqslant 1$

So , the range of cos²x will be :-

$0 \leqslant \cos^2(x) \leqslant 1$

→ 2 sin²x +3 cos²x

→ 2 sin²x +2 cos²x + cos²x

→ 2( sin²x +cos²x ) + cos²x

→ 2 (1) +cos²x

→ 2 + cos²x

Now we know that 2 is constant but here we can replace cos²x by its minimum value and that is 0 ,So

→ 2 + 0

$\huge\mathfrak{ answer\:=\:2}$