The correct option is A 2 1≤cos2x ≤1 ⇒ 10≤10cos2x≤10 8 10 ≤8+10cos2x ≤10+8 2≤y ≤18 ∴ (3) is correct ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Condition for Monotonically Increasing Function

If y=8+10co...

Question

If $y = 8 + 10 cos 2 x$ , then

0 < y < 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1 < y < 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2 < y < 18

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 4 < y < 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 2 < y < 18$
$- 1 \leq cos 2 x \leq 1$
$\Rightarrow$ $- 10 \leq 10 cos 2 x \leq 10$
$8 - 10 \leq 8 + 10 cos 2 x \leq 10 + 8$
$- 2 \leq y \leq 18$
$∴$ (3) is correct

Suggest Corrections

Similar questions

Factorise: $1 - 16 (x - 2 y)^{2}$ .

Q. If

(y^{2} + 5 y) (y^{2} + 5 y - 2) - 24 = 0

, then

y = 1, - 6, - 1, - 4

$I f 2 x^{2} - 3 x y + y^{2} + x + 2 y - 8 = 0 t h e n \frac{d y}{d x} =$

Find the vertex,focus,axis,directrix and latus-rectum of the following parabolas

$\\(i)~y^2=8x\\(ii)~4x^2+y=0\\(iii)~y^2-4y-3x+1=0\\(iv)~y^2-4y+4x=0\\(v)~~y^2+4x+4y-3=0\\(vi)~y^2=8x+8y\\(vii)~4(y-1)^2=-7(x-3)\\(viii)~y^2=5x-4y-9\\(ix)~x^2+y=6x-14$

If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $x y = c^{2}$ in four points $P (x_{1}, y_{1}), Q (x_{2}, y_{2}) R (x_{3}, y_{3}), S (x_{4}, y_{4})$ , then which of the following need not hold?

Join BYJU'S Learning Program

If y=8+10cos2x, then

The correct option is A −2<y<18−1≤cos2x≤1⇒ −10≤10cos2x≤108−10≤8+10cos2x≤10+8−2≤y≤18∴ (3) is correct

If $y = 8 + 10 cos 2 x$ , then

The correct option is A $- 2 < y < 18$
$- 1 \leq cos 2 x \leq 1$
$\Rightarrow$ $- 10 \leq 10 cos 2 x \leq 10$
$8 - 10 \leq 8 + 10 cos 2 x \leq 10 + 8$
$- 2 \leq y \leq 18$
$∴$ (3) is correct