Match the items of List-I to the items of List-II List- I List-II A)The period of ² x is I) 2π/3 B)Maximum value of π/3({3} 3x+ 3x) II) 12π C) The period of /3+/2 is III) π/2 D) Intersection points of y=| x| and y=1 in (0,π) IV) 3π/2 V) π The correct match is

Correct Answer is : ABCDVIIIIII{array}{cccc} {A} & {B} & {C} & {D} \\ {V} & {I} & {II} & {III} {array}AVBICIIDIII

{array}{cccc} {A} & {B} & {C} & {D} \\ {V} & {I} & {II} & {III} {array}

{array}{cccc} {A} & {B} & {C} & {D} \\ {IV} & {I} & {II} & {III} {array}

{array}{cccc} {A} & {B} & {C} & {D} \\ {III} & {I} & {IV} & {V} {array}

{array}{cccc} {A} & {B} & {C} & {D} \\ {IV} & {III} & {II} & {V} {array}

Correct Answer is : ABCDVIIIIII{array}{cccc} {A} & {B} & {C} & {D} \\ {V} & {I} & {II} & {III} {array}AVBICIIDIII

Test Index

TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 21 of 160

Marks: +1, -0

List- I	List-II
A)The period of $\sin^{2} x$ is	I) $\frac{2\pi}{3}$
B)Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x+\sin 3x)$	II) $12\pi$
C) The period of $\sin\frac{x}{3}+\cos\frac{x}{2}$ is	III) $\frac{\pi}{2}$
D) Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0,\pi)$	IV) $\frac{3\pi}{2}$
	V) $\pi$

The correct match is

$\begin{array}{cccc} \text{A} & \text{B} & \text{C} & \text{D} \\ \text{V} & \text{I} & \text{II} & \text{III} \end{array}$
$\begin{array}{cccc} \text{A} & \text{B} & \text{C} & \text{D} \\ \text{IV} & \text{I} & \text{II} & \text{III} \end{array}$
$\begin{array}{cccc} \text{A} & \text{B} & \text{C} & \text{D} \\ \text{III} & \text{I} & \text{IV} & \text{V} \end{array}$
$\begin{array}{cccc} \text{A} & \text{B} & \text{C} & \text{D} \\ \text{IV} & \text{III} & \text{II} & \text{V} \end{array}$

Solution:

A. Period of

\sin^{2} x

\pi

f(x)=\sin^{2} x

f(\pi+x)=\sin^{2}(\pi+x)=(-\sin x)^{2}=\sin^{2}x=f(x)

\;\frac{\pi}{3}(\sqrt{3}\cos 3x+\sin 3x)

Maximum value

=\;\frac{\pi}{3}\sqrt{(\sqrt{3})^{2}+(1)^{2}}=\;\frac{\pi}{3}\times 2=\;\frac{2\pi}{3}

C. We have,

\sin\;\frac{x}{3}+\cos\;\frac{x}{2}

Period of

\sin x

6\pi

and period of

\cos\;\frac{x}{2}

4\pi

\therefore

Period of

\sin\;\frac{x}{3}+\cos\;\frac{x}{2}

is LCM of

6\pi

and

4\pi

i.e.,

12\pi

D. We have,

y=|\sin x|

and

y=1 \Rightarrow 1=|\sin x|

\Rightarrow \;\; \sin x=\pm 1 \Rightarrow x=\;\frac{\pi}{2}

[0,\pi]

\therefore

Intersection part is

\;\frac{\pi}{2}

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