Does Shiksha provide solutions for graph-based questions in trigonometry in NCERT solutions?

0 4 Views | Posted 5 months ago
Asked by Piyush Vimal

  • 1 Answer

  • E

    Answered by

    Esha Garg

    5 months ago

    Yes, Shiksha provides clear and detailed solutions for graph-based questions in trigonometry as part of the NCERT Class 11 Math solutions. We have provided clear instructions, how to sketch and interpret the graphs of standard trigonometric functions like sin (x), cos (x), tan (x), and their transformations in our NCERT Solutions. Students can use this step-by-step approach used in the solutions explains key features such as amplitude, period, phase shift, and symmetry, making it easier for students to visualize and solve related questions. These graphical solutions are not only help score in CBSE exam preparation but also strengthen t

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Similar Questions for you

A
alok kumar singh

π 2 π 2 ( x 2 c o s x 1 + π 2 + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) d x = π 4 ( π + α ) 2

0 π 2 { ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) + ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) } d x

= π 4 ( π + α ) 2

0 π 2 ( x 2 c o s x + 1 + s i n 2 x ) d x = π 4 ( π + α ) 2

0 π 2 x 2 c o s x d x + 0 π 2 ( 1 + s i n 2 x ) d x = π 4 ( π + α ) 2 ....(1)

Let I 1 = 0 π 2 ( 1 + s i n 2 x ) d x

I 1 = 0 π 2 1 d x + 0 π 2 ( 1 c o s 2 x 2 ) d x

I 1 = π 2 + 1 2 [ π 2 + 0 ]

I 1 = 3 π 4

Let I 2 = 0 π 2 x 2 c o s x d x

I 2 = [ x 2 ( s i n x ) 2 x c o s x d x ] 0 π 2

I 2 = [ x 2 ( s i n x ) 2 x s i n x ] 0 π 2

I 2 = [ x 2 s i n x 2 ( x ( c o s x ) + c o s x ) ] 0 π 2

I 2 = [ x 2 s i n x 2 ( x c o s x + s i n x ) ] 0 π 2

I 2 = ( π 2 4 2 )

Put l1 and l2 in (1)

π 2 4 2 + 3 π 4

π 2 4 + 3 π 4 2

π 4 ( π + 3 ) 2

α = 3

A
alok kumar singh

P (2 obtained on even numbered toss) = k (let)

P (2) = 1 6  

P (  2 ¯ )= 5 6  

k = 5 6 × 1 6 + ( 5 6 ) 3 × 1 6 + ( 5 6 ) 5 × 1 6 + . . .

= 5 6 × 1 6 1 ( 5 6 ) 2

= 5 1 1

A
alok kumar singh

I = 9 0 9 [ 1 0 x x + 1 ] d x

= 9 [ 0 1 / 9 0 d x + 1 / 9 2 / 3 d x + 2 / 3 9 2 d x ]

= 9 [ 2 3 1 9 + 2 [ 9 2 3 ] ]

= 9 [ 5 9 + 2 × 2 5 3 ]

= 5 + 6 × 25

= 5 + 150

= 155

A
alok kumar singh

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  = 3 0 1 1 * 1 1 = 3 0 1 2 1

A
alok kumar singh

Given | a | = 1 , | b | = 4 , a b = 2

c = 2 ( a × b ) 3 b  

Dot product with  a on both sides

c a = 6 ... (1)

Dot product with  b  on both sides

b c = 4 8 ... (2)

c c = 4 | a × b | 2 + 9 | b | 2

| c | 2 = 4 [ | a | 2 | b | 2 ( a b ) 2 ] + 9 | b | 2

| c | 2 = 4 [ ( 1 ) ( 4 ) 2 ( 4 ) ] + 9 ( 1 6 )

| c | 2 = 4 [ 1 2 ] + 1 4 4

| c | 2 = 4 8 + 1 4 4

| c | 2 = 1 9 2

c o s θ = b c | b | | c |

c o s θ = 4 8 1 9 2 4

c o s θ = 4 8 8 3 4

c o s θ = 3 2 3

c o s θ = 3 2 θ = c o s 1 ( 3 2 )

 

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