Q) An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

1. The shape of the poses shown is
a) Spiral              b) Ellipse              c) Linear                   d) Parabola

2. The graph of parabola opens downwards, if
a) a > = 0             b) a = 0                 c) a < 0                      d) a > 0

3. In the graph, how many zeroes are there for the polynomial?
a) 0                      b) 1                       c) 2                            d) 3

4. The two zeroes in the above shown graph are
a) 2, 4                  b) – 2, 4                 c) – 8, 4                     d) 2, – 8

5. The zeroes of the quadratic polynomial 4 √3 x² + 5 x – 2 √3 are
a)                b) –
c)         d) –  

Ans:

STEP BY STEP SOLUTION

1. Shapes of Asanas:

Let’s understand the given shapes by this diagram:

Here we can see that all the asanas postures shown above are parabolic and in downward direction.

Therefore, option d) is correct

2. Nature of a for downward parabola:  

We know that in a standard quadratic polynomial expression, a x² + b x + c, the value of function will increase exponentially due to 1st term x² . For every value of X, whether positive or negative, its value will be same. It is value of a which decides the direction of parabola.

Let’s understand this, by an example:

Case 1: Let’s take the equation as f(x) = x² + 1  (here a = 1, b = 0, c =1)

To plot it, we consider equation of graph as f(x) = x² + 1 and plot values of f(x) on y axis.

for x = 0,  f(x) = (0)² + 1 = 1 

for x = 1, f(x) = (1)² + 1 = 2

for x = 2,  f(x) = (2)² + 1 = 5

for x = – 1, f(x) = (- 1)² + 1 = 2

for x = – 2, f(x) = (- 2)² + 1 = 5

Case 2: Let’s take the equation as f(x) = – x² + 1  (here a = – 1, b = 0, c =1)

To plot it, we consider equation of graph as f(x) = – x² + 1 and plot values of f(x) on y axis.

for x = 0,  f(x) = – (0)² + 1 = 1 

for x = 1, f(x) = – (1)² + 1 = 0

for x = 2,  f(x) = – (2)² + 1 = – 3

for x = – 1, f(x) = – (- 1)² + 1 = 0

for x = – 2, f(x) = – (- 2)² + 1 = – 3

Conclusion: For positive values of a, parabola is upward and for negative values of a, parabola is downward.

Therefore, option c) is correct

3. Number of zeroes:

In the given diagram (shown in right side), it is a graph for upward parabola.

We know that in a standard quadratic polynomial expression, number of zeroes is equal to the number of times the graph intersects x-axis.

In the given diagram, the graph is intersecting X – axis 2 times, hence its number of roots are 2.

Therefore, option c) is correct

4. Values of Zeroes:

Since, Zeroes are values x when the graph intersects X – axis.

Here, in the given diagram, the graph intersects the X-axis at x = – 2 and x = 4.

Hence, its roots are – 2 and 4.

Therefore, option b) is correct

5. Zeroes of the Quadratic Polynomial: 

The given Quadratic Polynomial equation is: f(x) = 4 √3 x² + 5 x – 2 √3

To find zeroes of the polynomial, let’ find the values of x.

f(x) = 4 √3 x² + 5 x – 2 √3

= 4 √3 x² + 8 x – 3 x – 2 √3

= 4 x(√3 x + 2) – √3 (√3 x + 2)

= (√3 x + 2) (4 x – √3)

When the polynomial equation’s graph will intersect the x-axis, its value will be zero. Hence, for those values of x, f(x) = 0.

Let’s solve both factors for values of x, when f(x) = 0

By putting (√3 x + 2) = 0 , we get x = –

and by (4 x – √3) = 0, we get x =

Therefore, for these 2 values of x, polynomial’s graph will intersect X-axis and these values of x are its zeroes.

Therefore, option b) is correct

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