Case study (4 Marks)

Question 14 Marks

A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h=25 t-5 t^2$.

(i) Write zeroes of the given polynomial.
(ii) Find the maximum height achieved by ball.
(iii) (a) After throwing upward, how much time did the ball take to reach to the height of 30 m ?
$OR$
(b) Find the two different values of t when the height of the ball was 20 m .

Answer

(i) The zeroes of polynomial $h(t)=25 t-5 t^2$ are given by
$
h(t)=0 \Rightarrow 25 t-5 t^2=0 \Rightarrow 5 t(5-t)=0 \Rightarrow t=0,5
$
(ii) The maximum height achieved by ball $h\left(\frac{5}{2}\right)=25 \times \frac{5}{2}-5 \times\left(\frac{5}{2}\right)^2=\frac{125}{2}-\frac{125}{4}=\frac{125}{4}=31.25 m$
(iii) (a): The time taken by the ball to reach to the height of 30 m is given by
$
h(t)=30 \Rightarrow 25 t-5 t^2=30 \Rightarrow 5 t-t^2=6 \Rightarrow t^2-5 t+6=0 \Rightarrow(t-2)(t-3)=0 \Rightarrow t=2,3
$
So, the time taken by the ball to reach to the height of 30 m is 2 seconds.
OR
(b): When the height of the ball is 20 m
$
\text { i.e. } \quad h(t)=20 \Rightarrow 25 t-5 t^2=20 \Rightarrow 5 t-t^2=4 \Rightarrow t^2-5 t+4=0 \Rightarrow(t-1)(t-4)=0 \Rightarrow t=1,4$
Hence, the height of the ball was 20 m at $t=1$ second and at $t=4$ seconds.

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Question 24 Marks

In a pool at an aquarium, a dolphin jumps out of the water travelling at 20 cm per second. Its height above water level after $t$ seconds is given by $h=20 t-16 t^2$.

(i) Find zeroes of polynomial $p(t)=20 t-16 t^2$.
(ii) Which of the following type of graph represents $p(t)$ ?

(iii) What would be the value of $h$ at $t=\frac{3}{2}$ ? Interpret the result.
(iv) How much distance has the dolphin covered before hitting the water level again?

Answer

(i) We have,
$
p(t)=20 t-16 t^2=-16 t\left(t-\frac{5}{4}\right)=-16(t-0)\left(t-\frac{5}{4}\right)
$
The zeroes of $p(t)$ are given by
$
p(t)=0 \Rightarrow-16(t-0)\left(t-\frac{5}{4}\right)=0 \Rightarrow(t-0)\left(t-\frac{5}{4}\right)=0 \Rightarrow t=0, \frac{5}{4}
$
Hence, the zeroes of $p(t)$ are 0 and $\frac{5}{4}$.
(ii) In $p(t)$, we find that the coefficient of $t^2$ is negative integer -16 . So, graph of $h=p(t)$ is a parabola opening downward which crosses time axis at $t=0$ and $t=\frac{5}{4}$. So, the graph in option (a) is the graph of $p(t)$.
(iii) The value of $h$ at $t=\frac{3}{2}$ is given by $h=p\left(\frac{3}{2}\right)=20 \times \frac{3}{2}-16 \times \frac{9}{4}=-6$. This means that the dolphin will be inside the water at a depth of 6 cm from water level.
(iv) The dolphin hits the water level again at $t=\frac{5}{4}$ seconds. The distance travelled by the dolphin before hitting the water level again is twice the ordinate of the vertex of the parabola $h=20 t-16 t^2$.
The abscissae of the vertex is $t=\frac{5}{8}$ and the ordinate at $t=\frac{5}{8}$ is given by
$
h=20 \times \frac{5}{8}-16 \times\left(\frac{5}{8}\right)^2=\frac{25}{4} cm
$
Hence, distance covered by the dolphin $=2 \times \frac{25}{4} cm=12.5 cm$.

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Question 34 Marks

Polynomials are everywhere. They play a key role in the study of algebra, in analysis and on the whole many mathematical problems involving them. Since polynomials are used to describe curves of various types, engineers use polynomials to graph the curves of roller coasters.

Based on the given information answer the following questions:
(i) If the Roller Coaster is represented by the graph $y=p(x)$, shown in Fig then the type of the polynomial it traces, is
(a) linear $\qquad$ (b) quadratic $\qquad$ (c) cubic $\qquad$ (d) biquadratic
(ii) The Roller Coasters are represented by the following graphs $y=p(x)$ shown in Fig Which Roller Coaster graph has more than three distinct zeroes?

(iii) If the Roller Coaster is represented by the cubic polynomial $t(x)=p x^3+q x^2+r x+s$, then which of the following is always true
(a) $s \neq 0$ $\qquad$ (b) $r \neq 0$ $\qquad$ (c) $q \neq 0$ $\qquad$ (d) $p \neq 0$
(iv) If the path traced by the Roller Coaster is represented by the graph $y=p(x)$, shown in Fig., then the number of zeroes is
(a) 0 $\qquad$ (b) 1 $\qquad$ (c) 2 $\qquad$ (d) 3

Answer

(i) (c): The graph of the polynomial $y=p(x)$ representing the Roller Coaster cuts $x$-axis at three points. So, $p(x)$ has three distinct zeroes and hence $p(x)$ is a cubic polynomial.
(ii) (d): The graph of the polynomial represented by the curve given in option (d) cuts $x$-axis at four distinct points. Hence, graph given in option (d) has four distinct zeroes.
(iii) (d): If the Roller Coaster is represented by the cubic polynomial $t(x)=p x^3+q x^2+r x+s$, then $p \neq 0$.
(iv) (d): The graph $y=p(x)$ shown in Fig. has three distinct viz. $-2,0$ and 2 .

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Question 44 Marks

Figure shows the path of a diver, when she takes a jump from the diving board. Clearly, it is a parabola. Annie was standing on a diving board, 48 feet above the water level. She took a dive into the pool. Her height (in feet) above the water level at any time ' $t$ ' in seconds is given by the polynomial $h(t)$ such that $h(t)=-16 t^2+8 t+k$.

(i) What is the value of $k$ ?
(a) 0 $\qquad$ (b) -48 $\qquad$ (c) 48 $\qquad$ (d) $48 /-16$
(ii) At what time will she touch the water in the pool?
(a) 30 seconds $\qquad$ (b) 2 seconds $\qquad$ (c) 1.5 seconds $\qquad$ (d) 0.5 seconds
(iii) Rita's height (in feet) above the water level is given by another polynomial $p(t)$ with zeroos -1 and 2 . Then, $p(t)$ is given by
(a) $t^2+t-2$
(b) $t^2+2 t-1$
(c) $24 t^2-24 t+48$
(d) $-24 t^2+24 t+48$
(iv) A polynomial $q(t)$ with sum of zeroes as 1 and the product as -6 is modelling Anu's height in feet above the water at any time $t$ (in seconds). Then $q(t)$ is given by
(a) $t^2+t+t$
(b) $t^2+t-6$
(c) $-8 t^2+8 t+48$
(d) $8 t^2-8 t+48$

Answer

(i) (c): Annie's height $h$ at any time $t$ is given by $h(t)=-16 t^2+8 t+k$.
Initially i.e. at $t=0$, it is given that Annie is standing on a diving board, 48 feet above the water level.
Therefore, $h=48$ at $t=0$. Putting $t=0, h=48$ in $h(t)=-16 t^2+8 t+k$, we obtain $48=k$.
(ii) (b): When Annie touches the water her height from the water level is 0 i.e. $h=0$. Putting $h=0$ in $h(t)=-16 t^2+8 t+k$, we obtain
$
-16 t^2+8 t+k=0 \Rightarrow-16 t^2+8 t+48=0 \Rightarrow 2 t^2-t-6=0 \Rightarrow(t-2)(2 t+3) \Rightarrow t=2
$
(iii) (d): A polynomial having $t=-1$ and $t=2$ as its zeros is given by $p(t)=k(t+1)(t-2)$...
When $t=0$, it is gievn that $p(0)=48$. Putting $t=0, p(0)=48$ in (i), we obtain:
$
48=k(0+1)(0-2) \Rightarrow k=-24
$
Putting $k=-24$ in (i), we obtain: $p(t)=-24(t+1)(t-2)$ or, $p(t)=-24 t^2+24 t+48$.
(iv) (c): A polynomial $q(t)$ with sum of zeroes as 1 and the produced as -6 is given by
$
q(t)=k\left(t^2-t-6\right)
$
At $t=0$, it is given that $q(0)=48$. Putting $t=0, q(0)=48$ in (ii), we obtain: $48=k(-6) \Rightarrow k=-8$
Putting $k=-8$ in (ii), we obtain: $q(t)=-8 t^2+8 t+48$

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Question 54 Marks

Basketball and Soccer are played with a spherical ball. Even through an athlete dribbles the ball in both sports, a basketball player uses his hands, a soccer player uses his feet. Usually, a soccer is played outdoors on a large field, the basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing a quadratic polynomial.

(i) The shape of the path traced shown in Fig. is
(a) Spiral $\qquad$ (b) Ellipse $\qquad$ (c) Linear $\qquad$ (d) Parabola
(ii) If $p(x)=a x^2+b x+c$ is the parabola representing the graphs shown in Fig. then
(a) $a=0$ $\qquad$ (b) $a<0$ $\qquad$ (c) $a>0$ $\qquad$ (d) $a \geq 0$
(iii) The polynomial representing the graph shown in Fig. 2.8 has
(a) no zero $\qquad$ (b) one zero $\qquad$ (c) two zeroes $\qquad$ (d) three zeroes

(iv) The three zeroes of the polynomial representing the above graph in Fig. 2.8 are
(a) $2,3,-1$ $\qquad$ (b) $-2,3,1$ $\qquad$ (c) $-3,-1,2$ $\qquad$ (d) $-2,-3,-1$

Answer

(i) (d): The path is parabolic in shape, so the shape of the path traced is a parabola.
(ii) (b): The parabolic paths shown in Fig. open downward. Therefore, $a<0$.
(iii) (d): The graph of the polynomial crosses $x$-axis at $(-3,0),(-1,0)$ and $(2,0)$. So, the polynomial has three zeroes.
(iv) (c): As the graph of the polynomial cuts $x$-axis at $(-3,0),(-1,0)$ and $(2,0)$. So, the three zeros of the polynomial are $-3,-1$ and 2 .

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Question 64 Marks

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 following figure, one can observe that poses can be related to representation of a quadratic polynomial.

(i) The shape of the poses shown is
(a) Spiral $\qquad$ (b) Ellipse $\qquad$ (c) Linear $\qquad$ (d) Parabola
(ii) The graph of the parabola representing polynomial $p(x)=a x^2+b x+c$ opens downward, if
(a) $a \geq 0$ $\qquad$ (b) $a=0$ $\qquad$ (c) $a<0$ $\qquad$ (d) $a>0$
(iii) The number of zeroes of the polynomial representing the graph in Fig. is
(a) 0 $\qquad$ (b) 1 $\qquad$ (c) 3 $\qquad$ (d) 2
(iv) The zeroes of the polynomial representing the graph in Fig. is
(a) $-2,3$ $\qquad$ (b) $2,-3$ $\qquad$ (c) $-2,6$ $\qquad$ (d) 2,3

Answer

(i)(d): It is evident from the given Figures that various poses are parabolic in shape. So, option (d) is correct.
(ii)(c): The curve represented by $p(x)=a x^2+b x+c$ is a parabola opening upward or downward according as $a>0$ or $a<0$. Hence, option (c) is correct.
(iii)(d): The graph shown in Fig. cuts $x$-axis at two points. So, the polynomial representing it has two zeros. Hence, option (d) is correct.
(iv)(a): The graph shown in Fig. cuts $x$-axis at $(-2,0)$ and $(3,0)$. So, the zeros of the polynomial are -2 and 3 .

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Question 74 Marks

In Fig. various structures are parabolic in shape. A parabola is a curve represented by a quadratic polynomial $p(x)=a x^2+b x+c$.

(i) In the standard form of quadratic polynomial $a x^2+b x+c$
(a) $a, b$ and $c$ all are rational numbers
(b) $a, b$ and $c$ all are integers
(c) $a, b$, and $c$ all are real numbers
(d) $a$ is a non-zero real number and $b$ and $c$ are real numbers
(ii) If the roots of the quadratic polynomial $a x^2+b x+c$ are equal, then its discriminant $D=b^2-4 a c$ satisfies
(a) $D>0$ $\qquad$ (b) $D<0$ $\qquad$ (c) $D \geq 0$ $\qquad$ (d) $D=0$
(iii) If $\alpha$ and $\frac{1}{\alpha}$ are zeroes of the quadratic polynomial $2 x^2-x+8 k$, then the value of $k$ is
(a) 4 $\qquad$ (b) $\frac{1}{4}$ $\qquad$ (c) $-\frac{1}{4}$ $\qquad$ (d) 2
(iv) The graph of the polynomial $p(x)=x^2+1$
(a) intersects $x$-axis at two distinct points
(b) touches $x$-axis at a point
(c) neither touches nor intersects $x$-axis
(d) never intersects $y$-axis

Answer

(i) (d): The general form of a quadratic polynomial is $a x^2+b x+c$, where $a, b$, and $c$ are real numbers such that $a \neq 0$.
(ii) (d): If the roots of a quadratic equation are equal, then its discriminant $D$ is equal to zero.
(iii) (b): If $\alpha$ and $\frac{1}{\alpha}$ are zeros of the quadratic polynomial $2 x^2-x+8 k$, then
$
\text { Product of zeroes }=\frac{8 k}{2} \Rightarrow \alpha \times \frac{1}{\alpha}=\frac{8 k}{2} \Rightarrow 1=4 k \Rightarrow k=\frac{1}{4}
$
(iv) (c): The discriminant $D$ of the polynomial $p(x)=x^2+0 x+1$ is given by $D=0-4=-4<0$. So, it has no real zeroes. Consequently, its graph neither touches nor intersects $x$-axis.

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Question 84 Marks

The figure shows the bridge with hanging wires showing a mathematical shape. Answer the following questions:.