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• Please check that this question paper consists of 11 pages.
• Code number given on the right hand side of the question paper should be written on the title page  of the answer book by the candidate.
• Please check that this question paper consists of 30 questions.
• Please write down the serial number of the question before attempting it.
• 15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer  on the answer book during this period.

SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: If HCF $\displaystyle (336, 54) = 6$, find LCM $\displaystyle (336, 54)$.

We know $\displaystyle HCF \times LCM =$ product of the two numbers

$\displaystyle \therefore 6 \times LCM = 336 \times 54$

$\displaystyle \Rightarrow LCM = \frac{336 \times 54}{6} = 3024$

$\displaystyle \\$

Question 2: Find the nature of roots of the quadratic equation $\displaystyle 2x^2 - 4x + 3 = 0$.

Given equation $\displaystyle 2x^2 - 4x + 3 = 0$

$\displaystyle \text{Therefore Roots } = \frac{4 \pm \sqrt{16 - 4 \times 2 \times 3}}{4} = \frac{4 \pm \sqrt{-8}}{4}$

Therefore the nature of the roots is imaginary

$\displaystyle \\$

Question 3: Find the common difference of the Arithmetic Progression $\displaystyle \text{(A.P.) } \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)$

$\displaystyle \text{Given AP: } \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)$

$\displaystyle \text{Common difference } d = a_2 - a_1, a_3 - a_2, \cdots$

$\displaystyle \therefore d = \frac{3-a}{3a} - \frac{1}{a} = \frac{3 -a - 3}{3a} = - \frac{1}{3}$

$\displaystyle \text{To confirm, we also do } a_3 - a_2 = \frac{3-2a}{3a} - \frac{3-a}{3a} = \frac{3 - 2a - 3 + a}{3a} = - \frac{1}{3}$

$\displaystyle \text{Hence } d = - \frac{1}{3}$

$\displaystyle \\$

Question 4: Evaluate : $\displaystyle \sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o$

Or

$\displaystyle \text{If } \sin A = \frac{3}{4} , \text{ calculate} \sec A$

$\displaystyle \sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o$

$\displaystyle = \Big( \frac{\sqrt{3}}{2} \Big)^2 + 2 (1) - \Big( \frac{\sqrt{3}}{2} \Big)^2$

$\displaystyle = 2$

Or

$\displaystyle \text{If } \sin A = \frac{3}{4}$

$\displaystyle \therefore AB = \sqrt{4^2 - 3^2} = \sqrt{7}$

$\displaystyle \therefore \sec A = \frac{4}{\sqrt{7}}$

$\displaystyle \\$

Question 5: Write the coordinates of a point $\displaystyle P$ on $\displaystyle x-axis$ which is equidistant from the points $\displaystyle A(- 2, 0)$ and $\displaystyle B(6, 0)$.

Let $\displaystyle P(x, 0)$ be equidistant from $\displaystyle A$ and $\displaystyle B$

$\displaystyle AP = BP \Rightarrow AP^2 = BP^2$

$\displaystyle \Rightarrow (-2-x)^2 +(0-0)^2 = (6-x)^2 + (0-0)^2$

$\displaystyle \Rightarrow 4 + x^2 + 4x = 36 + x^2 - 12x$

$\displaystyle \Rightarrow 16x = 32 \Rightarrow x = 2$

$\displaystyle \therefore P$ is $\displaystyle (2, 0)$

$\displaystyle \\$

Question 6: In Figure 1, $\displaystyle ABC$ is an isosceles triangle right angled at $\displaystyle C$ with $\displaystyle AC = 4$ cm. Find the length of $\displaystyle AB$.

Or

In Figure 2, $\displaystyle DE \parallel BC$. Find the length of side $\displaystyle AD$, given that $\displaystyle AE = 1.8$ cm, $\displaystyle BD = 7.2$ cm and $\displaystyle CE = 5.4$ cm.

Since $\displaystyle \triangle ABC$ is isosceles

$\displaystyle \Rightarrow AC = BC$

(hypotenuse $\displaystyle AB$ has to be the largest side, hence the other two sides are equal)

$\displaystyle \therefore AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}$ cm

Or

Since $\displaystyle DE \parallel BC$ by Thales Theorem

$\displaystyle \frac{AD}{BD} = \frac{AE}{EC}$

$\displaystyle \Rightarrow \frac{AD}{7.2} = \frac{1.8}{5.4}$

$\displaystyle \Rightarrow AD = \frac{1.8}{5.4} \times 7.2 = 2.4 \ cm$

$\displaystyle \\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Write the smallest number which is divisible by both $\displaystyle 306$ and $\displaystyle 657$.

$\displaystyle 306 = 2 \times 3^2 \times 17$

$\displaystyle 657 = 3^2 \times 73$

LCM of $\displaystyle 306$ and $\displaystyle 657 = 2 \times 3^2 \times 17 \times 73 = 22338$

Hence the smallest number which is divisible by $\displaystyle 306$ and $\displaystyle 657$ is $\displaystyle 22338$

$\displaystyle \\$

Question 8: Find a relation between $\displaystyle x$ and $\displaystyle y$ if the points $\displaystyle A(x, y), B( - 4, 6)$ and $\displaystyle C( - 2, 3)$ are collinear.

Or

Find the area of a triangle whose vertices are given as $\displaystyle (1, - 1) (- 4, 6)$ and $\displaystyle (- 3, - 5)$.

If $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$ are collinear, then the area of the $\displaystyle \triangle ABC$ should be $\displaystyle 0$.

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} \times |x_1(y_2-y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$

$\displaystyle \Rightarrow 0 = \frac{1}{2} \times |x(6-3) + (-4)(3-y) + (-2)(y-6) |$

$\displaystyle \Rightarrow 0 = | 3x - 12 + 4y - 2y + 12 |$

$\displaystyle \Rightarrow 0 = |3x + 2y |$

$\displaystyle \Rightarrow 3x + 2y = 0$

Or

Let the points be $\displaystyle A(1, -1), B(-4, 6)$ and $\displaystyle C(-3, -5)$

$\displaystyle \text{Area of triangle } ABC = \frac{1}{2} [x_1(y_2-y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]$

Here $\displaystyle x_1 = 1, y_1 = -1, x_2 = -4, y_2 = 6, x_3 = -3, y_3 = -5$

Substituting in the formula above

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} [1(6-(-5)) + (-4) (-5 - (-1)) + (-3)(-1-6) ]$

$\displaystyle = \frac{1}{2} [ 11+ 16 + 21 ] = \frac{1}{2} \times 48 = 24 \text { sq. units.}$

$\displaystyle \\$

Question 9: The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is $\displaystyle \frac{1}{5}$. The probability of selecting a black marble at random from the same jar is $\displaystyle \frac{1}{4}$. If the jar contains $\displaystyle 11$ green marbles, find the total number of marbles in the jar.

$\displaystyle \text{P(selecting Blue marble) } = \frac{1}{5}$

$\displaystyle \text{P(selecting Black marble) } = \frac{1}{4}$

$\displaystyle \text{P(selecting Green marble) } = 1 - \frac{1}{5} - \frac{1}{4} = \frac{20-9}{20} = \frac{11}{20}$

Therefore there are a total of $\displaystyle 20$ marbles in the Jar.

$\displaystyle \\$

Question 10: Find the value(s) of $\displaystyle k$ so that the pair of equations $\displaystyle x + 2y = 5$ and $\displaystyle 3x + ky + 15 = 0$ has a unique solution.

Pair of equations $\displaystyle x + 2y = 5$ and $\displaystyle 3x + ky + 15 = 0$

If the intersecting lines have a unique solution, then

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\displaystyle \Rightarrow \frac{1}{3} \neq \frac{2}{k} \neq \frac{-5}{15}$

From the first two terms, $\displaystyle k \neq 6$

From the second and the third terms, $\displaystyle k \neq -6$

Hence all values of $\displaystyle k$ except $\displaystyle 6$ and $\displaystyle -6$

$\displaystyle \\$

Question 11: The larger of two supplementary angles exceeds the smaller by $\displaystyle 18^o$. Find the angles.

Or

Sumit is $\displaystyle 3$ times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?

Let $\displaystyle x$ and $\displaystyle y$ be the two supplementary angles

$\displaystyle \therefore x + y = 180^o$

Also given $\displaystyle x = y + 18^o$ (given one angle is greater than the other by $\displaystyle 18^o$ )

Solving the two equations

$\displaystyle 2y + 18 = 180^o \Rightarrow 2y = 162^o \Rightarrow y = 81^o$

$\displaystyle \therefore x = 81^o + 18^o = 99^o$

Hence the two angles are $\displaystyle 81^o$ ad $\displaystyle 99^o$

Or

Let the age of son be $\displaystyle x$ years

Therefore are of Sumit is $\displaystyle 3x$

Age of son $\displaystyle 5$ years later $\displaystyle = x + 5$

Age of Sumit $\displaystyle 5$ years later $\displaystyle = 3x + 5$

Give: $\displaystyle 3x+5 = 2.5 (x + 5)$

$\displaystyle \Rightarrow 3x + 5 = 2.5 x + 12.5$

$\displaystyle \Rightarrow 0.5 x = 7.5$

$\displaystyle \Rightarrow x = 15$ years

Therefore age of Sumit $\displaystyle = 3 \times 15 = 45$ years

$\displaystyle \\$

Question 12: Find the mode of the following frequency distribution :

 Class Interval: 25-30 30-35 35-40 40-45 45-50 50-55 Frequency: 25 34 50 42 38 14

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that $2 + 5 \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Or

Using Euclid’s Algorithm, find the HCF of $2048$ and $960$.

Let us assume that $5 +2 \sqrt{3}$ is a rational number.

$\displaystyle \text{The rational number is in the form of } \frac{p}{q}$

$\displaystyle \therefore 5 + 2\sqrt{3} = \frac{p}{q}$

$\displaystyle 2\sqrt{3} = \frac{p}{q} - 5$

$\displaystyle 2\sqrt{3} = \frac{p-5q}{q}$

$\displaystyle \sqrt{3} = \frac{p-5q}{2q}$

$\displaystyle \therefore \sqrt{3} \text{ is a rational no as } \frac{p-5q}{2q} \text{ is a rational number }$

But it is given that $\sqrt{3}$ is an irrational number which contradicts our initial assumption.

Hence $5 + 2 \sqrt{3}$ is an irrational number.

Or

$960 ) \overline{2048} ( 2 \\ \hspace*{1cm} \underline{1920} \\ \hspace*{1cm} 128 ) \overline{960} ( 7 \\ \hspace*{2cm} \underline{896} \\ \hspace*{2cm} 64 ) \overline{128} ( 2 \\ \hspace*{3cm} \underline{128} \\ \hspace*{3cm} \times$

Hence HCF $= 64$

$\\$

Question 14: Two right triangles $ABC$ and $DBC$ are drawn on the same hypotenuse $BC$ and on the same side of $BC$. If $AC$ and $BD$ intersect at $P$, prove that $AP \times PC = BP \times DP$.

Or

Diagonals of a trapezium $PQRS$ intersect each other at the point $O, PQ \parallel RS$ and $PQ = 3RS$. Find the ratio of the areas of triangles $POQ$ and $ROS$.

To prove: $AP \times PC = BP \times PD$

Consider $\triangle ABP$ and $\triangle CDP$

$\angle BAP = \angle CDP = 90^o$ (given)

$\angle BPA = \angle CPD$ ( vertically opposite angles)

$\therefore \triangle ABP \sim \triangle CDP$ (By AA criterion)

$\displaystyle \therefore \frac{BP}{PC} = \frac{AP}{PD}$

$\Rightarrow AP \times PC = BP \times PD$

Hence proved.

Or

Given $PQRS$ is a trapezium

$PQ \parallel RS$ and $PQ = 3 RS$

Consider $\triangle POQ$ and $\triangle SRO$

$\angle SOR = \angle POQ$ (vertically opposite angles)

$\angle RSO = \angle OQP$ (alternate angles)

$\angle SRO = \angle OPQ$ (alternate angles)

$\therefore \triangle POQ \sim \triangle SRO$ (By AAA criterion)

By property of similar triangles

$\displaystyle \frac{ar (\triangle POQ)}{ar(\triangle SOR)} = \frac{PQ^2}{RS^2} = \Big( \frac{PQ}{RS} \Big)^2 = \Big( \frac{3}{1} \Big)^2 = \frac{9}{1}$

$\\$

Question 15: In Figure 3, $PQ$ and $RS$ are two parallel tangents to a circle with center $O$ and another tangent $AB$ with point of contact $C$ intersecting $PQ$at $A$ and $RS$ at $B$. Prove that $\angle AOB = 90^o$.

Given: $PQ \parallel RS$, $AB$ is a tangent

To prove: $\angle AOB = 90^o$

Join $DE$ through $O$. $DOE$ is diameter of the circle

$\therefore \angle ADO = \angle BEO = 90^o$

$\Rightarrow \angle ADO + \angle BEO = 180^o$

$\Rightarrow DA \parallel EB$

We know that tangents to a circle from an external point are equally inclined to the line segment joining this point to the center

$\therefore \angle 1 = \angle 2$ and $\angle 3 = \angle 4$

Now $DA \parallel RB$ and $AB$ is transversal

$\therefore (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = 180^o$

$\Rightarrow 2 \angle 1 + 2 \angle 3 = 180^o$

$\Rightarrow \angle 1 + \angle 3 = 90^o$

From $\triangle AOB$

$\angle AOB + \angle 1 + \angle 3 = 180^o$

$\Rightarrow \angle AOB = 180^o - 90^0 = 90^o$

Hence $\angle AOB = 90^o$

$\\$

Question 16: Find the ratio in which the line $\displaystyle x - 3y = 0$ divides the line segment joining the points $\displaystyle (- 2, - 5)$ and $\displaystyle (6, 3)$. Find the coordinates of the point of intersection.

Let $\displaystyle P(x, y)$ divides point $\displaystyle A(-2, -5) \text{ and } B (6, 3)$ in the ratio $\displaystyle k : 1$

$\displaystyle \text{Using section formula: } P(x, y) = \Big( \frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} \Big)$

$\displaystyle A(-2, -5) \text{ and } B(6, 3)$ in the ratio of $\displaystyle k:1$

$\displaystyle \Rightarrow x = \frac{6k-2}{k+1} \text{ and } \Rightarrow y = \frac{3k-5}{k+1}$

Since $\displaystyle P(x, y)$ lies on the line $\displaystyle x-3y = 0$

$\displaystyle \frac{6k-2}{k+1} - 3 \Big( \frac{3k-5}{k+1} \Big) = 0$

$\displaystyle \Rightarrow 6k - 2 - 9k + 15 = 0$

$\displaystyle \Rightarrow -3k +13 = 0$

$\displaystyle \Rightarrow k = \frac{13}{3}$

Therefore $\displaystyle P$ divided $\displaystyle A \text{ and } B$ in the ratio of $\displaystyle 13:3$

$\displaystyle \therefore x = \frac{6(\frac{13}{3})-2}{\frac{13}{3}+1} = \frac{(26-2)\times 3}{16} = \frac{9}{2}$

$\displaystyle y = \frac{3(\frac{13}{3})-5}{(\frac{13}{3})+1} = \frac{(13-5) \times 3}{16} = \frac{3}{2}$

$\displaystyle \text{Hence } P( \frac{9}{2} , \frac{3}{2} ) \text{which is the point of intersection.}$

$\displaystyle \\$

$\displaystyle \text{Question 17: Evaluate: } \Big( \frac{3 \sin 43^{\circ}}{\cos 47^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \mathrm{cosec} 53^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

$\displaystyle \Big( \frac{3 \sin 43^{\circ}}{\cos 47^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \mathrm{cosec} 53^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

$\displaystyle = \Big( \frac{3 \sin 43^{\circ}}{\cos (90-43)^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \frac{1}{\sin (90-37)^{\circ} } }{\tan 5^{\circ} \tan 25^{\circ} .1. \tan (90-25)^{\circ} \tan (90-5)^{\circ}}$

$\displaystyle = \Big( \frac{3 \sin 43^{\circ}}{\sin 43^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \frac{1}{\cos 37^{\circ} } }{\tan 5^{\circ} \tan 25^{\circ} .1. \cot 25^{\circ} \cot 5^{\circ}}$

$\displaystyle = 3^2 -1 = 9-1 = 8$

$\displaystyle \\$

Question 18: In Figure 4, a square $OABC$ is inscribed in a quadrant $OPBQ$. If $OA = 15$ cm, find the area of the shaded region. (Use $\pi = 3.14$)

Or

In Figure 5, $ABCD$ is a square with side $2 \sqrt{2}$ cm and inscribed in a circle. Find the area of the shaded region. (Use $\pi = 3.14$)

$OA = OB = 15$ cm (since $OABC$ is a square)

$\therefore OB = \sqrt{15^2 + 15^2} = 15 \sqrt{2} =$ Radius

Area of square $OABC = 15 \times 15 = 225 \ cm^2$

$\displaystyle \text {Area of quadrant } = \frac{1}{4} (\pi (15 \sqrt{2})^2 ) = 353.25 \ cm^2$

Shaded area $= 353.25 - 225 = 128.25 \ cm^2$

Or

Area of square $= 2\sqrt{2} \times 2\sqrt{2} = 8 \ cm^2$

Diameter $AC = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8+8} = 4 \ cm$

Therefore Radius of circle $= 2 \ cm$

Area of circle $= \pi (2)^2 = 3.14 \times 4 = 12.56 \ cm^2$

Therefore shaded area $= 12.56 - 8 = 4.56 \ cm^2$

$\\$

Question 19: A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $20$ cm and the diameter of the cylinder is $7$ cm. Find the total volume of the solid. (Use $\pi = \frac{22}{7}$ )

Radius of hemisphere $= 3.5 \ cm$

Therefore height of cylinder $= 20-3.5 -3.5 = 13 \ cm$

$\displaystyle \text{Therefore total volume }= 2 \times ( \frac{1}{2} \times \frac{4}{3} \times \pi (3.5)^2 ) + \pi (3.5)^2 \times 13$

$= 179.67 + 500.5 = 680.17 \ cm^2$

$\\$

Question 20: The marks obtained by $100$ students in an examination are given below :

 Marks: 30-35 35-40 40-45 45-50 50-55 55-60 60-65 Number of Students: 14 16 28 23 18 8 3

 Marks ( Class interval) No. of Students (Frequency) Mean Value $latex (x)$ $fx$$fx$ 30-35 14 32.5 455 35-40 16 37.5 600 40-45 28 42.5 1190 45-50 23 47.5 1092.5 50-55 18 52.5 945 55-60 8 57.5 460 60-65 3 62.5 187.5 $\Sigma f = 110$$\Sigma f = 110$ $\Sigma fx = 4930$$\Sigma fx = 4930$

$\displaystyle \text{Mean } = \frac{\Sigma fx}{\Sigma f} = \frac{4930}{110} = 44.818$

$\displaystyle \\$

Question 21: For what value of k, is the polynomial $\displaystyle f(x) = 3x^4 - 9x^3 + x^2 + 15x + k$ completely divisible by $\displaystyle 3x^2 - 5$ ?

Or

$\displaystyle \text{Find the zeroes of the quadratic polynomial } 7y^2 - \frac{11}{3} y - \frac{2}{3} \text{and verify the}$  relationship between the zeroes and the coefficients.

$\displaystyle f(x) \text{ divisible by } P(x) \Rightarrow \text{ the remainder of } \frac{f(x)}{P(x)} = 0$

$\displaystyle 3x^2-5 ) \overline{3x^4 - 9x^3 + x^2 + 15x + k} ( x^2-3x+2 \\ \hspace*{1cm}(-) { 3x^4-5x^2} \\ \hspace*{2cm} \overline{-9x^3 + 6x^2 + 15x+ k} \\ \hspace*{2cm}(-) \underline{ -9x^3+15x} \\ \hspace*{3.5cm} {6x^2+k} \\ \hspace*{3cm} (-) \underline{6x^2-10} \\ \hspace*{4cm} k+10$

Remainder should be $\displaystyle 0 \Rightarrow k+10 = 0 \Rightarrow k = -10$

$\displaystyle (3x^2-5) \text{ and } (x^2 - 3x+2)$ are factors of $\displaystyle f(x)$

$\displaystyle \Rightarrow (\sqrt{3} x - \sqrt{5}) \text{ and } (\sqrt{3} x + \sqrt{5})$ are factors

$\displaystyle \Rightarrow x = \frac{\sqrt{5}}{\sqrt{3}} \text{ or } - \frac{\sqrt{5}}{\sqrt{3}}$

Also $\displaystyle x^2 - 3x + 2 = (x-2)(x-1)$ are factors $\displaystyle \Rightarrow x = 2, 1$

$\displaystyle \text{Hence } \frac{\sqrt{5}}{\sqrt{3}} , - \frac{\sqrt{5}}{\sqrt{3}} , 2, 1 \text{ are zeros of } f(x)$

Or

$\displaystyle \text{Given polynomial } P(y) = y^2 - \frac{11}{3} y - \frac{2}{3} = 0$

$\displaystyle \Rightarrow 21y^2 - 11y - 2 = 0$

$\displaystyle \Rightarrow 7y(3y-2) + (3y-2) = 0$

$\displaystyle \Rightarrow (3y-2)(7y+1) = 0$

$\displaystyle \Rightarrow y = \frac{2}{3} \text{ or } y = - \frac{1}{7}$

$\displaystyle \text{Comparing } P(y) \text{ with } ax^2 + bx + c =0 \text{ we get } a = 7, b = - \frac{11}{3} \text{ and } c = - \frac{2}{3}$

$\displaystyle \text{Sum of zeros } = - \frac{b}{a} = - (- \frac{11}{3} ) ( \frac{1}{7} ) = \frac{11}{21}$ … … … … … i)

$\displaystyle \text{Sum of roots } = \frac{2}{3} + \Big( \frac{-1}{7} \Big) = \frac{11}{21}$ … … … … … ii)

Therefore i) = ii)

$\displaystyle \text{Products of zeros } = \frac{c}{a} = \frac{-\frac{2}{3}}{7} = \frac{-2}{21}$ … … … … … iii)

$\displaystyle \text{Product of roots } = \frac{2}{3} \times \Big( \frac{-1}{7} \Big) = \frac{-2}{21}$ … … … … … iv)

Therefore iii) = iv)

$\displaystyle \\$

Question 22: Write all the values of $p$ for which the quadratic equation $x^2 + px + 16 = 0$ has equal roots. Find the roots of the equation so obtained.

Given quadratic equation is: $x^2 + px + 16 = 0$

If the roots are equal then the discriminant is $0$

i.e. $b^2 - 4ac = 0$

$\Rightarrow p^2 - 4 (1) (16) = 0$

$\Rightarrow p^2 = 64$

$\Rightarrow p = \pm 8$

When $p = 8, x^2 + 8x + 16 = 0$

$\Rightarrow (x+4)(x+4) = 0$

$\Rightarrow x = -4$

When $p = -8, x^2 - 8x + 16 = 0$

$\Rightarrow (x-4)(x-4) = 0$

$\Rightarrow x = 4$

Therefore the roots of equation are $4, -4$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Given: In $\triangle ABC, DE$ intersects $AB$ and $AC$ at $D$ and $E$ respectively.

$\displaystyle \text{To Prove: } \frac{AD}{DB} = \frac{AE}{EC}$

Construct: $DM \perp AC$ and $EN \perp AB$

Proof:

$\displaystyle \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times EN$

$\displaystyle \text{Area of } \triangle BDE = \frac{1}{2} \times BD \times EN$

$\displaystyle \therefore \frac{ar(\triangle ADE)}{ar(\triangle BDE)} = \frac{AD}{BD}$   … … … … … i)

Now $ar(\triangle BDE) = ar(\triangle CDE)$   … … … … … ii)

(Because they have the same base $DE$ and are between the same parallel)

$\displaystyle \text{Similarly, } \frac{ar(\triangle ADE)}{ar(\triangle CDE)} = \frac{AE}{EC}$   … … … … … iii)

$\displaystyle \text{From i) , ii) and iii) we get } \frac{AD}{DB} \frac{AE}{EC} \text{. Hence proved.}$

$\\$

Question 24: Amit, standing on a horizontal plane, finds a bird flying at a distance of $200$ m from him at an elevation of $30^o$. Deepak standing on the roof of a $50$ m high building, finds the angle of elevation of the same bird to be $45^o$. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.

$\displaystyle \frac{h}{200} = \sin 30^o = \frac{1}{2}$

$\Rightarrow h = 100$ m

$\displaystyle \therefore \frac{100-50}{BD} = \sin 45^o = \frac{1}{\sqrt{2}}$

$\therefore BD = 50\sqrt{2}$

$\\$

Question 25: A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, which is surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find the mass of the pole, given that $1 \ cm^3$ of iron has approximately $8$ gm mass. (Use $\pi = 3.14$)

Volume of pole $= \pi (12)^2 \times 220 + \pi (8)^2 \times 60$

$= 31680 \pi + 3840 \pi$

$= 35520 \times 3.14$

$= 111532.8 \ cm^3$

$\displaystyle \text{Weight of the pole } = 111532.8 \ cm^3 \times 8 \frac{gm}{cm^3} = 892262.4 \ gm = 892.26 \ kg$

$\\$

Question 26: Construct an equilateral $\triangle ABC$ with each side $5$ cm. Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\triangle ABC$.

Or

Draw two concentric circles of radii $2$ cm and $5$ cm. Take a point $P$ on the outer circle and construct a pair of tangents $PA$ and $PB$ to the smaller circle. Measure $PA$.

$\\$

Question 27: Change the following data into ‘less than type’ distribution and draw its ogive :

 Class Interval: 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency: 7 5 8 10 6 6 8

$\\$

$\displaystyle \text{Question 28: Prove that: } \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \ \mathrm{cosec} \theta$

Or

$\displaystyle \text{Prove that: } \frac{\sin \theta}{\cot \theta + \mathrm{cosec} \theta } = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}$

$\displaystyle \text{LHS } = \frac{\tan \theta}{1- \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

$\displaystyle = \frac{\tan \theta}{1- \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}$

$\displaystyle = \frac{\tan^2 \theta}{\tan \theta -1 } + \frac{1}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = - \frac{\tan^2 \theta}{1- \tan \theta } + \frac{1}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = \frac{1 - \tan^3 \theta}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = \frac{1 + \tan^2 \theta + \tan \theta}{\tan \theta }$

$\displaystyle = \frac{\sec^2 \theta + \tan \theta}{\tan \theta }$

$\displaystyle = \sec^2 \theta \cot \theta + 1$

$\displaystyle = \frac{1}{\cos^2 \theta} \times \frac{\cos \theta}{\sin \theta} + 1$

$\displaystyle = \frac{1}{\cos \theta \sin \theta} + 1$

$\displaystyle = 1 + \sec \theta \ \mathrm{cosec} \theta$

Or

$\displaystyle \text{LHS } = \frac{\sin \theta}{\cot \theta+ \mathrm{cosec} \theta}$

$\displaystyle = \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta+1}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta+1} \times \frac{1 - \cos \theta}{1 - \cos \theta}$

$\displaystyle = \frac{\sin^2 \theta (1 - \cos \theta)}{1- \cos^2 \theta}$

$\displaystyle = \frac{\sin^2 \theta (1 - \cos \theta)}{\sin^2 \theta}$

$\displaystyle = 1- \cos \theta$

$\displaystyle \text{RHS } = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}$

$\displaystyle = 2 + \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}}$

$\displaystyle = 2 + \frac{\sin^2 \theta}{\cos \theta-1}$

$\displaystyle = 2 - \frac{\sin^2 \theta}{1-\cos \theta}\frac{(1 + \cos \theta)}{(1 + \cos \theta)}$

$\displaystyle = 2 - \frac{\sin^2 \theta}{1- \cos^2 \theta} (1 + \cos \theta)$

$\displaystyle = 2 - ( 1 + \cos \theta)$

$\displaystyle = 1 - \cos \theta$

$\displaystyle =$ LHS

Hence Proved.

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Question 29: Which term of the Arithmetic Progression $-7, -12, -17, -22, \cdots$ will be $- 82$ ? Is $-100$ any term of the A.P. ? Give reason for your answer.

Or

How many terms of the Arithmetic Progression $45, 39, 33, \cdots$ must be taken so that their sum is $180$ ? Explain the double answer.

Given AP: $-7, -12, -17, -22, \cdots$

$a_1 = -7$

Common difference $= -12 - (-7) = -5$

The $n^{th}$ term of AP

$T_n = a + (n-1)d$

$\Rightarrow -82 = -7 + (n-1)(-5)$

$\Rightarrow -82 = -7 - 5n + 5$

$\Rightarrow -82 = -2 - 5n$

$\Rightarrow -80 = -5n$

$\Rightarrow n = 16$

Hence $-82$ is the $16^{th}$ term

Since the numbers in the given $AP$ are not factors of $5, - 100$ will not be part of the AP.

Or

Given AP is $45, 39, 33, \cdots$

$a_1 = 45$

Common difference $d = 39 - 45 = - 6$

$S_n = 180$

$\frac{n}{2}$ $(2a+(n-1)d) = 180$

$\frac{n}{2}$ $( 2 \times 45 + (n-1)(-6)) = 180$

$n(90-6n + 6) = 360$

$n(96-6n) = 360$

$n(16-n) = 60$

$n^2 - 16n + 60 = 0$

$(n-10)(n-6) = 0$

$n = 10$ or $n = 6$

We get double answer because the sum of the $7^{th}$ to $10^{th}$ term is $0$.

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Question 30: In a class test, the sum of Arun’s marks in Hindi and English is $30$. Had he got $2$ marks more in Hindi and $3$ marks less in English, the product of the marks would have been $210$. Find his marks in the two subjects.

Let Arun’s marks in Hindi $= x$

Let Arun’s marks in English $= y$

$\therefore x + y = 30$   … … … … … i)

Also $(x+2)(y-3) = 210$   … … … … … ii)

From i) $y = 30 - x$   … … … … … iii)

Substituting in ii)

$(x+2)(30-x-3) = 210$

$\Rightarrow (x+2)(27-x) = 210$

$\Rightarrow 27x + 54 - x^2 - 2x = 210$

$\Rightarrow 25x-x^2= 156$

$\Rightarrow x^2 - 25x + 156 = 0$

$\Rightarrow (x-12)(x-13) = 0$

$x = 12$ or $13$

Substituting in iii), when $x = 12, y = 30-12 = 18$

When $x = 13 , y = 30-13 = 17$

Hence his marks in Hindi are $12$ or $13$. When he gets $12$ in Hindi, he gets $18$ in English. When he gets $13$ in Hindi, he gets $17$ in English.

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