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#### Question 1:

Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q and r, such that a = bq + r, where
(a) 0 < r < b
(b) 0 ≤ r < b
(c) 0 < rb
(d) 1 < r < b

(b) 0 ≤ r < b
On dividing a by b, let q be the quotient and r be the remainder.
Then, we have:
a = bq + r, where 0 ≤  r  <  b

#### Question 2:

In the given figure, the graph of the polynomial p(x) is shown. The number of zeros of p(x) is

(a) 1
(b) 3
(c) 2
(d) 4

(c) 2
Here, the number of zeros is two, as the graph intersects the x - axis at two points.

#### Question 3:

In ΔABC, DEBC. If AD = 3 cm, DB = 2 cm and DE = 6 cm, then BC = ?

(a) 9 cm
(b) 10 cm
(c) 12 cm
(d) 18 cm

(b) 10 cm
In Δ ABC, DEBC. Then,
$\frac{AD}{DB}=\frac{AE}{EC}\phantom{\rule{0ex}{0ex}}$      ( By Thales' theorem)
$⇒\frac{BD}{AD}=\frac{EC}{AE}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{BD}{AD}+1=\frac{EC}{AE}+1\phantom{\rule{0ex}{0ex}}$
$⇒\frac{BD+AD}{AD}=\frac{EC+AE}{AE}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{AB}{AD}=\frac{AC}{AE}$   (Since BD + AD = AB and EC + AE = AC)
and $\angle A=\angle A$              (Common)
$△ABC~△ADE$      (SAS similarity)
Thus,
$\frac{AB}{AD}=\frac{BC}{DE}$     [In similar triangles, corresponding sides are similar]
$⇒\frac{5}{3}=\frac{BC}{6}\phantom{\rule{0ex}{0ex}}$
$⇒BC=\left(\frac{6×5}{3}\right)=10$ cm

#### Question 4:

If sin 3θ and (θ − 2°), where 3θ and (θ − 2°) are both acute angles, then θ = ?
(a) 44°
(b) 22°
(c) 46°
(d) 23°

(d) 23°
Disclaimer: In the question instead of [sin 3θ and (θ − 2°)] it should be [sin 3θ = cos (θ − 2°)]

sin 3θ = cos (θ − 2°)
⇒ cos (90° - 3θ) = cos (θ − 2°)     [ Since cos ( 90° - θ) = sin θ]
⇒ (90° - 3θ) = (θ − 2°)
⇒ 92° = 4θ
⇒ θ = $\frac{92}{4}=23°$

#### Question 5:

If tan
(a) −1
(b) 1
(c) $-\frac{1}{2}$
(d) $\frac{1}{2}$

(d) $\frac{1}{2}$
Here,
$\mathrm{tan}\theta =\sqrt{3}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\sqrt{3}}{1}$
By Pythagoras' theorem, we get AB = 1.

Now, $\mathrm{sec\theta }=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{2}{1}$ and $\mathrm{cosec\theta }=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2}{\sqrt{3}}$
$\frac{se{c}^{2}\theta -\mathrm{cos}e{c}^{2}\theta }{se{c}^{2}\theta +\mathrm{cos}e{c}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left(2\right)}^{2}-{\left(\frac{2}{\sqrt{3}}\right)}^{2}}{{\left(2\right)}^{2}+{\left(\frac{2}{\sqrt{3}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4-\frac{4}{3}}{4+\frac{4}{3}}\phantom{\rule{0ex}{0ex}}=\frac{12-4}{12+4}\phantom{\rule{0ex}{0ex}}=\frac{8}{16}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

#### Question 6:

After how many places of decimal, will the decimal expansion of $\frac{49}{40}$ terminate ?
(a) 1
(b) 2
(c) 3
(d) It will not terminate

(c) 3

We have:
$\frac{49}{40}×\frac{25}{25}=\frac{1225}{1000}=\frac{1225}{{\left(10\right)}^{3}}$
So, it will terminate after 3 decimal places.

#### Question 7:

The pair of linear equations has
(a) one solution
(b) two solutions
(c) many solutions
(d) no solution

(d) no solution
The given equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where
a1 = 6, b1 = - 3, c1 = 10 and a2 = 2, b2 = - 1, c2 = 9
Since,
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$

$⇒\frac{6}{2}=\frac{-3}{-1}\ne \frac{10}{9}\phantom{\rule{0ex}{0ex}}$

$⇒\frac{3}{1}=\frac{3}{1}\ne \frac{10}{9}$

Hence, the given system of equations has no solution.

#### Question 8:

For a given set of data with 60 observations, the 'less than ogive' and 'more that ogive' intersect at (18.5, 30). The median of the data is
(a) 18
(b) 30
(c) 60
(d) 18.5

(d) 18.5
We know that the abscissa of the point of intersection of the two ogives gives the median.
Hence, the required median is 18.5.

#### Question 9:

Is (7 × 5 × 3 × 2 + 3) a composite number? Justify your answer.

We have:
(7 × 5 × 3 × 2 + 3) = 210 + 3 = 213
2 + 1 + 3 = 6
So, 213 is divisible by 3.   (If the sum of digits of a number is divisible by 3, then the number is divisible by 3)
Hence, (7 × 5 × 3 × 2 + 3) is a composite number.

#### Question 10:

When a polynomial p(x) is divided by (2x + 1), is it possible to have (x − 1) as a remainder? Justify your answer.

When we divide a polynomial by another polynomial of degree, say 'm', then the remainder left is always of a degree less than m, i.e. the remainder can be at most of degree '(m – 1)'.
Here, if we divide p(x) by (2x + 1), i.e. we divide p(x) by a polynomial of degree 1, then the remainder can be of degree '0', i.e. a constant polynomial.
Since degree of (x – 1) is 1, which is greater than '0', it is not possible to have (x – 1) as a remainder when p(x) is divided by (2x + 1).

#### Question 11:

If 3 cos2θ + 7 sin2θ = 4, show that cot θ = $\sqrt{3}$.
Or, if tan

We have:

$⇒3{\mathrm{cos}}^{2}\theta +3{\mathrm{sin}}^{2}\theta +4{\mathrm{sin}}^{2}\theta =4\phantom{\rule{0ex}{0ex}}$
$⇒3\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+4{\mathrm{sin}}^{2}\theta =4\phantom{\rule{0ex}{0ex}}$     [ Since cos2θ +  sin2θ = 1]
$⇒3×1+4{\mathrm{sin}}^{2}\theta =4\phantom{\rule{0ex}{0ex}}$
$⇒4{\mathrm{sin}}^{2}\theta =1⇒{\mathrm{sin}}^{2}\theta =\frac{1}{4}$
${\mathrm{cos}}^{2}\theta =\left(1-{\mathrm{sin}}^{2}\theta \right)=\left(1-\frac{1}{4}\right)=\frac{3}{4}\phantom{\rule{0ex}{0ex}}$
${\mathrm{tan}}^{2}\theta =\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }=\left(\frac{1}{4}×\frac{4}{3}\right)=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{tan}\theta =\frac{1}{\sqrt{3}}⇒cot\theta =\frac{1}{\mathrm{tan}\theta }=\sqrt{3}$

OR
We have:
$\frac{\left(2+2\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}{\left(1+\mathrm{cos}\theta \right)\left(2-2\mathrm{cos}\theta \right)}=\frac{2\left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}{2\left(1+\mathrm{cos}\theta \right)\left(1-\mathrm{cos}\theta \right)}=\frac{\left(1-{\mathrm{sin}}^{2}\theta \right)}{\left(1-{\mathrm{cos}}^{2}\theta \right)}=\frac{{\mathrm{cos}}^{2}\theta }{{\mathrm{sin}}^{2}\theta }=co{t}^{2}\theta =\left(cot\theta \right)$2
Given that,
$\mathrm{tan}\theta =\frac{8}{15}⇒cot\theta =\frac{15}{8}$
${\left(cot\theta \right)}^{2}={\left(\frac{15}{8}\right)}^{2}=\left(\frac{15}{8}×\frac{15}{8}\right)=\frac{225}{64}$
Hence,
$\frac{\left(2+2\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}{\left(1+\mathrm{cos}\theta \right)\left(2-2\mathrm{cos}\theta \right)}=\frac{225}{64}$

#### Question 12:

In the given figure, DE ∥ AC and DF AE.
Prove that $\frac{EC}{BE}=\frac{FE}{BF}.$

In ΔBAE, $\mathrm{DF}\parallel \mathrm{AE}$
∴  $\frac{AD}{DB}=\frac{EF}{FB}$ ............(i)   [by Thale's theorem]
In ΔBAC, $DE\parallel AC$
$\frac{AD}{DB}=\frac{CE}{EB}$ ..............(ii)     [by Thale's theorem]
From (i) and (ii), we get:
$\frac{\mathrm{EF}}{\mathrm{FB}}=\frac{\mathrm{CE}}{\mathrm{EB}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{EC}}{\mathrm{BE}}=\frac{\mathrm{FE}}{\mathrm{BF}}$

#### Question 13:

In the given figure, AD ⊥ BC and $\mathrm{BD}=\frac{1}{3}\mathrm{CD}.$
Prove that $2C{A}^{2}=2A{B}^{2}+B{C}^{2}.$

Given: In ΔABC, AD⊥BC and $\mathrm{BD}=\frac{1}{3}\mathrm{CD}$.
To prove: 2CA2 = 2AB2 + BC2
Proof:
$\mathrm{BD}=\frac{1}{3}\mathrm{CD}$ ............(i)
∴ BC =  BD + CD
$\mathrm{BC}=\frac{1}{3}\mathrm{CD}+\mathrm{CD}=\frac{4}{3}\mathrm{CD}$
or,
and $BD=\frac{1}{3}CD=\frac{1}{3}×\frac{3}{4}BC=\frac{1}{4}BC$

#### Question 14:

Find the mode of the following distribution of marks obtained by 80 students:

 Marks obtained 0-10 10-20 20-30 30-40 40-50 Number of students 6 10 12 32 20

As the class 30 - 40 has maximum frequency, it is the modal class.
Therefore,
xk = 30, h = 10, fk = 32, fk-1 = 12, fk+1 = 20
$\mathrm{Mode},{\mathrm{M}}_{0}={\mathrm{x}}_{\mathrm{k}}+\left\{\mathrm{h}×\frac{\left({\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}-1}\right)}{\left(2{\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}-1}-{\mathrm{f}}_{\mathrm{k}+1}\right)}\right\}\phantom{\rule{0ex}{0ex}}$
$=30+\left\{10×\frac{\left(32-12\right)}{\left(2×32-12-20\right)}\right\}\phantom{\rule{0ex}{0ex}}$
$=30+\left\{10×\frac{20}{\left(64-32\right)}\right\}\phantom{\rule{0ex}{0ex}}=30+\left\{10×\frac{20}{32}\right\}\phantom{\rule{0ex}{0ex}}=\left(30+\frac{25}{4}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{145}{4}\right)\phantom{\rule{0ex}{0ex}}=36.25$

#### Question 15:

Show that any positive odd integer is of the form (4q + 1) or (4q + 3), where q is a positive integer.

Let a be any odd positive integer. we have to prove that a is of the form 4q + 1 or 4q + 3, where q is some integer.
Since a is an integer, consider b = 4 as another integer.
Applying Euclid's division lemma, we get:
a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2 and 3, since 0 ≤ r < 4.
Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
However, since a is odd, it cannot take the values 4q or 4q + 2 (since all these are divisible by 2).
Hence, any odd integer can be expressed in the form 4q + 1 or 4q + 3, where q is some integer.

#### Question 16:

Prove that $\left(5-\sqrt{3}\right)$ is irrational.
Or, prove that $\frac{3\sqrt{3}}{5}$ is irrational.

Let us assume that $\left(5-\sqrt{3}\right)$ is rational. That is, we can find integers p and q(≠ 0) such that

$⇒\sqrt{3}=5-\frac{\mathrm{p}}{\mathrm{q}}$
Since, p and q are integers, $5-\frac{p}{q}$ is rational; so, $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
So, we can conclude that $\left(5-\sqrt{3}\right)$ is irrational.

OR
Let us assume that $\frac{3\sqrt{3}}{5}$ is rational. That is, we can find co -prime integers p and q(≠ 0) such that
$⇒\sqrt{3}=\frac{5p}{3q}$
Since, p and q are integers, $\frac{5p}{3q}$ is rational; so, $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
So, we can conclude that$\frac{3\sqrt{3}}{5}$ is irrational.

#### Question 17:

A man can row a boat at the rate of 4 km/hour in still water. He takes thrice as much time in going 30 km upstream as in going 30 km downstream. Find the speed of the stream.

Or, in a competitive examination, 5 marks are awarded for each correct answer, while 2 marks are deduced for each wrong answer. Jayant answered 120 questions and got 348 marks. How many questions did he answer correctly?

The speed of the boat in still water is 4 km/hr.
Let the speed of the stream be ‘y’ km/hr. Then,
Speed of the boat with the stream = downstream speed = (4 + y) km/hr
Speed of the boat against the stream = Upstream speed= (4 - y) km/hr
According to question,
time taken by boat to cover 30 km downstream = t = $\frac{30}{\left(4+y\right)}\mathrm{hr}$
⇒ 30 =(4 + y)t
⇒ 30 = 4t + yt ......(i)
Time taken by boat to cover 30 km upstream = 3t = $\frac{30}{\left(4-y\right)}\mathrm{hr}$
⇒ 30 = (4 - y)3t
⇒ 30 = 12t - 3yt ......(ii)
On multiplying (i) by 3, we get:
90 = 12t + 3yt ......(iii)
On solving (ii) and (iii), we get:
t = 5 and  y = 2

Hence, the speed of the stream is 2 km/hr.

OR

Let the number of correct answers be x and number of wrong answer be y. Then,
x + y = 120 .............(i)
5x - 2y = 348 .............(ii)
On multiplying (i) by 2, we get:
2x + 2y = 240 ...........(iii)
On adding (ii) and (iii), we get:
7x = 588 ⇒ x = 84
Hence, Jayant answered 84 questions correctly.

#### Question 18:

If α and β are the zeros of the polynomial $2{x}^{2}+x-6,$ then form a quadratic equation whose zeros are 2α and 2β.

f(x) = 2x2 + x - 6 is the given polynomial.
It is given that,α and β are the zeros of the polynomial.

Let 2α and 2β be the zeros of the polynomial g(x).
Thus,
Sum of roots = 2α and 2β = 2(α + β) = $\left(2×\frac{-1}{2}\right)=-1$
Product of roots = 2α × 2β = 4αβ = 4(- 3) = - 12

The required polynomial g(x) = x2 - (Sum of roots)x + Product of roots
= x2 - (- 1)x - 12 = x2 + x - 12
= x2 + x - 12

#### Question 19:

Prove that (cosec θ − sin θ)(sec θ − cos θ) = $\frac{1}{\mathrm{tan\theta }+\mathrm{cot\theta }}.$

LHS = (cosec θ − sin θ)(sec θ − cos θ)
$=\left(\frac{1}{\mathrm{sin\theta }}-\mathrm{sin\theta }\right)\left(\frac{1}{\mathrm{cos\theta }}-\mathrm{cos}\mathrm{\theta }\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(1-{\mathrm{sin}}^{2}\mathrm{\theta }\right)}{\mathrm{sin\theta }}×\frac{\left(1-{\mathrm{cos}}^{2}\mathrm{\theta }\right)}{\mathrm{cos\theta }}\phantom{\rule{0ex}{0ex}}$
$=\frac{{\mathrm{cos}}^{2}{\mathrm{\theta sin}}^{2}\mathrm{\theta }}{\mathrm{sin\theta cos\theta }}\phantom{\rule{0ex}{0ex}}$      [Since (1 - sin2θ) =  cos2θ and (1 - cos2θ) = sin2θ]
$=\mathrm{cos}\mathrm{\theta sin\theta }$
$\mathrm{RHS}=\frac{1}{\mathrm{tan\theta }+\mathrm{cot\theta }}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos\theta sin\theta }}{\left({\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{\mathrm{cos\theta sin\theta }}{1}=\mathrm{cos\theta sin\theta }$   [ Since sin2θ + cos2θ = 1]
Hence, LHS =  RHS

#### Question 20:

If cos θ + sin θ = $\sqrt{2}\mathrm{cos\theta }$, prove that cos θ − sin θ = $\sqrt{2}\mathrm{sin\theta }.$

We have:
(cos θ + sin θ)2 + (cos θ - sin θ)2 = 2(cos2θ + sin2θ)
Given that:
$\mathrm{cos\theta }+\mathrm{sin\theta }=\sqrt{2}\mathrm{cos\theta }$
Thus,
${\left(\sqrt{2}\mathrm{cos\theta }\right)}^{2}+{\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}^{2}=2$            [since cos2θ + sin2θ = 1]
$⇒{\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}^{2}=2-2{\mathrm{cos}}^{2}\mathrm{\theta }=2\left(1-{\mathrm{cos}}^{2}\mathrm{\theta }\right)=2{\mathrm{sin}}^{2}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}$      [Since (1 - cos2θ) = sin2θ]
Taking square root of both the sides, we get:
$⇒\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)=\sqrt{2}\mathrm{sin\theta }$

#### Question 21:

ΔABC and ΔDBC are on the same base BC and on opposite sides of BC. If O is the point of intersection of BC and AD, prove that $\frac{ar\left(∆ABC\right)}{ar\left(∆DBC\right)}=\frac{AO}{DO}.$

Given: ΔABC and ΔDBC are on the same base BC and AD intersects BC at O.
To prove: $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}$
Construction : Draw AL⊥BC and DM⊥BC.

Proof: $\angle$ALO = $\angle$DMO = 90°
$\angle$AOL = $\angle$DOM (vertically-opposite angles)
∴ Δ ALO ∼Δ DMO  [ By AA - similarity]
$⇒\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$ ...................(i)
Therefore,

Hence,$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}$

#### Question 22:

In ΔABC, AD is a median and E is the mid-point of AD. If BE is produced to meet AC at F, show that $AF=\frac{1}{3}AC.$

Given: In Δ ABC, AD is a median and E is the mid-point of AD. Also, BE is produced to meet AC at F.
To Prove: $\mathrm{AF}=\frac{1}{3}\mathrm{AC}$
Construction:
From D, draw $DG\parallel EF$, meeting AC at G.

Proof:
In ΔBCF, D is the mid-point of BC and $DG\parallel BF$.
∴ G  is the mid point of CF.
So, FG =  GC
In ΔADG, E is the mid-point of AD and $EF\parallel DG$.
∴ F  is the mid-point of AG.
So, AF =  FG
Thus, AF = FG = GC
∴ AC = (AF + FG + GC) = 3AF
Hence, $\mathrm{AF}=\frac{1}{3}\mathrm{AC}$

#### Question 23:

Find the mean of the following frequency distribution using step deviation method:

 Class interval 0-50 50-100 100-150 150-200 200-250 250-300 Frequency 17 35 43 40 21 24

Or, The mean of the following frequency distribution is 24. Find the value of P.
 Class 0-10 10-20 20-30 30-40 40-50 Frequency 15 20 35 p 10

Let us assumed mean A be 125; h = 50
For calculating the mean, we prepare the table as shown below.

$\mathrm{Mean},\overline{\mathrm{x}}=\mathrm{A}+\left\{\mathrm{h}×\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\right\}\phantom{\rule{0ex}{0ex}}$
$=125+\left\{50×\frac{85}{180}\right\}=125+23.61=148.61$
Hence, the mean of the given frequency is 148.61.

OR
We have:

Therefore,
$\mathrm{Mean},\overline{\mathrm{x}}=\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\left(1700+35p\right)}{\left(80+p\right)}=24\phantom{\rule{0ex}{0ex}}$
$⇒\left(1700+35p\right)=1920+24p\phantom{\rule{0ex}{0ex}}$
$⇒11p=\left(1920-1700\right)=220\phantom{\rule{0ex}{0ex}}$
$⇒p=\frac{220}{11}=20$
Hence, the value of p is 20.

#### Question 24:

Find the median of the following data:

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8

The cumulative frequency table for the given data is shown below.

$\frac{N}{2}=\frac{53}{2}=26.5$
Cumulative frequency just greater than 26.5 is 29.
The corresponding class interval is 60 - 70.
Thus, median class is 60 - 70.
Cumulative frequency just before this class = 22
So, l = 60, f = 7 , $\frac{N}{2}=26.5$, h = 10 and cf = 22
$\mathrm{Median},{M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-cf\right)}{f}\right\}\phantom{\rule{0ex}{0ex}}$
$=60+\left\{10×\left(\frac{26.5-22}{7}\right)\right\}=60+\frac{10×4.5}{7}=60+6.43=66.43$
Hence, median = 66.43

#### Question 25:

p(x)=2x4 − 3x3 − 5x2 + 9x − 3 and two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$. Find the other two zeros.

The given polynomial is f(x) = 2x4 - 3x3 - 5x2 + 9x - 3.
Since, $\sqrt{3}\mathrm{and}-\sqrt{3}$ are are two zeros of f(x), it follows that each one of  is a factor of f(x).
Consequently, $\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=\left({x}^{2}-3\right)$ is a factor of f(x).
On dividing f(x) by (x2 - 3), we get:

f(x) = 0 ⇒ (x2 - 3)(2x2 - 3x + 1)
⇒ (x2 - 3)(2x2 - 3x + 1) = 0
⇒ (x2 - 3)(2x - 1)(x - 1) = 0
$\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(2x-1\right)\left(x-1\right)=0$

Hence, the other two zeros are .

#### Question 26:

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Or, prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, the angle opposite the first side is a right angle.

Given: Δ ABC ∼ Δ DEF
To Prove: $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{DE}}^{2}}=\frac{{\mathrm{AC}}^{2}}{{\mathrm{DF}}^{2}}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}$

Construction:

Proof:
Since Δ ABC ∼ Δ DEF, it follows that they are equiangular and their sides are proportional.
$\angle$A = $\angle$D, $\angle$B = $\angle$E, $\angle$C = $\angle$F
and $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ ..................(i)
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\left(\frac{1}{2}×\mathrm{BC}×\mathrm{AL}\right)\phantom{\rule{0ex}{0ex}}$
$\mathrm{ar}\left(△\mathrm{DEF}\right)=\left(\frac{1}{2}×\mathrm{EF}×\mathrm{DM}\right)\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{\left(\frac{1}{2}×\mathrm{BC}×\mathrm{AL}\right)}{\left(\frac{1}{2}×\mathrm{EF}×\mathrm{DM}\right)}=\frac{\mathrm{BC}}{\mathrm{EF}}×\frac{\mathrm{AL}}{\mathrm{DM}}$ ...............(ii)
In Δ ALB and Δ DME, we have:
∠ALB = ∠DME = 90° and $\angle$B = $\angle$E    [ from (i)]
∴∠ALB ∼ ∠DME

Consequently, $\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AB}}{\mathrm{DE}}$
[from (i)]
$\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{BC}}{\mathrm{EF}}$..................(iii)
Using (iii) in (ii), we get:
$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}×\frac{\mathrm{BC}}{\mathrm{EF}}\right)=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}\phantom{\rule{0ex}{0ex}}$
Similarly,
Hence, $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{DE}}^{2}}=\frac{{\mathrm{AC}}^{2}}{{\mathrm{DF}}^{2}}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}$

OR

Given: In Δ ABC, AC2 = AB2 + BC2
To prove: $\angle$B = 90°
Construction: Draw Δ DEF, such that DE = AB, EF = BC and $\angle$E = 90°.

Proof:
In Δ DEF, we have $\angle$E = 90°
So, by Pythagoras' theorem, we have:
DF2 = DE2 + EF2
⇒ DF2 = AB2 + BC2 ..............(i)  [Since DE =  AB and EE =  BC]
But,  AC2  =  AB2 + BC2 .............(ii)   [given]

From (i) and (ii), we get:
AC2 = DF2 ⇒ AC =  DF
Now in Δ ABC and Δ DEF, we have:
AB =  DE, BC = EF and AC =  DF
∴ Δ ABC ∼ Δ DEF
Hence, $\angle$B=$\angle$E = 90°.

#### Question 27:

Prove that $\frac{\mathrm{sin\theta }-\mathrm{cos\theta }+1}{\mathrm{sin\theta }+\mathrm{cos\theta }-1}=\frac{1}{\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)}.$
Or, evaluate:

$=\frac{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}-1+\frac{1}{\mathrm{cos\theta }}}{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+1-\frac{1}{\mathrm{cos\theta }}}\phantom{\rule{0ex}{0ex}}$                    [On dividing numerator and denominator by cos θ]
$=\frac{\mathrm{tan\theta }-1+\mathrm{sec\theta }}{\mathrm{tan\theta }+1-\mathrm{sec\theta }}=\left(\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}\right)\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)-\left({\mathrm{sec}}^{2}\mathrm{\theta }-{\mathrm{tan}}^{2}\mathrm{\theta }\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}\phantom{\rule{0ex}{0ex}}$        [Since 1 = sec2θ -  tan2θ]
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left[1-\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)\right]}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}=\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)$

$\mathrm{RHS}=\frac{1}{\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)}×\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}{\left({\mathrm{sec}}^{2}\mathrm{\theta }-{\mathrm{tan}}^{2}\mathrm{\theta }\right)}\phantom{\rule{0ex}{0ex}}$
$=\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)$    [Since sec2θ -  tan2θ = 1]
Hence, LHS  =  RHS

OR
$\frac{\mathrm{sec\theta cosec}\left(90°-\mathrm{\theta }\right)-\mathrm{tan\theta cot}\left(90°-\mathrm{\theta }\right)+{\mathrm{sin}}^{2}65°+{\mathrm{sin}}^{2}25°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°\mathrm{tan}70°\mathrm{tan}80°}\phantom{\rule{0ex}{0ex}}$

$=\frac{{\mathrm{sec}}^{2}\mathrm{\theta }-{\mathrm{tan}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\left(90°-25°\right)+{\mathrm{sin}}^{2}25°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°\mathrm{tan}\left(90°-20°\right)\mathrm{tan}\left(90°-10°\right)}\phantom{\rule{0ex}{0ex}}$

$=\frac{1+{\mathrm{cos}}^{2}25°+{\mathrm{sin}}^{2}25°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°×\mathrm{cot}20°×\mathrm{cot}10°}\phantom{\rule{0ex}{0ex}}$        [Since sec2θ -  tan2θ = 1]

$=\frac{1+1}{\mathrm{tan}10°\mathrm{tan}20°×\sqrt{3}×\frac{1}{\mathrm{tan}20°}×\frac{1}{\mathrm{tan}10°}}\phantom{\rule{0ex}{0ex}}$     [Since cos2θ + sin2θ = 1]
$=\frac{2}{\sqrt{3}}$

#### Question 28:

If sec θ + tan θ = x, prove that sin θ = $\frac{{\mathit{x}}^{\mathit{2}}\mathit{-}\mathit{1}}{{\mathit{x}}^{\mathit{2}}\mathit{+}\mathit{1}}\mathit{.}$

We have,
$\frac{{x}^{2}-1}{{x}^{2}+1}=\frac{{\left(sec\theta +\mathrm{tan}\theta \right)}^{2}-1}{{\left(sec\theta +\mathrm{tan}\theta \right)}^{2}+1}\phantom{\rule{0ex}{0ex}}$

2θ-1=tan2θ

$=\frac{2\mathrm{tan}\theta \left(\mathrm{tan}\theta +sec\theta \right)}{2sec\theta \left(\mathrm{tan}\theta +sec\theta \right)}=\frac{\mathrm{tan}\theta }{sec\theta }=\left(\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }×\mathrm{cos}\theta \right)=\mathrm{sin}\theta$

Hence, $\frac{{x}^{2}-1}{{x}^{2}+1}=\mathrm{sin}\theta$

#### Question 29:

Solve the following system of linear equations graphically:

Shade the region bounded by these lines and the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y - axis respectively.
Graph of 2xy = 1
2xy = 1
y = (2x − 1) ........(i)
Putting x = 1, we get y = 1.
Putting x = 2, we get y = 3.
Putting x = 0, we get y = −1.
Thus, we have the following table for the equation 2xy = 1:

 x 1 2 0 y 1 3 −1

Now, plots the points A(1, 1) , B( 2, 3) and C(0, −1) on the graph paper.
Join AB and AC to get the graph line BC. Extend it  both ways.
Thus, line BC is the graph of  2xy = 1.

Graph of xy = −1
xy = −1
y = (x + 1) ............(ii)
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Putting x = 0, we get y = 1.
Thus, we have the following table for the equation xy = −1:
 x 1 2 0 y 2 3 1
Now, plot the points P(1, 2), Q(0, 1). The point B(2, 3) has already been plotted. Join PB and PQ to get the graph line BQ. Extend it both ways.
Then, the line BQ is the graph of the equation xy = 1.

The two graph lines intersect at B(2, 3).
x = 2 and y = 3 is the solution of the given system of equations.
These graph lines intersect the y-axis at C and Q.
Hence, the region bounded by these lines and the y - axis has been shaded.

#### Question 30:

The following table gives the yield per hectare of wheat of 100 farms of a village:

 Yield in kg/hectare 50-55 55-60 60-65 65-70 70-75 75-80 Number of farms 2 8 12 24 38 16

Change the above distribution to 'more than type' distribution and draw its ogive.

We may prepare the more the series as shown below.

 Production yield (lower class limits) Cumulative frequency more than or equal to 50 100 more than or equal to 55 100 − 2 = 98 more than or equal to 60 98 − 8 = 90 more than or equal to 65 90 − 12 = 78 more than or equal to 70 78 − 24 = 54 more than or equal to 75 54 − 38 = 16

Taking the lower class limits on x - axis and their respective cumulative frequencies on y - axis, its o-give can be obtained as follows:

#### Question 31:

Solve for x and y:

The equations may be written as
ax + by = a - b ..............(i)
bx - ay = a + b ..............(ii)
Multiplying (i) by a and (ii) by b, we get:
a2x + aby = a2 - ab .................(iii)
b2x - aby = ab + b2 ...................(iv)
On adding (iii) and (iv), we get:
(a2 + b2)x = a2 + b2
$⇒x=\frac{\left({a}^{2}+{b}^{2}\right)}{\left({a}^{2}+{b}^{2}\right)}=1$
Putting x = 1 in (i), we get:
(a × 1) + by = a - b
a + by = a - b
by = - b y = - 1
x = 1 and  y = - 1 is the required solution.

#### Question 33:

ΔABC is right-angled at B and D is the mid-point of BC.
Prove that: $A{C}^{2}=\left(4A{D}^{2}-3A{B}^{2}\right).$

Given: In Δ ABC, $\angle$B = 90° and D is the mid-point of BC.
To prove: AC2 = (4AD2 - 3AB2)
Proof: In Δ ABC , $\angle$B = 90°

∴ AC2 = AB2 + BC2  ( By Pythagoras' theorem)
= AB2 + (2BD)2  [ Since BC = 2BD]
= AB2 + 4BD2
= AB2 + 4(AD2 - AB2)   [ Since AB2 + BD2 = AD2]
Hence, AC2 = 4AD2 - 3AB2

#### Question 34:

Find the mean, mode and median of the following data:

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 5 10 18 30 20 12 5

We have,

(i) Mean:
Let the assumed mean, A, be 35; h = 10
$\mathrm{Mean},\overline{\mathrm{x}}=\mathrm{A}+\left\{\mathrm{h}×\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\right\}\phantom{\rule{0ex}{0ex}}$
$=35+\left\{10×\frac{6}{100}\right\}=35+0.6=35.6$
(ii) Median:

$\frac{N}{2}=\frac{100}{2}=50$
Cumulative frequency just greater than 50 is 63.
Corresponding class interval is 30 - 40.
Thus, median class is 30 - 40.
Cumulative frequency  just before this class = 33.
So, l = 30, f = 30 , $\frac{N}{2}=50$, h = 10 and cf = 33
$\mathrm{Median},{M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-cf\right)}{f}\right\}\phantom{\rule{0ex}{0ex}}$
$=30+\left\{10×\left(\frac{50-33}{30}\right)\right\}=30+\left(10×\frac{17}{30}\right)=30+\frac{17}{3}=30+5.67=35.67$
(iii) Mode:
Mode = (3 × median) - (2 × mean)
= (3 × 35.67) - (2 × 35.6)
= (107.01 - 71.2) = 35.81

Hence, Mean = 35.6, Median = 35.67 and Mode = 35.81

#### Question 32:

Prove that: $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}=\left(\mathrm{cosec\theta }-\mathrm{cot\theta }{\right)}^{2}.$

$\mathrm{RHS}={\left(\mathrm{cosec\theta }-\mathrm{cot\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}$
$={\left(\frac{1}{\mathrm{sin\theta }}-\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)}^{2}\phantom{\rule{0ex}{0ex}}$
$=\frac{{\left(1-\mathrm{cos\theta }\right)}^{2}}{{\mathrm{sin}}^{2}\mathrm{\theta }}\phantom{\rule{0ex}{0ex}}$
$=\frac{{\left(1-\mathrm{cos\theta }\right)}^{2}}{\left(1-{\mathrm{cos}}^{2}\mathrm{\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1-\mathrm{cos\theta }\right)\left(1-\mathrm{cos\theta }\right)}{\left(1-\mathrm{cos\theta }\right)\left(1+\mathrm{cos\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1-\mathrm{cos\theta }\right)}{\left(1+\mathrm{cos\theta }\right)}$

LHS = $\frac{\left(1-\mathrm{cos\theta }\right)}{\left(1+\mathrm{cos\theta }\right)}$

Hence, LHS =  RHS

#### Question 1:

What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?
(a) 15
(b) 16
(c) 9
(d) 5

(b) 16
Clearly the required number divides (245 - 5), i.e. 240 and (1029 - 5), i.e. 1024 exactly.
So, the required number is H.C.F. (240, 1024).
Now,
240 = 2 × 2 × 2 × 2 × 3 × 5 = 24 × 3 × 5
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2  = 210
Therefore, H.C.F. of 240 and 1024 = 2 × 2 × 2 × 2 = 16
Hence, 16 is the largest number that divides 245 and 1029 and leaves remainder 5 in each case.

#### Question 2:

If the product of zeros of the polynomial ax2−6x−6 is 4, then a = ?
(a) $\frac{2}{3}$
(b) $\frac{-2}{3}$
(c) $\frac{3}{2}$
(d) $\frac{-3}{2}$

(d) $\frac{-3}{2}$
Let be the zeros of ax2 - 6x - 6.
Then,

⇒ 4a = - 6
$a=\frac{-6}{4}=\frac{-3}{2}$
Hence, $a=\frac{-3}{2}$

#### Question 3:

The areas of two similar triangles ΔABC  and ΔPQR are 25 cm2 and 49 cm2, respectively, and QR = 9.8 cm. Find BC.
(a) 5 cm
(b) 8 cm
(c) 7 cm
(d) 6.3 cm

(c) 7 cm
It is given that $△ABC~△PQR$.

Also, QR = 9.8 cm. We have to find BC.
We know that the ratio of the  areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
$\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{B{C}^{2}}{Q{R}^{2}}\phantom{\rule{0ex}{0ex}}$
$⇒{\left(\frac{BC}{QR}\right)}^{2}=\frac{ar\left(△ABC\right)}{ar\left(△PQR\right)}=\frac{25}{49}={\left(\frac{5}{7}\right)}^{2}\phantom{\rule{0ex}{0ex}}$

$⇒\frac{BC}{QR}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}⇒\frac{BC}{9.8}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}$
$⇒BC=\frac{\left(5×9.8\right)}{7}=\frac{49}{7}=7$
Hence, BC = 7 cm.

#### Question 4:

If sin (θ + 34°) = cos θ and θ + 34° is acute, find θ.
(a) 56°
(b) 28°
(c) 17°
(d) 14°

(b) 28°
sin (θ + 34°) = cos θ
⇒ sin (θ + 34°) = cos (90° - θ)     [ Since cos θ  = sin (90° - θ)]
⇒ (θ + 34°) = (90° - θ)
⇒ 2θ = 56°
⇒ θ = $\frac{56}{2}=28°$

#### Question 5:

If cos θ = 0.6, find 5 sin θ − 3 tan θ.
(a) 0.4
(b) 0.2
(c) 0.3
(d) 0

(d) 0
We have:
cos θ = $\frac{6}{10}=\frac{3}{5}$
$\frac{AB}{AC}=\frac{3}{5}=\frac{3k}{5k}$

BC2 = (AC2 - AB2)
= (25k2 - 9k2)
= 16k2
$BC=\sqrt{16{k}^{2}}=4k$
∴ sin θ = $\frac{BC}{AC}=\frac{4k}{5k}=\frac{4}{5}$,  tan θ = $\frac{BC}{AB}=\frac{4k}{3k}=\frac{4}{3}$
Hence,
5 sin θ − 3 tan θ
$=5×\frac{4}{5}-3×\frac{4}{3}\phantom{\rule{0ex}{0ex}}=4-4\phantom{\rule{0ex}{0ex}}=0$

#### Question 6:

The simplest form of $\frac{1095}{1168}$ is
(a) $\frac{17}{26}$
(b) $\frac{25}{26}$
(c) $\frac{15}{16}$
(d) $\frac{13}{16}$

(c) $\frac{15}{16}$

$\frac{1095}{1168}=\frac{3×5×73}{2×2×2×2×73}=\frac{15}{16}$

#### Question 7:

The pair of linear equations 4x − 5y −20 = 0 and 3x + 5y − 15 = 0 has
(a) a unique solution
(b) two solutions
(c) many solutions
(d) no solution

(a) a unique solution
The given equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where
a1 = 4, b1 = - 5, c1 = - 20 and a2 = 3, b2 = 5 and c2 = - 15
$\frac{{a}_{1}}{{a}_{2}}=\frac{4}{3}$, $\frac{{b}_{1}}{{b}_{2}}=\frac{-5}{5}$ and $\frac{{c}_{1}}{{c}_{2}}=\frac{-20}{-15}=\frac{4}{3}$
∴ These graph lines will intersect at a unique point, when $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.
Hence, the given system has a unique solution.

#### Question 8:

If mode = x(median) −y (mean), then
(a) x = 2, y = 3
(b) x = 3, y = 2
(c) x = 4, y = 3
(d) x = 3, y = 4

(b) x = 3, y = 2
The relationship among mean, mode and median:
Mode = 3(Median) - 2(Mean)
Hence, by the above formula, x = 3 and  y = 2.

#### Question 9:

Check whether 6n  can end with the digit 0? Justify your answer.

If any number ends with the digit 0, it should be divisible by 10. In other words, its prime factorisation must include primes 2 and 5.
Prime factorisation of 6n = (2 x 3)n
By the Fundamental Theorem of Arithmetic, prime factorisation of a number is unique.
5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Hence, 6n cannot end with the digit 0 for any natural number n.

#### Question 10:

Find the zeros of the polynomial 9x2 − 5 and verify the relation between zeros and coefficients.

$f\left(x\right)=9{x}^{2}-5={\left(3x\right)}^{2}-{\left(\sqrt{5}\right)}^{2}=\left(3x+\sqrt{5}\right)\left(3x-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}$
$f\left(x\right)=0⇒\left(3x+\sqrt{5}\right)\left(3x-\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}$

So, the zeros of f(x) are $-\frac{\sqrt{5}}{3}\mathrm{and}\frac{\sqrt{5}}{3}$.

#### Question 11:

If 2 sin 2θ = $\sqrt{3}$, find the value of θ.
Or, If 7 sin2θ + 3 cos2θ = 4, show that tan θ = $\frac{1}{\sqrt{3}}$.

$2\mathrm{sin}2\theta =\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}2\theta =\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}2\theta =\mathrm{sin}60°\phantom{\rule{0ex}{0ex}}$

$⇒2\theta =60°\phantom{\rule{0ex}{0ex}}⇒\theta =30°$

OR
7 sin2θ + 3 cos2θ = 4
⇒ 4 sin2θ + 3 sin2θ + 3 cos2θ = 4
⇒ 4 sin2θ + 3(sin2θ + cos2θ) = 4
⇒ 4 sin2θ + 3 = 4    [ since sin2θ + cos2θ = 1]
⇒ 4 sin2θ = 1
$⇒{\mathrm{sin}}^{2}\theta =\frac{1}{4}$
${\mathrm{cos}}^{2}\theta =\left(1-{\mathrm{sin}}^{2}\theta \right)=\left(1-\frac{1}{4}\right)=\frac{3}{4}\phantom{\rule{0ex}{0ex}}$
${\mathrm{tan}}^{2}\theta =\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }=\left(\frac{1×4}{4×3}\right)=\frac{1}{3}\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{tan}\theta =\frac{1}{\sqrt{3}}$

#### Question 12:

In ΔABC, D and E are points on AB and AC, respectively, such that AD = 5 cm, DB = 8 cm and DEBC. If AC = 6.5 cm, find AE.

Let AE be  x cm.
Then, EC = (AC - AE) cm = (6.5 - x) cm
In Δ ABC, $\mathrm{DE}\parallel \mathrm{BC}$
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$    [By Thales' theorem]
$⇒\frac{5}{8}=\frac{x}{\left(6.5-x\right)}\phantom{\rule{0ex}{0ex}}$
$⇒5\left(6.5-x\right)=8x\phantom{\rule{0ex}{0ex}}$
$⇒13x=32.5\phantom{\rule{0ex}{0ex}}$
$⇒x=\frac{32.5}{13}=2.5$
Hence, AE = 2.5 cm.

#### Question 13:

D is a point on the side BC of ΔABC, such that ∠ADC and ∠BAC are equal.
Prove that: CA2 = DC × CB.

Given: In ΔABC, D is a point on BC, such that $\angle$ADC = $\angle$BAC.
To prove: CA2 = DC × CB.
Proof: In ΔABC and ΔDAC, we have:
$\angle$BAC = $\angle$ADC  (given)
$\angle$ACB = $\angle$DCA  (common)
∴ Δ ABC ∼  Δ DAC   [ By AA similarity]
So, the sides of ΔABC and ΔDAC are proportional.
$\frac{CA}{CB}=\frac{CD}{CA}$
Hence, CA2 = DC × CB

#### Question 14:

Calculate the mode for the following frequency distribution:

 Class interval 0-4 4-8 8-12 12-16 Frequency 4 8 5 6

As the class 4 - 8 has maximum frequency, it is the modal class.
Therefore,
xk = 4, h = 4, fk = 8, fk-1 = 4, fk+1 = 5
$\mathrm{Mode},{\mathrm{M}}_{0}={\mathrm{x}}_{\mathrm{k}}+\left\{\mathrm{h}×\frac{\left({\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}-1}\right)}{\left(2{\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}-1}-{\mathrm{f}}_{\mathrm{k}+1}\right)}\right\}\phantom{\rule{0ex}{0ex}}$
$=4+\left\{4×\frac{\left(8-4\right)}{\left(2×8-4-5\right)}\right\}\phantom{\rule{0ex}{0ex}}$
$=4+\left\{4×\frac{4}{16-9}\right\}\phantom{\rule{0ex}{0ex}}=4+\left\{4×\frac{4}{7}\right\}\phantom{\rule{0ex}{0ex}}=\left(4+\frac{16}{7}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{28+16}{4}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{44}{7}\right)\phantom{\rule{0ex}{0ex}}=6.29$

#### Question 15:

Show that any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5), where q is some integer.

Let a be any odd positive integer. We have to prove that a is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Since a is an integer, consider b = 6 as another integer.
By applying Euclid's division lemma, we get:
a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However, since a is odd, it cannot take the values 6q, 6q + 2 and 6q + 4 (since these are divisible by 2)
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer.
6q + 3 = (6q + 2) + 1 = 2(3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.
Hence, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

#### Question 16:

Prove that $\left(3-\sqrt{5}\right)$ is irrational.
Or, prove that $\frac{2\sqrt{2}}{3}$ is irrational.

Let us assume that $\left(3-\sqrt{5}\right)$ is rational.
That is, we can find integers p and q(≠ 0) such that:

$⇒\sqrt{5}=3-\frac{\mathrm{p}}{\mathrm{q}}$
Since, p and q are integers, we get $3-\frac{p}{q}$ is rational; so, $\sqrt{5}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational.
So, we conclude that$\left(3-\sqrt{5}\right)$ is irrational.

OR
Let us assume that $\frac{2\sqrt{2}}{3}$ is rational. That is, we can find co -prime integers p and q(≠ 0), such that

$⇒\sqrt{2}=\frac{3p}{2q}$
Since, p and q are integers, $\frac{3p}{2q}$ is rational; so $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
So, we conclude that$\frac{2\sqrt{2}}{3}$ is irrational.

#### Question 17:

What number must be added to each of 5, 9, 17, 27 so that the new numbers are in proportion?
Or, the sum of two numbers is 18 and the sum of their reciprocals is $\frac{1}{4}$. Find the numbers.

Let the required number be x.

Then, according to proportionality rule, we get:
$\frac{5+x}{9+x}=\frac{17+x}{27+x}\phantom{\rule{0ex}{0ex}}$
⇒ (5 + x)(27 + x) = (17 + x)(9 + x)
⇒ 135 + 5x + 27x + x2 = 153 + 17x + 9x + x2
⇒ 135 + 32x = 153 + 26x
⇒ 32x - 26x = 153 - 135
⇒ 6x = 18
x = 3
Hence, the required number is 3.

OR
Let the required numbers be x and y.
Then, x + y = 18
and $\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\left(x+y\right)}{xy}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{18}{xy}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒xy=72\phantom{\rule{0ex}{0ex}}$
$\left(x-y\right)=\sqrt{{\left(x+y\right)}^{2}-4xy}\phantom{\rule{0ex}{0ex}}$
$=\sqrt{{18}^{2}-4×72}\phantom{\rule{0ex}{0ex}}=\sqrt{324-288}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=±6$
x + y = 18 .....(i)
and x - y = 6 ......(ii)
Or
x + y = 18 .....(iii)
and x - y = - 6 ......(iv)
On solving (i) and (ii), we get: x = 12 and  y = 6.
On solving (iii) and (iv), we get: x = 6 and  y = 12.
Hence, the required numbers are 12 and 6.

#### Question 18:

If α, β are the zeros of the polynomial (x2x − 12), then form a quadratic equation whose zeros are 2α and 2β.

f(x) = x2 - x - 12 is the given polynomial.
It is given that,α and β are the zeros of the polynomial.

Let 2α and 2β be the zeros of the polynomial g(x).
Thus,
Sum of roots = 2α and 2β = 2(α + β) = (2 × 1) = 2
Product of roots = 2α × 2β = 4αβ = 4(-12) = - 48
Now,
The required polynomial g(x) = x2 - (Sum of roots)x + Product of roots
= x2 - (2)x - 48
= x2 - 2x - 48
Hence, the required polynomial is g(x) = x2 - 2x - 48.

#### Question 19:

Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2θ + cot2θ.

LHS = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2sinθcosecθ) + (cos2θ + sec2θ + 2cosθsecθ)
= (sin2θ + cosec2θ + 2) + (cos2θ + sec2θ + 2)    [since sinθcosecθ = 1 and cosθsecθ = 1]
= (sin2θ + cos2θ) + 4 + (cosec2θ + sec2θ)
= 1 + 4 + (1 + cot2θ) + ( 1 + tan2θ)     [Since (sin2θ + cos2θ) = 1 , cosec2θ = (1 + cot2θ) and sec2θ = ( 1 + tan2θ)
= 7 + cot2θ + tan2θ
RHS = 7 + cot2θ + tan2θ
Hence, LHS = RHS

#### Question 20:

If sec θ tan θ = m, show that $\left(\frac{{m}^{2}-1}{{m}^{2}+1}\right)=\mathrm{sin\theta }.$

$\frac{{\mathrm{m}}^{2}-1}{{\mathrm{m}}^{2}+1}=\frac{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^{2}-1}{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^{2}+1}\phantom{\rule{0ex}{0ex}}$

$=\frac{2{\mathrm{tan}}^{2}\theta +2sec\theta \mathrm{tan}\theta }{2se{c}^{2}\theta +2sec\theta \mathrm{tan}\theta }$   [Since, sec2θ-1=tan2θ and tan2θ+1=sec2θ ]

$=\frac{2\mathrm{tan}\theta \left(\mathrm{tan}\theta +sec\theta \right)}{2sec\theta \left(\mathrm{tan}\theta +sec\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{tan}\theta }{sec\theta }\phantom{\rule{0ex}{0ex}}=\left(\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }×\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}\theta$
Hence, $\frac{{\mathrm{m}}^{2}-1}{{\mathrm{m}}^{2}+1}=\mathrm{sin\theta }$

#### Question 21:

In a trapezium ABCD, O is the point of intersection of AC and BD, ABCD and AB = 2 × CD. If area of AOB = 84 cm2, find the area of ΔCOD.

In Δ AOB and Δ COD, we have:
$\angle$OAB = $\angle$OCD    [ Alternate interior angles]
$\angle$OBA = $\angle$ODC    [ Alternate interior angles]
∴ ΔAOB ∼ Δ COD   [ By AA similarity]
$\frac{ar△AOB}{ar△COD}=\frac{A{B}^{2}}{C{D}^{2}}=\frac{{\left(2CD\right)}^{2}}{C{D}^{2}}$   [ Since AB = 2 × CD]
$=\frac{4×C{D}^{2}}{C{D}^{2}}=4\phantom{\rule{0ex}{0ex}}$

Hence, area of ΔCOD is 21 cm2.

#### Question 22:

In the given figure, ABBC, GFBC and DEAC. Prove that ∆ADE ∼ ∆GCE.

We have:

In ΔABC,
$\angle$1 + $\angle$4 = 90°
In ΔGFC,
$\angle$1 + $\angle$2 =90°
Hence, $\angle$1 + $\angle$4 = $\angle$1 + $\angle$2
$\angle$4 = $\angle$2
$\angle$A = $\angle$G ..........(i)
Also $\angle$E = $\angle$F     ( Each measures 90°) ................(ii)
From (i) and (ii), we get AA similarity for ΔADE and ΔGCF.

#### Question 23:

Find the mean of the following distribution, using the step deviation method:

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 7 12 13 10 8

Or, the mean of the following distribution is 78. Find the value of P.
 Class 50-60 60-70 70-80 80-90 90-100 Frequency 8 6 12 11 p

Let the assumed mean A be 25; h = 10
For calculating the mean, we prepare the table given below.

$\mathrm{Mean},\overline{\mathrm{x}}=\mathrm{A}+\left\{\mathrm{h}×\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\right\}\phantom{\rule{0ex}{0ex}}$
$=25+\left\{10×\frac{0}{50}\right\}=25+0=25$
Hence, the mean of the given frequency is 25

OR
We have:

Therefore,
$\mathrm{Mean},\overline{\mathrm{x}}=\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\left(2665+95p\right)}{\left(37+p\right)}=78\phantom{\rule{0ex}{0ex}}$
$⇒\left(2665+95p\right)=2886+78p\phantom{\rule{0ex}{0ex}}$
$⇒17p=\left(2886-2665\right)=221\phantom{\rule{0ex}{0ex}}$
$⇒p=\frac{221}{17}=13$
Hence, the value of p is 13.

#### Question 24:

Find the median of the following data:

 Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 Number of students 2 3 8 6 6 3 2

We prepare the cumulative frequency table, as given below:

$\frac{N}{2}=\frac{30}{2}=15$
Cumulative frequency just greater than 15 is 19.
Corresponding class interval is 55 - 60.
Thus, median class is 55 - 60.
Cumulative frequency  just before this class = 13
So, l = 55, f = 6 , $\frac{N}{2}=15$, h = 5 and cf = 13
$\mathrm{Median},{M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-cf\right)}{f}\right\}\phantom{\rule{0ex}{0ex}}$
$=55+\left\{5×\left(\frac{15-13}{6}\right)\right\}\phantom{\rule{0ex}{0ex}}=55+\frac{5×2}{6}\phantom{\rule{0ex}{0ex}}=55+1.67\phantom{\rule{0ex}{0ex}}=56.67$
Hence, median = 56.67 kg

#### Question 25:

If two zeros of the polynomial p(x) = then find the other two zeroes.

The given polynomial is f(x) = 2x4 + 7x3 - 19x2 - 14x + 30
Since, $\sqrt{2}\mathrm{and}-\sqrt{2}$ are are two zeros of f(x), it follows that each one of  is a factor of f(x).
Consequently, $\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=\left({x}^{2}-2\right)$ is a factor of f(x).
On dividing f(x) by (x2 - 2), we get:

f(x) = 0
⇒ (x2 - 2)(2x2 + 7x - 15)=0
⇒ (x2 - 2)(2x2 + 7x - 15) = 0
⇒ (x2 - 2)(2x - 3)(x + 5) = 0
$\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(2x-3\right)\left(x+5\right)=0$

Hence, the other two zeros are .

#### Question 26:

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of an equilateral triangle described on one of its diagonals.
Or, prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Let ABCD be a square of side a.
Therefore, its diagonal is $\sqrt{2}a$.

Two equilateral triangles Δ ABE and Δ DBF are drawn.
Side of equilateral triangle, ΔABE, described on one of its sides = a
Side of equilateral triangle, ΔDBF, described on one of its diagonals= $\sqrt{2}a$
We know that all the angles of an equilateral triangle measure 60º and all its sides are equal.
Therefore, all equilateral triangles are similar to each other.
Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
$\frac{\mathrm{Areaof}△\mathrm{ABE}}{\mathrm{Areaof}△\mathrm{DBF}}=\frac{\frac{\sqrt{3}}{4}A{B}^{2}}{\frac{\sqrt{3}}{4}B{D}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{a}}^{2}}{2{\mathrm{a}}^{2}}=\frac{1}{2}$

OR
Given: Δ ABC ∼ Δ DEF
To Prove: $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{DE}}^{2}}=\frac{{\mathrm{AC}}^{2}}{{\mathrm{DF}}^{2}}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}$

Construction: .

Proof:
Since Δ ABC ∼ Δ DEF, it follows that they are equiangular and their sides are proportional.
$\angle$A = $\angle$D, $\angle$B = $\angle$E, $\angle$C = $\angle$F
and $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ ..................(i)
$\mathrm{ar}\left(△\mathrm{ABC}\right)=\left(\frac{1}{2}×\mathrm{BC}×\mathrm{AL}\right)\phantom{\rule{0ex}{0ex}}$
$\mathrm{ar}\left(△\mathrm{DEF}\right)=\left(\frac{1}{2}×\mathrm{EF}×\mathrm{DM}\right)\phantom{\rule{0ex}{0ex}}$
$⇒\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{\left(\frac{1}{2}×\mathrm{BC}×\mathrm{AL}\right)}{\left(\frac{1}{2}×\mathrm{EF}×\mathrm{DM}\right)}=\frac{\mathrm{BC}}{\mathrm{EF}}×\frac{\mathrm{AL}}{\mathrm{DM}}$ ...............(ii)
In Δ ALB and Δ DME, we have:
∠ALB = ∠DME = 90° and $\angle$B = $\angle$∠E    [ from (i)]
$\angle$ALB ∼ $\angle$DME

Consequently, $\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AB}}{\mathrm{DE}}$
[from (i)]

$\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{BC}}{\mathrm{EF}}$..................(iii)
Using (iii) in (ii), we get:
$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}×\frac{\mathrm{BC}}{\mathrm{EF}}\right)=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}\phantom{\rule{0ex}{0ex}}$
Similarly,

Hence, $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DEF}\right)}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{DE}}^{2}}=\frac{{\mathrm{AC}}^{2}}{{\mathrm{DF}}^{2}}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}$

#### Question 27:

Prove that: $\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}=\frac{\mathrm{cos\theta }}{1-\mathrm{sin\theta }}$
Or, Evaluate:

$\mathrm{LHS}=\left(\frac{\mathrm{sec\theta }+\mathrm{tan\theta }-1}{\mathrm{tan\theta }-\mathrm{sec\theta }+1}\right)\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)-\left({\mathrm{sec}}^{2}\mathrm{\theta }-{\mathrm{tan}}^{2}\mathrm{\theta }\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}\phantom{\rule{0ex}{0ex}}$        [Since, 1 = sec2θ -  tan2θ]
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left[1-\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)\right]}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}{\left(\mathrm{tan\theta }-\mathrm{sec\theta }+1\right)}=\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)$
$=\left(\frac{1}{\mathrm{cos\theta }}+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\mathrm{sin\theta }\right)}{\mathrm{cos\theta }}×\frac{\left(1-\mathrm{sin\theta }\right)}{\left(1-\mathrm{sin\theta }\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(1-{\mathrm{sin}}^{2}\mathrm{\theta }\right)}{\mathrm{cos\theta }\left(1-\mathrm{sin\theta }\right)}=\frac{{\mathrm{cos}}^{2}\mathrm{\theta }}{\mathrm{cos\theta }\left(1-\mathrm{sin\theta }\right)}=\frac{\mathrm{cos\theta }}{\left(1-\mathrm{sin\theta }\right)}$

$\mathrm{RHS}=\frac{\mathrm{cos\theta }}{\left(1-\mathrm{sin\theta }\right)}$
Hence, LHS = RHS

OR

$\frac{\mathrm{sec\theta cosec}\left(90°-\mathrm{\theta }\right)-\mathrm{tan\theta cot}\left(90°-\mathrm{\theta }\right)+{\mathrm{sin}}^{2}55°+{\mathrm{sin}}^{2}35°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°\mathrm{tan}70°\mathrm{tan}80°}\phantom{\rule{0ex}{0ex}}$

$=\frac{{\mathrm{sec}}^{2}\mathrm{\theta }-{\mathrm{tan}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\left(90°-35°\right)+{\mathrm{sin}}^{2}35°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°\mathrm{tan}\left(90°-20°\right)\mathrm{tan}\left(90°-10°\right)}\phantom{\rule{0ex}{0ex}}$

$=\frac{1+{\mathrm{cos}}^{2}35°+{\mathrm{sin}}^{2}35°}{\mathrm{tan}10°\mathrm{tan}20°\mathrm{tan}60°×\mathrm{cot}20°×\mathrm{cot}10°}\phantom{\rule{0ex}{0ex}}$        [Since, sec2θ -  tan2θ = 1]

$=\frac{1+1}{\mathrm{tan}10°\mathrm{tan}20°×\sqrt{3}×\frac{1}{\mathrm{tan}20°}×\frac{1}{\mathrm{tan}10°}}\phantom{\rule{0ex}{0ex}}$     [Since, cos2θ + sin2θ = 1]
$=\frac{2}{\sqrt{3}}$

#### Question 28:

If sec θ + tan θ = m, prove that sin $\mathrm{\theta }=\frac{\left({m}^{2}-1\right)}{\left({m}^{2}+1\right)}.$

$\frac{{\mathrm{m}}^{2}-1}{{\mathrm{m}}^{2}+1}=\frac{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^{2}-1}{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^{2}+1}\phantom{\rule{0ex}{0ex}}$
$=\frac{{\mathrm{sec}}^{2}\mathrm{\theta }+{\mathrm{tan}}^{2}\mathrm{\theta }+2\mathrm{sec\theta tan\theta }-1}{{\mathrm{sec}}^{2}\mathrm{\theta }+{\mathrm{tan}}^{2}\mathrm{\theta }+2\mathrm{sec\theta tan\theta }+1}\phantom{\rule{0ex}{0ex}}$      [Since sec2θ - 1 = tan2θ and tan2θ + 1 = sec2θ]
$=\frac{2{\mathrm{tan}}^{2}\mathrm{\theta }+2\mathrm{sec\theta tan\theta }}{2{\mathrm{sec}}^{2}\mathrm{\theta }+2\mathrm{sec\theta tan\theta }}\phantom{\rule{0ex}{0ex}}$
$=\frac{2\mathrm{tan\theta }\left(\mathrm{tan\theta }+\mathrm{sec\theta }\right)}{2\mathrm{sec\theta }\left(\mathrm{tan\theta }+\mathrm{sec\theta }\right)}=\frac{\mathrm{tan\theta }}{\mathrm{sec\theta }}\phantom{\rule{0ex}{0ex}}$
$=\left(\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}×\mathrm{cos}\mathrm{\theta }\right)=\mathrm{sin}\mathrm{\theta }$
Hence, $\frac{{\mathrm{m}}^{2}-1}{{\mathrm{m}}^{2}+1}=\mathrm{sin\theta }$

#### Question 29:

Draw the graph of the following equations:

Shade the region bounded by these lines and the y-axis.

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis respectively.
Graph of 3x + y - 11= 0

3x + y = 11
y = (11- 3x) ...........(i)
Putting x = 2, we get y = 5.
Putting x =  3, we get y = 2.
Putting x = 5, we get y = - 4.
Thus, we have the following table for the equation 3x + y = 11:

 x 2 3 5 y 5 2 -4

Now, plots the points A(2, 5) , B(3, 2) and C(5, - 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it both ways.
Thus, line AC is the graph of  3x + y = 11.

Graph of x  y = 1
x y  = 1
y = (x - 1) .............(ii)
Putting x = −3, we get y = −4.
Putting x = 0, we get y = - 1.
Putting x = 3, we get y = 2.
Thus, we have the following table for the equation x y  = 1:
 x −3 0 3 y −4 -1 2
Now, plots the points P(-3, -4) and Q(0, −1). The point B(3, 2) has already been plotted. Join PQ and QB and extend it both ways.
Thus, line PB is the graph of  x y  = 1.

The two graph lines intersect at B(3 , 2).
x = 3 and y = 2 is the solution of the given system of equations.
The region bounded by these lines and the y - axis has been shaded.

#### Question 30:

The table given below shows the frequency distribution of the scores obtained by 200 candidates in a BCA entrance examination:

 Scores 200-250 250-300 300-350 350-400 400-450 450-500 500-550 550-600 No. of candidates 30 15 45 20 25 40 10 15

Draw a cumulative frequency curve using 'Less than' series.

Less than Series:
We may prepare the 'Less than' series as shown:

 Score Number of candidates Less than 250 30 Less than 300 45 Less than 350 90 Less than 400 110 Less than 450 135 Less than 500 175 Less than 550 185 Less than 600 200

Scale:
Along the x - axis, 1 big division = 50
Along the y - axis, 1 big division = 20
We plot the points A(250,30) , B(300, 45), C(350, 90), D(400, 110), E(450, 135), F(500, 175), G(550, 185) and H(600, 200) on a graph paper.
Join, AB, BC, CD, DE, EF, FG and GH with a free hand to get the curve representing the 'Less than' series.

#### Question 31:

For what value of k will the following pair of linear equations have infinitely many solutions?

The given equations are:
2x - 3y - 7 = 0
(k + 1)x + (1 - 2k)y - (- 5k + 4) = 0
The given equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where
a1 = 2, b1 = - 3, c1 = - 7 and a2 = (k + 1), b2 = (1 - 2k) and c2 = (4 - 5k)
The given pair of equations can have infinitely many solutions if
$\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$
Or
$\frac{2}{k+1}=\frac{-3}{\left(1-2k\right)}=\frac{-7}{\left(4-5k\right)}\phantom{\rule{0ex}{0ex}}$

⇒ 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7
k = 5 and k = 5
Hence, k = 5.

#### Question 32:

Prove that: $\left(\mathrm{sin\theta }-\mathrm{cosec\theta }\right)\left(\mathrm{cos\theta }-\mathrm{sec\theta }\right)=\frac{1}{\left(\mathrm{tan\theta }+\mathrm{cot\theta }\right)}.$

$\mathrm{LHS}=\left(\mathrm{sin\theta }-\mathrm{cosec\theta }\right)\left(\mathrm{cos\theta }-\mathrm{sec\theta }\right)\phantom{\rule{0ex}{0ex}}$
$=\left(\mathrm{sin\theta }-\frac{1}{\mathrm{sin\theta }}\right)\left(\mathrm{cos\theta }-\frac{1}{\mathrm{cos\theta }}\right)\phantom{\rule{0ex}{0ex}}=\frac{\left({\mathrm{sin}}^{2}\mathrm{\theta }-1\right)}{\mathrm{sin\theta }}×\frac{\left({\mathrm{cos}}^{2}\mathrm{\theta }-1\right)}{\mathrm{cos\theta }}\phantom{\rule{0ex}{0ex}}$

Now, (1- sin2θ) =  cos2θ and (1- cos2θ) = sin2θ

$\mathrm{RHS}=\frac{1}{\mathrm{tan\theta }+\mathrm{cot\theta }}$
$=\frac{1}{\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}$

$=c\mathrm{os\theta sin\theta }$
Hence, LHS =  RHS

#### Question 33:

ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that
ABC is a right triangle.

Given that
AB2 = 2AC2
⇒ AB2 = AC2 + AC2
⇒ AB2 = AC2 + BC2   [ Since BC =  AC (given)]

Thus, the triangle satisfies Pythagoras' theorem.
Hence, the given triangle is a right-angled triangle.

#### Question 34:

The table given below shows the daily expenditure on food of 30 households in a locality:

 Daily expenditure (in Rs) Number of households 100-150 6 150-200 7 20-250 12 250-300 3 300-350 2

Find the mean and median of daily expenditure on food.

Let the assumed mean A be 225; h = 50.
$\mathrm{Mean},\overline{\mathrm{x}}=\mathrm{A}+\left\{\mathrm{h}×\frac{\sum \left({\mathrm{f}}_{\mathrm{i}}×{\mathrm{u}}_{\mathrm{i}}\right)}{\sum {\mathrm{f}}_{\mathrm{i}}}\right\}\phantom{\rule{0ex}{0ex}}$
$=225+\left\{50×\frac{-12}{30}\right\}\phantom{\rule{0ex}{0ex}}=225-20\phantom{\rule{0ex}{0ex}}=205$
Now,
$\frac{N}{2}=\frac{30}{2}=15$
Cumulative frequency just after 15 is 25.
Corresponding class interval is 200 - 250.
Median class is 200 - 250.
Cumulative frequency  just before this class = 13
So, l = 200, f = 12 , $\frac{N}{2}=15$, h = 50 and cf = 13
$\mathrm{Median},{M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-cf\right)}{f}\right\}\phantom{\rule{0ex}{0ex}}$
$=200+50\left(\frac{15-13}{12}\right)\phantom{\rule{0ex}{0ex}}=200+\frac{50×2}{12}\phantom{\rule{0ex}{0ex}}=200+\frac{25}{3}\phantom{\rule{0ex}{0ex}}=200+8.33\phantom{\rule{0ex}{0ex}}=208.33$
Hence, mean = 205 and median = 208.33

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