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Page No 422:

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

Answer:

(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

Page No 422:

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

Answer:

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

Page No 422:

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

Answer:

(b) 10 cm
In Δ ABC, DEBC. Then,
ADDB=AEEC      ( By Thales' theorem)
BDAD=ECAE
BDAD+1=ECAE+1
BD+ADAD=EC+AEAE
ABAD=ACAE   (Since BD + AD = AB and EC + AE = AC)
and A=A              (Common)
ABC~ADE      (SAS similarity)
Thus,
ABAD=BCDE     [In similar triangles, corresponding sides are similar]
53=BC6
BC=6×53=10 cm



Page No 423:

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°

Answer:

(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 = 924=23°

Page No 423:

Question 5:

If tan θ=3, thensec2θ-cosec2θsec2θ+cosec2θ=?
(a) −1
(b) 1
(c) -12
(d) 12

Answer:

(d) 12
Here,
tanθ=3
ACBC=31
By Pythagoras' theorem, we get AB = 1.

Now, secθ=ABBC=21 and cosecθ=ABAC=23
sec2θ-cosec2θsec2θ+cosec2θ=22-23222+232=4-434+43=12-412+4=816=12

Page No 423:

Question 6:

After how many places of decimal, will the decimal expansion of 4940 terminate ?
(a) 1
(b) 2
(c) 3
(d) It will not terminate

Answer:

(c) 3

We have:
4940×2525=12251000=1225103
So, it will terminate after 3 decimal places.

Page No 423:

Question 7:

The pair of linear equations 6x-3y+10=0, 2x-y+9=0 has
(a) one solution
(b) two solutions
(c) many solutions
(d) no solution

Answer:

(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,
a1a2=b1b2c1c2

62=-3-1109

31=31109

Hence, the given system of equations has no solution.

Page No 423:

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

Answer:

(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.

Page No 423:

Question 9:

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

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.

Page No 423:

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.

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).

Page No 423:

Question 11:

If 3 cos2θ + 7 sin2θ = 4, show that cot θ = 3.
Or, if tan θ=815, evaluate(2+2sinθ)(1-sinθ)(1+cosθ)(2-2cosθ).

Answer:

We have:
3cos2θ+ 7sin2θ=4
3cos2θ+3sin2θ+4sin2θ=4
3cos2θ+sin2θ+4sin2θ=4     [ Since cos2θ +  sin2θ = 1]
3×1+4sin2θ=4
4sin2θ=1sin2θ=14
cos2θ=1-sin2θ=1-14=34
tan2θ=sin2θcos2θ=14×43=13
tanθ=13cotθ=1tanθ=3

OR
We have:
2+2sinθ1-sinθ1+cosθ2-2cosθ=21+sinθ1-sinθ21+cosθ1-cosθ=1-sin2θ1-cos2θ=cos2θsin2θ=cot2θ=cotθ2
Given that,
tanθ=815cotθ=158
cotθ2=1582=158×158=22564
Hence,
2+2sinθ1-sinθ1+cosθ2-2cosθ=22564

Page No 423:

Question 12:

In the given figure, DE ∥ AC and DF AE.
Prove that ECBE=FEBF.

Answer:

In ΔBAE, DFAE
∴  ADDB=EFFB ............(i)   [by Thale's theorem]
In ΔBAC, DEAC
ADDB=CEEB ..............(ii)     [by Thale's theorem]
From (i) and (ii), we get:
EFFB=CEEB     Each equal toADDB
ECBE=FEBF



Page No 424:

Question 13:

In the given figure, AD ⊥ BC and BD=13CD.
Prove that 2CA2=2AB2+BC2.

Answer:

Given: In ΔABC, AD⊥BC and BD=13CD.
To prove: 2CA2 = 2AB2 + BC2
Proof:
BD=13CD ............(i)
∴ BC =  BD + CD
BC=13CD+CD=43CD
or,CD = 34BC
and BD=13CD=13×34BC=14BC
In Δ ADC, ADC= 90°.AC2=AD2+DC2...(i)    (By Pythagoras' theroem)In Δ ADB,ADB = 90°.AB2=AD2+DB2AD2=AB2-DB2By substituting the value of AD2 in (i) we get:AC2=AB2-DB2+DC2Again substituting the values of DB and DC, we get:AC2=AB2-116BC2+916BC2AC2=AB2+916-116BC2AC2=AB2+12BC22AC2=2AB2+BC2

Page No 424:

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

Answer:

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
Mode,M0=xk+h×fk-fk-12fk-fk-1-fk+1
=30+10×32-122×32-12-20
=30+10×2064-32=30+10×2032=30+254=1454=36.25

Page No 424:

Question 15:

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

Answer:

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.

Page No 424:

Question 16:

Prove that (5-3) is irrational.
Or, prove that 335 is irrational.

Answer:

Let us assume that 5-3 is rational. That is, we can find integers p and q(≠ 0) such that
5-3=pq or 5-pq=3
3=5-pq
Since, p and q are integers, 5-pq is rational; so, 3 is rational.
But this contradicts the fact that 3 is irrational.
So, we can conclude that 5-3 is irrational.

OR
Let us assume that 335 is rational. That is, we can find co -prime integers p and q(≠ 0) such that 335=pq or 5p3q=3
3=5p3q
Since, p and q are integers, 5p3q is rational; so, 3 is rational.
But this contradicts the fact that 3 is irrational.
So, we can conclude that335 is irrational.

Page No 424:

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?

Answer:

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 = 304+yhr
⇒ 30 =(4 + y)t
⇒ 30 = 4t + yt ......(i)
Time taken by boat to cover 30 km upstream = 3t = 304-yhr
⇒ 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.

Page No 424:

Question 18:

If α and β are the zeros of the polynomial 2x2+x-6, then form a quadratic equation whose zeros are 2α and 2β.

Answer:

f(x) = 2x2 + x - 6 is the given polynomial.
It is given that,α and β are the zeros of the polynomial.
α+β=-12 and αβ=-62=-3
Let 2α and 2β be the zeros of the polynomial g(x).
Thus,
Sum of roots = 2α and 2β = 2(α + β) = 2×-12=-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

Page No 424:

Question 19:

Prove that (cosec θ − sin θ)(sec θ − cos θ) = 1tanθ+cotθ.

Answer:

LHS = (cosec θ − sin θ)(sec θ − cos θ)
 =1sinθ-sinθ1cosθ-cosθ=1-sin2θsinθ×1-cos2θcosθ
=cos2θsin2θsinθcosθ      [Since (1 - sin2θ) =  cos2θ and (1 - cos2θ) = sin2θ]
=cosθsinθ
RHS=1tanθ+cotθ
=1sinθcosθ+cosθsinθ=cosθsinθsin2θ+cos2θ
=cosθsinθ1=cosθsinθ   [ Since sin2θ + cos2θ = 1]
Hence, LHS =  RHS

Page No 424:

Question 20:

If cos θ + sin θ = 2cosθ, prove that cos θ − sin θ = 2sinθ.

Answer:

We have:
(cos θ + sin θ)2 + (cos θ - sin θ)2 = 2(cos2θ + sin2θ)
Given that:
cosθ+sinθ=2cosθ
Thus,
2cosθ2+cosθ-sinθ2=2            [since cos2θ + sin2θ = 1]
cosθ-sinθ2=2-2cos2θ=21-cos2θ=2sin2θ      [Since (1 - cos2θ) = sin2θ]
Taking square root of both the sides, we get:
cosθ-sinθ=2sinθ

Page No 424:

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 ar(ABC)ar(DBC)=AODO.

Answer:

Given: ΔABC and ΔDBC are on the same base BC and AD intersects BC at O.
To prove: arABCarDBC=AODO
Construction : Draw AL⊥BC and DM⊥BC.

Proof: ALO = DMO = 90°
AOL = DOM (vertically-opposite angles)
∴ Δ ALO ∼Δ DMO  [ By AA - similarity]
ALDM=AODO ...................(i)
Therefore,
arABCarDBC=12×BC×AL12×BC×DM=ALDM=AODO      [ using(i)]
Hence,arABCarDBC=AODO



Page No 425:

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=13AC.

Answer:

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: AF=13AC
Construction:
From D, draw DGEF, meeting AC at G.

Proof:
In ΔBCF, D is the mid-point of BC and DGBF.
∴ G  is the mid point of CF.
So, FG =  GC
In ΔADG, E is the mid-point of AD and EFDG.
∴ F  is the mid-point of AG.
So, AF =  FG
Thus, AF = FG = GC
∴ AC = (AF + FG + GC) = 3AF
Hence, AF=13AC

Page No 425:

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

Answer:


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

Mean,x¯=A+h×fi×uifi
=125+50×85180=125+23.61=148.61
Hence, the mean of the given frequency is 148.61.

OR
We have:

Therefore,
Mean,x¯=fi×uifi
1700+35p80+p=24
1700+35p=1920+24p
11p=1920-1700=220
p=22011=20
Hence, the value of p is 20.

Page No 425:

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

Answer:

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

N2=532=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 , N2=26.5, h = 10 and cf = 22
Median,Me=l+h×N2-cff
=60+10×26.5-227=60+10×4.57=60+6.43=66.43
Hence, median = 66.43

Page No 425:

Question 25:

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

Answer:

The given polynomial is f(x) = 2x4 - 3x3 - 5x2 + 9x - 3.
Since, 3and-3 are are two zeros of f(x), it follows that each one of  x-3 and x+3 is a factor of f(x).
Consequently, x-3x+3=x2-3 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
x-3x+32x-1x-1=0
x-3= 0, or x+3= 0, or 2x-1= 0, or x-1=0
x=3or x=-3, or x=12, or x=1
Hence, the other two zeros are 12and 1.

Page No 425:

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.

Answer:

Given: Δ ABC ∼ Δ DEF
To Prove: arABCarDEF=AB2DE2=AC2DF2=BC2EF2

Construction: Draw ALBC and DMEF

Proof:
Since Δ ABC ∼ Δ DEF, it follows that they are equiangular and their sides are proportional.
A = D, B = E, C = F
and ABDE=BCEF=ACDF ..................(i)
arABC=12×BC×AL
arDEF=12×EF×DM
arABCarDEF=12×BC×AL12×EF×DM=BCEF×ALDM ...............(ii)
In Δ ALB and Δ DME, we have:
∠ALB = ∠DME = 90° and B = E    [ from (i)]
∴∠ALB ∼ ∠DME

Consequently, ALDM=ABDE
But ABDE=BCEF  [from (i)]
ALDM=BCEF..................(iii)
Using (iii) in (ii), we get:
arABCarDEF=BCEF×BCEF=BC2EF2
Similarly, arABCarDEF=AB2DE2 and arABCarDEF=AC2DF2
Hence, arABCarDEF=AB2DE2=AC2DF2=BC2EF2

OR

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

Proof:
In Δ DEF, we have 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, B=E = 90°.

Page No 425:

Question 27:

Prove that sinθ-cosθ+1sinθ+cosθ-1=1(secθ-tanθ).
Or, evaluate: secθ cosec(90°-θ)-tan θ cot(90°-θ)+sin265°+sin225°tan10° tan20° tan60° tan70° tan80°

Answer:

LHS =sinθ-cosθ+1sinθ+cosθ-1

=sinθcosθ-1+1cosθsinθcosθ+1-1cosθ                    [On dividing numerator and denominator by cos θ]
=tanθ-1+secθtanθ+1-secθ=secθ+tanθ-1tanθ-secθ+1
=secθ+tanθ-sec2θ-tan2θtanθ-secθ+1        [Since 1 = sec2θ -  tan2θ]
=secθ+tanθ1-secθ-tanθtanθ-secθ+1
=secθ+tanθtanθ-secθ+1tanθ-secθ+1=secθ+tanθ

RHS=1secθ-tanθ
=1secθ-tanθ×secθ+tanθsecθ+tanθ=secθ+tanθsec2θ-tan2θ
=secθ+tanθ    [Since sec2θ -  tan2θ = 1]
Hence, LHS  =  RHS

OR
secθcosec90°-θ-tanθcot90°-θ+sin265°+sin225°tan10°tan20°tan60°tan70°tan80°

=sec2θ-tan2θ+sin290°-25°+sin225°tan10°tan20°tan60°tan90°-20°tan90°-10°

=1+cos225°+sin225°tan10°tan20°tan60°×cot20°×cot10°        [Since sec2θ -  tan2θ = 1]

=1+1tan10°tan20°×3×1tan20°×1tan10°     [Since cos2θ + sin2θ = 1]
=23

Page No 425:

Question 28:

If sec θ + tan θ = x, prove that sin θ = x2-1x2+1.

Answer:

We have,
x2-1x2+1=secθ+tanθ2-1secθ+tanθ2+1

=sec2θ+tan2θ+2secθtanθ-1sec2θ+tan2θ+2secθtanθ+1 

=2tan2θ+2secθtanθ2sec2θ+2secθtanθ since sec2θ-1=tan2θ

=2tanθtanθ+secθ2secθtanθ+secθ=tanθsecθ=sinθcosθ×cosθ=sinθ

Hence, x2-1x2+1=sinθ

Page No 425:

Question 29:

Solve the following system of linear equations graphically:
2x-y=1, x-y=-1.
Shade the region bounded by these lines and the y-axis.

Answer:

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.



Page No 426:

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.

Answer:

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:

Page No 426:

Question 31:

Solve for x and y:
ax+by-a+b=0, bx-ay-a-b=0.

Answer:

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=a2+b2a2+b2=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.

Page No 426:

Question 32:

Prove that: 1-cosθ1+cosθ=(cosecθ-cotθ)2.

Answer:

RHS=cosecθ-cotθ2
=1sinθ-cosθsinθ2
=1-cosθ2sin2θ
=1-cosθ21-cos2θ=1-cosθ1-cosθ1-cosθ1+cosθ=1-cosθ1+cosθ

LHS = 1-cosθ1+cosθ

Hence, LHS =  RHS

Page No 426:

Question 33:

ΔABC is right-angled at B and D is the mid-point of BC.
Prove that: AC2=(4AD2-3AB2).

Answer:

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

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

Page No 426:

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

Answer:

We have,

(i) Mean:
Let the assumed mean, A, be 35; h = 10
Mean,x¯=A+h×fi×uifi
=35+10×6100=35+0.6=35.6
(ii) Median:

N2=1002=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 , N2=50, h = 10 and cf = 33
Median,Me=l+h×N2-cff
=30+10×50-3330=30+10×1730=30+173=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



Page No 427:

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

Answer:

(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.

Page No 427:

Question 2:

If the product of zeros of the polynomial ax2−6x−6 is 4, then a = ?
(a) 23
(b) -23
(c) 32
(d) -32

Answer:

(d) -32
Let α and β be the zeros of ax2 - 6x - 6.
Then,
αβ=-6a-6a=4    [ since αβ=4]

⇒ 4a = - 6
a=-64=-32
Hence, a=-32

 

Page No 427:

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

Answer:

(c) 7 cm
It is given that ABC~PQR.
 arABC=25 cm2 and  arPQR=49 cm2 
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.
arABCarPQR=BC2QR2
BCQR2=arABCarPQR=2549=572

BCQR=57BC9.8=57
BC=5×9.87=497=7
Hence, BC = 7 cm.

Page No 427:

Question 4:

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

Answer:

(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 = 562=28°

Page No 427:

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

Answer:

(d) 0
We have:
cos θ = 610=35
ABAC=35=3k5k

BC2 = (AC2 - AB2)
= (25k2 - 9k2)
= 16k2
BC=16k2=4k
∴ sin θ = BCAC=4k5k=45,  tan θ = BCAB=4k3k=43
Hence,
5 sin θ − 3 tan θ
 =5×45-3×43=4-4=0



Page No 428:

Question 6:

The simplest form of 10951168 is
(a) 1726
(b) 2526
(c) 1516
(d) 1316

Answer:

(c) 1516

10951168=3×5×732×2×2×2×73=1516

Page No 428:

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

Answer:

(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
a1a2=43, b1b2=-55 and c1c2=-20-15=43
∴ These graph lines will intersect at a unique point, when a1a2b1b2.
Hence, the given system has a unique solution.

Page No 428:

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

Answer:

(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.

Page No 428:

Question 9:

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

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.

Page No 428:

Question 10:

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

Answer:

fx=9x2-5=3x2-52=3x+53x-5
fx=03x+53x-5=0
3x+5=0 or 3x-5=0
x=-53 or x=53
So, the zeros of f(x) are -53and53.
Sum of zeros=-53+53=0=09=-coefficient of xcoefficient of x2
Product of zeros=-53×53=-59=constant termcoefficient of x2

Page No 428:

Question 11:

If 2 sin 2θ = 3, find the value of θ.
Or, If 7 sin2θ + 3 cos2θ = 4, show that tan θ = 13.

Answer:

2sin2θ=3sin2θ=32sin2θ=sin60°

2θ=60°θ=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
sin2θ=14
cos2θ=1-sin2θ=1-14=34
tan2θ=sin2θcos2θ=1×44×3=13
tanθ=13

Page No 428:

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.

Answer:

Let AE be  x cm.
Then, EC = (AC - AE) cm = (6.5 - x) cm
In Δ ABC, DEBC
ADDB=AEEC    [By Thales' theorem]
58=x6.5-x
56.5-x=8x
13x=32.5
x=32.513=2.5
Hence, AE = 2.5 cm.

Page No 428:

Question 13:

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

Answer:

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

Page No 428:

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

Answer:

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
Mode,M0=xk+h×fk-fk-12fk-fk-1-fk+1
=4+4×8-42×8-4-5
=4+4×416-9=4+4×47=4+167=28+164=447=6.29

Page No 428:

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.

Answer:

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.



Page No 429:

Question 16:

Prove that (3-5) is irrational.
Or, prove that 223 is irrational.

Answer:

Let us assume that 3-5 is rational.
That is, we can find integers p and q(≠ 0) such that:
3-5=pq or 3-pq=5
5=3-pq
Since, p and q are integers, we get 3-pq is rational; so, 5 is rational.
But this contradicts the fact that 5 is irrational.
So, we conclude that3-5 is irrational.

OR
Let us assume that 223 is rational. That is, we can find co -prime integers p and q(≠ 0), such that
223=pq or 3p2q=2
2=3p2q
Since, p and q are integers, 3p2q is rational; so 2 is rational.
But this contradicts the fact that 2 is irrational.
So, we conclude that223 is irrational.

Page No 429:

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 14. Find the numbers.

Answer:

Let the required number be x.

Then, according to proportionality rule, we get:
5+x9+x=17+x27+x
⇒ (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 1x+1y=14
x+yxy=14
18xy=14xy=72
x-y=x+y2-4xy
=182-4×72=324-288=36=±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.

Page No 429:

Question 18:

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

Answer:

f(x) = x2 - x - 12 is the given polynomial.
It is given that,α and β are the zeros of the polynomial.
α+β=-ba=--11=1 and αβ=-121=-12
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.

Page No 429:

Question 19:

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

Answer:

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

Page No 429:

Question 20:

If sec θ tan θ = m, show that m2-1m2+1=sinθ.

Answer:

m2-1m2+1=secθ+tanθ2-1secθ+tanθ2+1
=sec2θ+tan2θ+2secθtanθ-1sec2θ+tan2θ+2secθtanθ+1 

=2tan2θ+2secθtanθ2sec2θ+2secθtanθ   [Since, sec2θ-1=tan2θ and tan2θ+1=sec2θ ]

=2tanθtanθ+secθ2secθtanθ+secθ=tanθsecθ=sinθcosθ×cosθ=sinθ
Hence, m2-1m2+1=sinθ

Page No 429:

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.

Answer:

In Δ AOB and Δ COD, we have:
OAB = OCD    [ Alternate interior angles]
OBA = ODC    [ Alternate interior angles]
∴ ΔAOB ∼ Δ COD   [ By AA similarity]
arAOBarCOD=AB2CD2=2CD2CD2   [ Since AB = 2 × CD]
                                =4×CD2CD2=4
arCOD=14×arAOB=14×84cm2 = 21cm2
Hence, area of ΔCOD is 21 cm2.

Page No 429:

Question 22:

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

Answer:

We have:

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

Page No 429:

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

Answer:

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

Mean,x¯=A+h×fi×uifi
=25+10×050=25+0=25
Hence, the mean of the given frequency is 25

OR
We have:

Therefore,
Mean,x¯=fi×uifi
2665+95p37+p=78
2665+95p=2886+78p
17p=2886-2665=221
p=22117=13
Hence, the value of p is 13.



Page No 430:

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

Answer:

We prepare the cumulative frequency table, as given below:

N2=302=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 , N2=15, h = 5 and cf = 13
Median,Me=l+h×N2-cff
=55+5×15-136=55+5×26=55+1.67=56.67
Hence, median = 56.67 kg

Page No 430:

Question 25:

If two zeros of the polynomial p(x) = 2x4+7x3-19x2-14x+30 are 2 and -2, then find the other two zeroes.

Answer:

The given polynomial is f(x) = 2x4 + 7x3 - 19x2 - 14x + 30
Since, 2and-2 are are two zeros of f(x), it follows that each one of  x-2 and x+2 is a factor of f(x).
Consequently, x-2x+2=x2-2 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
x-2x+22x-3x+5=0
x-2= 0, or x+2= 0, or 2x-3= 0, or x+5=0
x=2or x=-2, or x=32, or x=-5
Hence, the other two zeros are 32and -5.

Page No 430:

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.

Answer:

Let ABCD be a square of side a.
Therefore, its diagonal is 2a.

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= 2a
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.
AreaofABEAreaofDBF=34AB234BD2=a2a2=a22a2 =12

OR
Given: Δ ABC ∼ Δ DEF
To Prove: arABCarDEF=AB2DE2=AC2DF2=BC2EF2

Construction: Draw ALBC and DMEF.

Proof:
Since Δ ABC ∼ Δ DEF, it follows that they are equiangular and their sides are proportional.
A = D, B = E, C = F
and ABDE=BCEF=ACDF ..................(i)
arABC=12×BC×AL
arDEF=12×EF×DM
arABCarDEF=12×BC×AL12×EF×DM=BCEF×ALDM ...............(ii)
In Δ ALB and Δ DME, we have:
∠ALB = ∠DME = 90° and B = ∠E    [ from (i)]
ALB ∼ DME

Consequently, ALDM=ABDE
But ABDE=BCEF  [from (i)]

ALDM=BCEF..................(iii)
Using (iii) in (ii), we get:
arABCarDEF=BCEF×BCEF=BC2EF2
Similarly, arABCarDEF=AB2DE2 and arABCarDEF=AC2DF2

Hence, arABCarDEF=AB2DE2=AC2DF2=BC2EF2

Page No 430:

Question 27:

Prove that: secθ+tanθ-1tanθ-secθ+1=cosθ1-sinθ
Or, Evaluate: secθ cosec(90°-θ)-tanθ cot(90°-θ)+sin255°+sin235°tan10° tan20° tan60° tan70° tan80°

Answer:

LHS=secθ+tanθ-1tanθ-secθ+1
=secθ+tanθ-sec2θ-tan2θtanθ-secθ+1        [Since, 1 = sec2θ -  tan2θ]
=secθ+tanθ1-secθ-tanθtanθ-secθ+1
=secθ+tanθtanθ-secθ+1tanθ-secθ+1=secθ+tanθ
=1cosθ+sinθcosθ=1+sinθcosθ×1-sinθ1-sinθ
=1-sin2θcosθ1-sinθ=cos2θcosθ1-sinθ=cosθ1-sinθ

RHS=cosθ1-sinθ
Hence, LHS = RHS

OR

secθcosec90°-θ-tanθcot90°-θ+sin255°+sin235°tan10°tan20°tan60°tan70°tan80°

=sec2θ-tan2θ+sin290°-35°+sin235°tan10°tan20°tan60°tan90°-20°tan90°-10°

=1+cos235°+sin235°tan10°tan20°tan60°×cot20°×cot10°        [Since, sec2θ -  tan2θ = 1]

=1+1tan10°tan20°×3×1tan20°×1tan10°     [Since, cos2θ + sin2θ = 1]
=23

Page No 430:

Question 28:

If sec θ + tan θ = m, prove that sin θ=(m2-1)(m2+1).

Answer:

m2-1m2+1=secθ+tanθ2-1secθ+tanθ2+1
=sec2θ+tan2θ+2secθtanθ-1sec2θ+tan2θ+2secθtanθ+1      [Since sec2θ - 1 = tan2θ and tan2θ + 1 = sec2θ]
=2tan2θ+2secθtanθ2sec2θ+2secθtanθ
=2tanθtanθ+secθ2secθtanθ+secθ=tanθsecθ
=sinθcosθ×cosθ=sinθ
Hence, m2-1m2+1=sinθ

Page No 430:

Question 29:

Draw the graph of the following equations:
3x+y-11=0 and x-y-1=0.
Shade the region bounded by these lines and the y-axis.

Answer:

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.

Page No 430:

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.

Answer:

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.



Page No 431:

Question 31:

For what value of k will the following pair of linear equations have infinitely many solutions?
                   2x-3y=7(k+1)x+(1-2k)y=(5x-4)

Answer:

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
a1a2=b1b2=c1c2
Or
2k+1=-31-2k=-74-5k
2k+1=32k-1 and 32k-1=75k-4
⇒ 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7
k = 5 and k = 5
Hence, k = 5.

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Question 32:

Prove that: (sinθ-cosecθ)(cosθ-secθ)=1(tanθ+cotθ).

Answer:

LHS=sinθ-cosecθcosθ-secθ
=sinθ-1sinθcosθ-1cosθ=sin2θ-1sinθ×cos2θ-1cosθ
=-1-sin2θ×-1-cos2θsinθcosθ   
  Now, (1- sin2θ) =  cos2θ and (1- cos2θ) = sin2θ
Therefore, we have:-cos2θ-sin2θsinθcosθ=sin2θcos2θsinθcosθ=cosθsinθ
RHS=1tanθ+cotθ
=1sinθcosθ+cosθsinθ
=cosθsinθsin2θ+cos2θ    Since sin2θ+cos2=1
=cosθsinθ
Hence, LHS =  RHS

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Question 33:

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

Answer:

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.

Page No 431:

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.

Answer:


Let the assumed mean A be 225; h = 50.
Mean,x¯=A+h×fi×uifi
=225+50×-1230=225-20=205
Now,
N2=302=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 , N2=15, h = 50 and cf = 13
Median,Me=l+h×N2-cff
=200+5015-1312=200+50×212=200+253=200+8.33=208.33
Hence, mean = 205 and median = 208.33



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