Given, sec \(\theta\) + tan \(\theta\) = l    ..........(i)
\(\Rightarrow \frac{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}=l\)
\(\Rightarrow \frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}=l \Rightarrow \sec \theta-\tan \theta=\frac{1}{l}\)          ...(ii)
Hence, find sec \(\theta\) using Eq. (i) and (ii)...

f₁=28, f₂=24

14.48km/L No, the manufacturer is claiming mileage 1.52km/L more than the average mileage.

109.92

35

Here, class interval are not continuous. But it does not affect mid-values. So, we will solve it without making it continuous.

Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 362 }{ 28 } =12.93\)

Rs.12000 in scheme A and Rs.10000 in scheme B

10 km/h, 40 km/h

Plot the lines as follows (0, 0), (4,4), (6,2)

Recall the identity a² - b² = (a - b)(a + b). Using it, we can write:
x² - 3 = (x - \(\sqrt{3}\))(x + \(\sqrt{3}\))
So, the value of x² - 3 is zero when x = \(\sqrt{3}\) or x = -\(\sqrt{3}\)
Therefore, the zeroes of x² - 3 are \(\sqrt{3}\) and -\(\sqrt{3}\)
Now,
sum of zeroes = \(\sqrt{3}\) - \(\sqrt{3}\) = 0 = \(\frac{-(Coefficient \quad of \quad x)}{Coefficient \quad of \quad x^{2}}\)
product of zeroes = (\(\sqrt{3}\))(-\(\sqrt{3}\)) = -3 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

Factors of 1 to 10 numbers
1 = 1
2 = 1 x 2
3 = 1 x 3
4 = 1 x 2 x 2
5 = 1 x 5
6 = 1 x 2 x 3
7 = 1 x 7
8 = 1 x 2 x 2 x 2
9 = 1 x 3 x 3
10 = 1 x 2 x 5
\(\therefore \)  LCM of numbers 1 to 10
= LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520

\(a={ x }^{ 3 }{ y }^{ 2 }=x\times x\times x\times y\times y\) and \(b={ x }{ y }^{ 3 }=x\times y\times y\times y\)
HCF (a,b) = xy²

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

No, because by Euclid's division lemma, 
we have, a = 4q + r,  \(0\le r<4\)             ...(i)
a can be in the form 4q, 4q + 1, 4q + 2 or 4q + 3

(a + 2) + (b + 1),(a + 2) + (b + 2),(a + 3) + (b + 2)

a₁₈ + a₁₉ + a₂₀ = 225
a₃₅ + a₃₆ + a₃₇ = 429
3, 7, 11, 15, ...

(i) \((i)\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } \)
(ii)\((i)-\frac { 1 }{ 2 } ,2\)

45 km/h.

42 cm²

1386 cm²

Lert height of rainfall be h cm
V _Cuboid = V_Cylinder
\(lbh=\pi r^{ 2 }h\)
\(22\times 20\times h=\frac { 22 }{ 7 } \times 1\times 1\times \frac { 35 }{ 10 } \)
\(\Rightarrow h=\frac { 11 }{ 22\times 20 } \)
\(\Rightarrow h=\frac { 1 }{ 40 } m=2.5cm\)

Possible outcomes on tossing a coin 3 times, are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
\(\therefore\)  Total number of outcomes = 8
(i) Let E₁ be the event that Sweta losses the entry fee
i.e. shen tosses tail three times i.e. TTT.
\(\therefore\) Number of outcomes favourable to E₁ = 1
Hence, required probability = P(E₁) = \(\frac{1}{8}\)
(ii) Let E₂, be the event that Sweta gets double entry fee
i.e. she tosses heads three times
i.e. HHH
\(\therefore\) Number of outcomes favourable to E₂ = 1
Hence, required probability \(=P\left(E_2\right)=\frac{1}{8}\)
(iii) Let E₃ be the event that Sweta gets her entry fee back
i.e. Sweta gets heads one or two times
i.e. event of getting
HTT, THT, TTH, HHT, HTH or THH
\(\therefore\) Number of outcomes favourable to E₃ = 6
Hence, required probability = P(E₃) = \(\frac{6}{8}=\frac{3}{4}\)

A.T.Q., S₅ + S₇ = 167
\(\Rightarrow\) \({5\over2}[2a+14d]+{7\over2}[2a+6d]=167\)
\(\Rightarrow\) 5 ( a + 2d ) + 7 ( a + 3d ) = 167
\(\Rightarrow\)  12a + 31d = 167                   ...(i)
and  S₁₀ = 235
\(\Rightarrow\)  \({10\over2}[2a+9d]=235\)
\(\Rightarrow\)  2a + 9d = \({235\over4}=47\)                ...(ii)
Multiplying equation (ii) by 6 and then subtractigfrom (i), we have
12a + 31d = 167
12a + 54d = 282
-     -          -      
     -23d = -115
d= \({-115\over-23}=5\)
\(\therefore\)  From (ii), 2a + 9d = 47
\(\Rightarrow\)   2a + 9 x 5 = 47
\(\Rightarrow\)    2a = 47 - 45
\(\Rightarrow\)   2a = 2 \(\Rightarrow\) a = 1
Hence, S₂₀ = \({20\over2}[2a+19d]\)
= 10 [ 2 x 1 + 19x 5 ]
= 10 [ 2 + 95 ] = 10 x 97 = 970

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

\(\because\)         S₆ = 36 and S₁₆ = 256.
\(\Rightarrow\)   S₆ = \({6\over2}\) [ 2a + 5d ]
   [ \(\because\) S_n = \({n\over2} [ 2a + ( n - 1)d ]\)
\(\Rightarrow\)    S₆ = 3 ( 2a + 5d )
\(\Rightarrow\)    \({36\over3}\) = 2a + 5d 
\(\Rightarrow\)    12 = 2a + 5d                       ...(i)
and   S₁₆ = \({16\over2}[2a+15d]\)
\(\Rightarrow\)       \({256\over8}=2a+15d\)
\(\Rightarrow\)   32 = 2a + 15d                     ...(ii)
Subtracting (i) and (ii), 2n + 5d = 12
2a + 15d = 32
-    -           -    
     -10d = -20   \(\Rightarrow\)  d = 2
\(\therefore\)  From (i), 12 = 2a + 5(2)
12 - 10 = 2a  \(\Rightarrow\)  2a = 2 \(\Rightarrow\) a = 1
Hence   S₁₀ = \({10\over2}[2a+9d]\)
= 5 ( 2 x 1 +  9 x 2  )
= 5 ( 2 + 18 )
\(\Rightarrow\)   S₁₀ = 5 x 20 = 100.