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MATHMAA

SECTION - B

9. Here, n is any natural number so let us plug

n = 1, we get,

$$4^{n}$$ = 4

n = 2, we get,

$$4^{2}$$ = 16

Ending with 6.

n = 3, We get,

$$4^{3}$$ = 64

Ending with 4

Therefore $$4^{n}$$ where 'n' is a natural number can end with either 4 or 6 but not 0.

10)

x² + 7x + 10 = x² + 2x + 5x + 10

= x(x+2) + 5(x+2)

= (x+2)(x+5)

Therefore the zeroes are -2 and -5

Sum of the roots = -2 - 5 = -7

Sum of the roots = $$\large \frac{-7}{1}$$ = -7

Product of the roots = -2(-5)

= 10

Product of the roots = $$\large \frac{10}{1}$$

= 10

11)

$$\large \frac{3}{2k-1}$$ $$\neq$$ $$\large \frac{1}{k-1}$$

3(k-1) $$\neq$$ 2k-1

3k - 3 $$\neq$$ 2k - 1

k $$\neq$$ 2

Therefore the value of k is 2.

12)

$$\angle B =90°$$

Using Pythagoras theorem,

AC² = AB² + BC²         ---------(1)

AD² = AB² + BC² + CD²

= AC² + CD²         ($$\because from (1)$$)

Using converse of Pythagoras theorem we get $$\angle ACD = 90°$$.

Hence proved.

PQRT forms a rectangle,So

PT = QR = 5cm
PR = 13cm
PS = 14cm
TS = PS - PT = 14 - 5 = 9cm

Using Pythagoras theorem we get

TR = √(PR² - PT²)

= √(169 - 25)

= √144

= 12

Tan S = $$\large \frac{TR}{TS}$$

= $$\large \frac{12}{9}$$

= $$\large \frac{4}{3}$$

13)

sinΘ = $$\large \frac{3}{5}$$ = $$\large \frac{AB}{AC}$$

AB = 3
AC = 5
BC = √(AC² - AB²)
= √(25 - 9)
= √16
= 4

cosΘ = $$\large \frac{4}{5}$$

14)

median = l + $$\large \left [ \frac{\frac{n}{2}-C.F}{f}\right]$$h

Here, l = 125 , f = 20 , C.F. = 22, median = 137, h = 145-125 = 20

175 = 125 + $$\large \left[ \frac{\frac{n}{2}-22}{20}\right]$$*20

175 = 125 + $$\large \frac{n}{2}$$ - 22

175 = 103 + $$\large \frac{n}{2}$$

$$\large \frac{n}{2}$$ = 175 - 103

= 72

n = 72*2

= 144

Therefore the sum of frequencies (n) is 144.