Section B

Solution. 5
Given that, A is a skew- symmetric matrix of order 3.
So, A' = -A ⇒ lA'l = l-Al
lAl = (-1)³ lAl
lA'l = lAl
lA'l = lAl and lkAl = kⁿlAl
lAl = -lAl
2lAl = 0
lAl = 0

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Solution. 6
Give that the function is continuous at x = 4 .
⇒ = = 4k + 1
⇒ 7 = 4k + 1 ⇒ 6 = 4k ⇒ k =

6/4

=

3/2

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Solution. 7
We know that , area of an equilateral triangle. A =

√3/4

a² (a = side).
. As per question

da/dt

= 4 cm/s
on differentiating both sides w.r.t.t, we get

dA/dt

=

√3/4

x 2a x

da/dt

=

√3/4

x 2 x 20 x 4
= 40√3 cm²/s
( ∴ Altitude =

√3/2

a)
a = 20

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Solution. 8

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Solution. 9
, and are coplaner .

So, [ ] = 0

= 0

1(0 + 9) - (x - 2)(-2 + 9) + 4(3 -0) = 0
9 - 7x + 14 + 12 = 0
35 = 7x
x = 5

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Solution. 10
We have ā = 6î + 3ĵ - 2k̂ and b̄ = 2î + 2ĵ - 5k̂

Therefore, the vector equation of the line is

r̄ = ā + μb̄

r̄ = 6î + 3ĵ - 2k̂ + μ( 2î + 2ĵ - 5k̂)

Now, r̄ is the position vector of any point P(x,y,z) on the line .

therefore , xî + yĵ - zk̂ = 6î + 3ĵ - 2k̂ + μ (2î + 2ĵ - 5k̂)

Eliminating μ, we get

x - 6/2

=

y - 3/2

=

z + 2/-5

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Solution. 11
May (z) = 9x + 8y

subject to 6x + 2y ≤ 100

4x + 5y ≤ 120

x ≥ 0, y ≥ 0 .

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Solution. 12
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

x is a random variable getting tail

x = 0, 1, 2, 3

x	0	1	2	3
P(x)	1/8	3/8	3/8	1/8

Mean = f(x) = ∑ x_ip_i

= 0 x

1/8

+ 1 x

3/8

+ 2 x

3/8

x 3 x

1/8

=

3/8

+

6/8

+

3/8

=

12/8

= 1.5

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