Mathematics 2010 | Study Mode

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Question 31
A
-1
B
14
C
4
D
1
Explanation
Given that y = -3 - 2x + X2

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1

Correct option: D
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Question 32
A
y + 4x + 11 = 0
B
y - 4x - 11 = 0
C
y + 4x - 11 = 0
D
y - 4x + 11 = 0
Explanation
By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0

Correct option: C
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Question 33
A
\(\frac{5P - MX + 5}{M}\)
B
\(\frac{5P - MX - 5}{M}\)
C
\(\frac{5P + MX + 5}{M}\)
D
\(\frac{5P + MX - 5}{M}\)
Explanation
p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)

Correct option: B
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Question 34
A
2y + 3x
B
2y - 3x
C
3x + 2y
D
3x - 2y
Explanation
27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y

Correct option: D
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Question 35
A
\(\frac{x(x - 5)}{2(x + 2)}\)
B
\(\frac{x(x + 5)}{2(x + 2)}\)
C
\(\frac{x(x - 5)}{2(x - 2)}\)
D
\(\frac{x ^2 + 5}{2x + 4}\)
Explanation
\(\frac{x^3 + 3x^2 - 10x}{2x^2 - 8}\) = \(\frac {x(x^2 + 3x - 10)}{2(x^2 - 4)}\)

= \(\frac {x(x^2 + 5x - 2x - 10)}{2(x + 2)(x - 2)}\)

= \(\frac {x(x - 2)(x + 5)}{2(x + 2)(x - 2)}\)

= \(\frac {x(x + 5)}{2(x + 2)}\)

Correct option: B
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Question 36
A
(-1, 3)
B
(3, 1)
C
(-3, 1)
D
(1, 3)
Explanation
x - y = 2 ...........(1)

x2 - y2 = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x2 - (x - 2)2 = 8

x2 - (x2 - 4x + 4) = 8

x2 - x2 + 4x - 4 = 8

4x = 8 + 4 = 12

x = \(\frac{12}{4}\)

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

Correct option: B
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Question 37
A
6
B
12
C
3
D
5
Explanation
y = \(\alpha\)x ........(1)

y = kx ........(2)

When y = 3, x = 16,

(2) becomes 3 = k16 or 3 = k x 4

giving k = \(\frac{3}{4}\)

from (2), y = \(\frac{3}{4}\)x

When x = 64, y = \(\frac{3}{4}\)64

y = \(\frac{3}{4}\) x 8

= 6

Correct option: A
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Question 38
A
4
B
5
C
1\(\frac{1}{4}\)
D
2\(\frac{1}{4}\)
Explanation
x \(\alpha\) \(\frac{1}{y}\) .........(1)

x = k x \(\frac{1}{y}\) .........(2)

When x = 2\(\frac{1}{2}\)

= \(\frac{5}{2}\), y = 2

(2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\)

giving k = 5

from (2), x = \(\frac{5}{y}\)

so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)

Correct option: C
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Question 39
A
x < \(\frac{3}{2}\)
B
x > \(\frac{3}{2}\)
C
x < -\(\frac{3}{2}\)
D
x > -\(\frac{3}{2}\)
Explanation
\(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)

Multiply through by through by the LCM of 2, 3 and 4

12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\)

6x + 3 > 4x + 6

6x - 4x > 6 - 3

2x > 3

\(\frac{2x}{2}\) > \(\frac{3}{2}\)

x > \(\frac{3}{2}\)

Correct option: B
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Question 40
A
-1 < x < 5
B
-1 < x \(\leq\) 5
C
-1 \(\leq\) x \(\leq\) 6
D
-1 \(\leq\) x < 6
Explanation
-6 \(\leq\) 4 - 2x < 5 - x
split inequalities into two and solve each part as follows:

-6 \(\leq\) 4 - 2x = -6 - 4 \(\leq\) -2x

-10 \(\leq\) -2x

\(\frac{-10}{-2}\) \(\geq\) \(\frac{-2x}{-2}\)

giving 5 \(\geq\) x or x \(\leq\) 5

4 - 2x < 5 - x

-2x + x < 5 - 4

-x < 1

\(\frac{-x}{-1}\) > \(\frac{1}{-1}\)

giving x > -1 or -1 < x

Combining the two results, gives -1 < x \(\leq\) 5

Correct option: B

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