Mathematics Test

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Mathematics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

A straight line passes through (1, –2, 3) and perpendicular to the plane 2x + 3y – z = 7.

What is the image of the point (1, –2, 3) in the plane?

A
(2, –1, 5)

B
(–1, 2, –3)

C
(5, 4, 1)

D
None of these

Solution

Let Q (x,y,z) is the image of (1, -2, 3) in the plane

$\begin{array}{lrlrlrlrlrl} \frac{x + 1}{2} = 3 \Rightarrow x = 5 \end{array}$

$\frac{y - 2}{2} = 1 \Rightarrow y = 4$

$\frac{z + 3}{2} = 3 \Rightarrow x = 5$

∴Image of (1, -2, 3) are (5, 4, 1)
Question 2
1 / -0

What is the slope of the tangent of the curve y = sin^-1 (sin² x) at x = 0?

A
0

B
1

C
2

D
None of these

Solution

y = sin^-1(sin² x)

$\frac{d y}{d x} = \frac{2 \sin x \cos x}{\sqrt{1 - \sin^{4} x}} \Rightarrow \frac{d y}{d x} = \frac{\sin 2 x}{\sqrt{1 - \sin^{4} x}}$

$x = 0, \frac{d y}{d x} = 0$
Question 3
1 / -0

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant:

A
Becomes zero

B
Remains the same

C
Changes its sign

D
Is doubled

Solution

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant changes its sign.
Question 4
1 / -0

What is the equation of a parabola whose vertex is at (0, 0) and at the centre point (0, –2)?

A
y²+ 8x = 0

B
y² -8x = 0

C
x² + 8y = 0

D
x² - 8y = 0

Solution

Focus (0, -2)

a = -2

$x^{2} = 4 a y$

$x^{2} = - 8 y$

$x^{2} + 8 y = 0$
Question 5
1 / -0

The sum of an infinite GP is x and the common ratio r is such that |r|<1. If the first term of the GP is 2, then which one of the following is correct ?

A
-1 < x < 1

B
-∞ < x < 1

C
1 < x < ∞

D
None of these

Solution

GP = x

$\frac{a}{1 - r}$ = x(where, a = Ist term and r = common ratio)

$\Rightarrow \frac{2}{1 - r} = x . . . . (i) (∵ G i e v n a = 2 a n d l r l < 1)$

$\Rightarrow - 1 < r < 1 \Rightarrow > - r > - 1$

$\Rightarrow 1 + 1 > - r > 1 - 1$

$\Rightarrow 0 < 1 - r < 2$

$\Rightarrow \frac{1}{1 - r} > \frac{1}{2}, \frac{2}{1 - r} > 1$

From equation (i) $x > 1$

Hence, $1 < x < \infty$
Question 6
1 / -0

Four points (0, 4, 1) (2, 3, -1) (4, 5, 0) (2, 6, 2) Which of the following are the vertices of the figure.

A
Rhombus

B
Rectangle

C
Square

D
Parallelogram

Solution

The constant of four points A, B, C and D is (0, 4, 1) (2, 3, -1) (4, 5, 0) (2, 6, 2), AB = BC = CD = DA = 3 unit

And AC = BD = $\sqrt{18}$ units

Therefore, the four sides and diagonals of the quadrilateral are equal,

Hence, quadrilateral ABCD will be square.
Question 7
1 / -0

A line passes through the points (6, –7, –1) and (2, –3, 1). What are the direction ratios of the line?

A
<4, –4, 2>

B
<4, 4, 2>

C
< –4, 4, 2>

D
<2, 1, 1>

Solution

Direction ratios < (2 - 6), (-3 + 7), (1 + 1)>

= < -4, 4, 2 >
Question 8
1 / -0

If P and Q are two complex numbers, then the modulus of the quotient of P and Q is:

A
Greater than the quotient of their moduli

B
Less than the quotient of their moduli

C
Less than or equal to the quotient of their moduli

D
Equal to the quotient of their moduli

Solution

The two complex number are

P = x + iy and Qα + iβ

Quotient $= \frac{P}{Q} = \frac{x + i y}{α + i β}, |\frac{P}{Q}| = |\frac{x + i y}{α + i β}|$

$= \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{α^{2} + β^{2}}} = \frac{\sqrt{x^{2} + y^{2}}}{\sqrt{α^{2} + β^{2}}} = |\frac{z_{1}}{z_{2}}| = |\frac{z_{1}}{z_{2}}|$

Hence, the quotient of their modulus is equal to the quotient of their moduli.
Question 9
1 / -0

For two variables x and y, the two regression coefficients are b_yx = $- \frac{3}{2}$ and b_xy =- $\frac{1}{6}$ .The correlation coefficient between x and y is :

A
$- \frac{1}{4}$

B
$\frac{1}{4}$

C
$- \frac{1}{2}$

D
$\frac{1}{2}$

Solution

$r = \sqrt{b_{x y} . b_{y x}}$

$= \sqrt{(- \frac{1}{6}) \times (- \frac{3}{2})} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \pm \frac{1}{2}$

b_xy and b_yx both have negative sign. Therefore we have to take negative sign Hence, correlation coefficient (r) = $- \frac{1}{2}$
Question 10
1 / -0

What is the mean of the squares of the first 20 natural numbers?

A
151.5

B
143.5

C
65

D
72

Solution

Mean of the squares of the first 20 natural number $= \frac{(n + 1) (2 n + 1)}{6} = \frac{21 \times 41}{6} = 143.5$

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Mathematics Test - 9

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Mathematics Test - 9

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Question 1

(2, –1, 5)

(–1, 2, –3)

(5, 4, 1)

None of these

Solution

Question 2

0

1

2

None of these

Solution

Question 3

Becomes zero

Remains the same

Changes its sign

Is doubled

Solution

Question 4

y2 + 8x = 0

y2 -8x = 0

x2 + 8y = 0

x2 - 8y = 0

Solution

Question 5

-1 < x < 1

-∞ < x < 1

1 < x < ∞

None of these

Solution

Question 6

Rhombus

Rectangle

Square

Parallelogram

Solution

Question 7

<4, –4, 2>

<4, 4, 2>

< –4, 4, 2>

<2, 1, 1>

Solution

Question 8

Greater than the quotient of their moduli

Less than the quotient of their moduli

Less than or equal to the quotient of their moduli

Equal to the quotient of their moduli

Solution

Question 9

-14

14

-12

12

Solution

Question 10

151.5

143.5

65

72

Solution

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y²+ 8x = 0

y² -8x = 0

x² + 8y = 0

x² - 8y = 0

$- \frac{1}{4}$

$\frac{1}{4}$

$- \frac{1}{2}$

$\frac{1}{2}$