Mathematics Test

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Mathematics Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The value of $\cos 1 ° \cos 2 ° \cos 3 ° \dots \cos 179 °$ is:

A
0

B
1

C

-1

D

2

Solution

To find $\cos 1 ° \cos 2 ° \cos 3 ° \dots \cos 179 ° = ?$

$\cos 1^{0} \times \cos 2^{0} \times \cos 3^{0} \times . . . . . . \times \cos 179^{0} = \cos 1^{0} \times \cos 2^{0} \times \cos 3^{0} \times . . . . . \times \cos 90^{0} \times . . . . . \times \cos 179^{0} = \cos 1^{0} \times \cos 2^{0} \times \cos 3^{0} \times . . . . . \times 0 \times . . . . . \times \cos 179^{0} = 0$ (We know that: cos 90° =0)
Question 2
1 / -0

A line with positive direction cosines passes through the point P(2,−1,2) and makes equal angles with the coordinate axes. The line meets the plane 2x+y+z=9 at point Q. The length of the line segment PQ equals

A

√3

B

1

C

√2

D

2

Solution

(Direction cosines) D.C of the lines are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

Equation of the line is

$\frac{x - 2}{1} = \frac{y + 1}{1} = \frac{z - 2}{1} = r$ $\Rightarrow x = r + 2, y = r - 1, z = r + 2$

The point $(r + 2, r - 1, r + 2)$ lies on the plane $2 x + y + z = 9$ is $2 (r + 2) + r - 1 + r + 2 = 9$ $\Rightarrow r = 1$

So, the coordinate of Q are $(3, 0, 3)$ and thus

PQ= $\sqrt{{(3 - 2)}^{2} + {(0 + 1)}^{2} + {(3 - 2)}^{2}}$

= $\sqrt{3}$
Question 3
1 / -0

Interval in which at least one of the points of local maximum of cos^2 x + sinx lies

A

(0, π/2)

B

(π/8, 5π/4)

C

(-π/2, π/2)

D

All of these
Question 4
1 / -0

A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The probability that X = 3 equals

A

25/216

B

5/36

C

25/36

D

125/216

Solution

Given: A fair die is rolled repeatedly until a six is obtained.

The probability of getting a 6 in the third toss is

$P (X = 3) = (1 - \frac{1}{6}) (1 - \frac{1}{6}) (\frac{1}{6})$

= $= \frac{5 \times 5 \times 1}{6^{3}} = \frac{25}{216}$
Question 5
1 / -0

Let ABC and ABC′ be two non−congruent triangles with sides AB = 4, AC = AC′ = 2√2 and angle B = 30°. The absolute value of the difference between the areas of these triangles is

A

4

B

5

C

3

D

6
Question 6
1 / -0

A

at least 4 but less than 7

B

at least 10

C

atleast 7 but less than 10

D

less than 4
Question 7
1 / -0

Points of discontinuity of the function f(x) = [6x/π]cos[3x/π]

A

π/2

B

π/6

C

π/3

D

All of the above
Question 8
1 / -0

The maximum value of the function $f (x) = 2 x^{3} - 15 x^{2} + 36 x - 48$ on the set A = $x | x^{2} + 20 \leq 9 x |$ is

A

7

B

8

C

6

D

9

Solution

Given:

$x ² + 20 \leq 9 x$ $\Rightarrow x ² - 9 x + 20 \leq 0$

$\Rightarrow x ² - 4 x - 5 x + 20 \leq 0$

$\Rightarrow x (x - 4) - 5 (x - 4) \leq 0$

$\Rightarrow (x - 4) (x - 5) \leq 0 \Rightarrow 4 \leq x \leq 5 \Rightarrow A = 4 \leq x \leq 5$

Now,

$f (x) = 2 x ³ - 15 x ² + 36 x - 48$

Differentiate w.r.t x
$f' (x) = 6 x ² - 30 x + 36 = 6 (x ² - 5 x + 6) = 6 (x - 2) (x - 3)$

So f(x) is increasing in (−∞,2)∪(3,∞)

maximum value of f(x) at x=5

$f (5) = 2 (5) ³ - 15 (5) ² + 36 (5) - 48$

$= 250 - 375 + 180 - 48 = 430 - 423 = 7$

$∴ f m a x = f (5) = 7$
Question 9
1 / -0

A

All of these

B

Option C

C

Option B

D

Option A
Question 10
1 / -0

A

option A

B

Option B

C

Option C

D

Option D

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 5

Result

Mathematics Test - 5

Score

-

Rank

-

SHARING IS CARING

Question 1

0

1

-1

2

Solution

Question 2

√3

1

√2

2

Solution

Question 3

(0, π/2)

(π/8, 5π/4)

(-π/2, π/2)

All of these

Question 4

25/216

5/36

25/36

125/216

Solution

Question 5

4

5

3

6

Question 6

at least 4 but less than 7

at least 10

atleast 7 but less than 10

less than 4

Question 7

π/2

π/6

π/3

All of the above

Question 8

7

8

6

9

Solution

Question 9

All of these

Option C

Option B

Option A

Question 10

option A

Option B

Option C

Option D

User

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My Perfomance

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-

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-

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