Mathematics Test

Result

Mathematics Test - 28

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Question 1
3 / -1

Let $c_{1}$ and $c_{2}$ be defined as follows:
$c_{1} : b^{2} - 4 a c \geq 0$
$c_{2} : a, - b, c$ are of the same sign.
The roots of $a x^{2} + b x + c = 0$ are real and positive if:

A
Both $c_{1}$ and $c_{2}$ are satisfied

B
Only $c_{2}$ is satisfied

C
Only $c_{1}$ is satisfied

D
None of these

Solution

For the roots to be real, we must have $D = b^{2} - 4 ac \geq 0$
Also, both the roots will be positive if and only if the sum of the roots and product of the roots are both positive,that is, $- \frac{b}{a} > 0 & \frac{c}{a} > 0$ .
In other words, $a, - b$ and $c$ are of the same sign.
Question 2
3 / -1

A
0

B
1 +π /4

C
1

D
3/2

Solution
Question 3
3 / -1

A
sin B/sin C

B
cos C/cos B

C
cos B/cos C

D
None of these

Solution
Question 4
3 / -1

The value of sin 10° + sin 20° + sin 30° + ........+ sin 360° is

A
1

B
0

C
-1

D
None of these

Solution

Since, sin 190° = - sin 10°, sin 200° = - sin 20°,
sin 210° = - sin 30°, sin 360° = sin 180° = 0 etc.
Hence all the terms in the expression cancels out, therefore answer is 0.
Question 5
3 / -1

Find the value of $\int \frac{\ln (x)}{x} d x$

A
$\frac{\ln^{2} (x)}{2}$

B
$\frac{\ln^{2} (x)}{3}$

C
$\frac{\ln^{2} (x)}{4}$

D
1

Solution

Given, $f (x) = \int \frac{\ln (x)}{x} d x$
Let, $z = \ln (x)$
=>d z= $\frac{d x}{x}$
=> $f (x) = \int z d z = \frac{z^{2}}{2}$
$= \frac{\ln^{2} (x)}{2}$
Question 6
3 / -1

If $A = {x \in R : x^{2} = x}$ and $B = {x \in R : x^{2} - 3 x + 2 = 0}$ , then find the value of $B - A$ .

A
${1, 2}$

B
${2}$

C
${0, 1, 2}$

D
None of these

Solution

It is given that, $A = {x \in R : x^{2} = x}$ and $B = {x \in R : x^{2} - 3 x + 2 = 0}$ .
Let us first consider $A = {x \in R : x^{2} = x}$
$∵ x^{2} = x$
$∴ x^{2} - x = 0$
$\Rightarrow x (x - 1) = 0$
$\Rightarrow x = 0$ or 1
So, $A = {0, 1}$
Now, $B = {x \in R : x^{2} - 3 x + 2 = 0}$
$\Rightarrow x^{2} - 3 x + 2 = 0$
$\Rightarrow x^{2} - x - 2 x + 2 = 0$
$\Rightarrow (x - 1) (x - 2) = 0$
$\Rightarrow x = 1$ or $2$
So, $B = {1, 2}$
As we know that, $B - A = {x : x \in B$ and $x \notin A}$
$∴ B - A = {2}$
Question 7
3 / -1

(tan 3x - tan 2x - tan x) is equal to

A
tan x tan 2x tan 3x

B
-tan x + tan 2x tan 3x

C
tan x tan x tan 3x - tan 2x tan 3x

D
None of these

Solution
Question 8
3 / -1

The area bounded by the curve y = 2x - x²and the straight line y = - x is given by

A
9/2

B
43/6

C
35/6

D
None of these

Solution
Question 9
3 / -1

The area under the curve y = sin 2x + cos 2x between x = 0 and x = π/4 is

A
2 sq. units

B
1 sq. unit

C
3 sq. units

D
4 sq. units

Solution
Question 10
3 / -1

If the two pairs of lines x²– 2mxy – y² = 0 and x² – 2nxy – y² = 0 are such that one of them represents the bisector of the angles between the other, then

A
mn + 1 = 0

B
mn – 1 = 0

C
1/m + 1/n = 0

D
1/m - 1/n = 0

Solution