Mathematics Test

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Mathematics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,

Then n $(A^{c} \cap B^{c})$ =

A
400

B
600

C
300

D
200

Solution

\begin{array}{l}
n\left(A^{C} \cap B^{C}\right)=n(U)-n(A \cup B) \\
=n(U)-[n(A)+n(B)-n(A \cap B)] \\
=700-[200+300-100]=300
\end{array}
Question 2
1 / -0

The set of values of x for which the inequality [x]² − 8[x] + 15≤0 (where [x] denote the greatest integer function) hold if

A
3 < [x] < 5

B
3 ≤ x < 6

C
3 ≤ [x] ≤ 5

D
(b) and (c) both

Solution

$[x]^{2}-8[x]+15 \leq 0$
$\Rightarrow([x]-3)([x]-5) \leq 0 \Rightarrow 3 \leq[x] \leq 5$
$\Rightarrow 3 \leq x<6$
Question 3
1 / -0

The value of log₃4log₄5log₅6log₆7log₇8log₈9 is

A
1

B
2

C
3

D
4

Solution
Question 4
1 / -0

The least positive integer n for which √(n+1) − √(n-1) < 0.2 is

A
24

B
25

C
26

D
27

Solution

$\sqrt{n+1}-\sqrt{n-1}<0.2$
\[
\begin{array}{l}
\sqrt{n+1}<\sqrt{n-1}+0.2 \\
\sqrt{n-1}+0.2>\sqrt{n+1}
\end{array}
\]
squaring both sides
\[
n-1+0.04+0.4 \sqrt{n-1}>n+1
\]
$0.4 \sqrt{n-1}>1.96$
$\sqrt{n-1}>4.9$
\[
n-1>24.01
\]
$n>25.01$
So the least positive integral value of $n$ is 26 Option $C$ is correct.
Question 5
1 / -0

1 + cos 56^∘ + cos 58^∘ − cos 66^∘ =

A
2cos28^∘cos29^∘cos33^∘

B
4cos28^∘cos29^∘cos33^∘

C
4cos28^∘ cos29^∘sin33^∘

D
2cos28^∘ cos29^∘ sin33^∘

Solution
Question 6
1 / -0

Two lines intersect at O. Points Ai and Bi (i=1,2,....,n) are taken on these two lines respectively, the number of triangles that can be drawn with the help of these 2n+1 points is

A
n

B
n²

C
n³

D
None of these

Solution
Question 7
1 / -0

Let f(x) = x² − 3x + 2 then area bounded by the curve f(|x|) (in square units) and x-axis is

A
1/3

B
5/6

C
5/3

D
None of these

Solution

If the area bounded by the curve $y=f(x),$ x-axis and $y$ -axis be $a$ square units then area bounded by $f(|x|)$ and $x$ -axis be twice of $a$ Shaded Area $=\int_{0}^{1} f(x) d x=\frac{5}{6} s q . \quad$ units
again Graph of $f(|x|)$
$\therefore$ Required Area bounded by $f(|x|)$ $=\int_{-1}^{1} f(x) d x=2 \int_{0}^{1} f(x) d x=2 \cdot \frac{5}{6}=\frac{5}{3} s q$ units
Question 8
1 / -0

With reference to a universal set, the inclusion of a subset in another, is relation, which is

A
Symmetric only

B
Equivalence relation

C
Reflexive only

D
None of these

Solution
Question 9
1 / -0

A five digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is

A
216

B
240

C
600

D
3125

Solution

We know that a five digit number is divisible by 3 , if and only if sum of its digits $(=15)$ is divisible by $3,$ therefore we should not use 0 or 3 while forming the five digit numbers.

Now

(i) In case we do not use 0 the five digit number can be formed (from the digit $1,2,3,4,5)$ in $^{5} P_{5}$ ways.

(ii) In case we do not use 3 , the five digit number can be formed (from the digit $0,1,2,4,5)$ in $^{5} P_{5}-^{4} P_{4}=5 !-4 !=120-24=96$ ways

$\therefore$ The total number of such 5 digit number $=$ $^5 P_{5}+\left(^{5} P_{5}-^{4} P_{4}\right)=120+96=216$
Question 10
1 / -0

The equation of the smallest circle passing through the points (2,2) and (3,3) is

A
x² + y²+ 5x + 5y + 12 = 0

B
x² + y² − 5x − 5y + 12 = 0

C
x² + y² + 5x − 5y + 12 = 0

D
x² + y²− 5x + 5y + 12 = 0

Solution

If the smallest circle is drawn through the points (2,2) and (3,3) then, the point have to be the opposite end of the diameter of the circle. Therefore, the centre of the circle is $C=\left(\frac{5}{2}, \frac{5}{2}\right)$ The radius of the circle is $R=\frac{\sqrt{2}}{2} \ldots .$ (using distance formula) Hence equation of the circle is $\left(x-\frac{5}{2}\right)^{2}+\left(y-\frac{5}{2}\right)^{2}=\frac{1}{2}$
$x^{2}+y^{2}-5 x-5 y+\frac{50}{4}=\frac{1}{2}$
$\mathrm{Or}$
$x^{2}+y^{2}-5 x-5 y+\frac{25}{2}-\frac{1}{2}=0$
$O r$
\[
x^{2}+y^{2}-5 x-5 y+12=0
\]

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 19

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Mathematics Test - 19

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SHARING IS CARING

Question 1

400

600

300

200

Solution

Question 2

3 < [x] < 5

3 ≤ x < 6

3 ≤ [x] ≤ 5

(b) and (c) both

Solution

Question 3

1

2

3

4

Solution

Question 4

24

25

26

27

Solution

Question 5

2cos28∘ cos29∘ cos33∘

4cos28∘ cos29∘ cos33∘

4cos28∘ cos29∘ sin33∘

2cos28∘ cos29∘ sin33∘

Solution

Question 6

n

n2

n3

None of these

Solution

Question 7

1/3

5/6

5/3

None of these

Solution

Question 8

Symmetric only

Equivalence relation

Reflexive only

None of these

Solution

Question 9

216

240

600

3125

Solution

Question 10

x2 + y2 + 5x + 5y + 12 = 0

x2 + y2 − 5x − 5y + 12 = 0

x2 + y2 + 5x − 5y + 12 = 0

x2 + y2 − 5x + 5y + 12 = 0

Solution

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2cos28^∘cos29^∘cos33^∘

4cos28^∘cos29^∘cos33^∘

4cos28^∘ cos29^∘sin33^∘

2cos28^∘ cos29^∘ sin33^∘

n²

n³

x² + y²+ 5x + 5y + 12 = 0

x² + y² − 5x − 5y + 12 = 0

x² + y² + 5x − 5y + 12 = 0

x² + y²− 5x + 5y + 12 = 0