Mathematics Test

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Mathematics Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Find all zeroes of $2 x^{4}-3 x^{3}-3 x^{2}+6 x-2,$ if you know that two of its zeroes are $\sqrt{2} \&-\sqrt{2}$.

A
$- \frac{1}{2}, - 1$

B
$1, - 1$

C
$\frac{1}{2^{'}} 1$

D
$\frac{1}{3}, \frac{1}{2}$

Solution

since two zeroes are $\sqrt{2}$ and $-\sqrt{2},(x-\sqrt{2})(x+\sqrt{2})=x^{2}-2$
is a factor of the given Polynomial. Now we divide the given polynomial by $x^{2}+2$

So, $2 x^{4}-3 x^{3}-3 x^{2}+6 x-2=\left(x^{2}-2\right)\left(2 x^{2}-3 x+1\right)$
Now, by splitting $-3 x,$ are factories $2 x^{2}-3 x+1$ as $(2 x-1)(x-1)$ So, its zeroes are given by $x=\frac{1}{2} \& x=1 .$ Therefore the zeroes of the given polynomial are $\sqrt{2},-\sqrt{2}, \frac{1}{2}$ and 1.
Question 2
1 / -0

A
$0$

B
$I$

C
-I

D
2I

Solution
Question 3
1 / -0

A
1/2

B
1/3

C
1/4

D
1

Solution
Question 4
1 / -0

The equation of the circle passing through (4, 5) having the centre at (2, 2) is?

A
$x^{2} + y^{2} + 4 x + 4 y - 5 = 0$

B
$x^{2} + y^{2} - 4 x - 4 y - 5 = 0$

C
$x^{2} + y^{2} - 4 x = 13$

D
$x^{2} + y^{2} - 4 x - 4 y + 5 = 0$

Solution

Here $r=$ Distance between (4,5) and (2,2) $\therefore \mathrm{r}^{2}=4+9=13$
\[
\begin{array}{l}
\Rightarrow(x-2)^{2}+(y-2)^{2}=13 \\
\Rightarrow x^{2}+y^{2}-4 x-4 y-5=0
\end{array}
\]
Question 5
1 / -0

Find the approximation tens of 56937.

A
56930

B
56940

C
56950

D
56900

Solution

56940

Hence correct answer is option B
Question 6
1 / -0

A

B

C

D

Solution
Question 7
1 / -0

Compute the modal value for the following frequency distribution.

x 95 105 115 125 135 145 155 165 175

f 4 2 18 22 21 19 10 3 2

A
115

B
125

C
22

D
120

Solution

From the given table, it is clear that 125 has the highest frequency 22. Hence modal value (mode) of the given frequency distribution is 125.
Question 8
1 / -0

In the given figure, PB & QA are perpendiculars to segment AB. If PO = 5cm, Q^∘ = 7cm and area ΔPOB = 150cm². Find the area of △QOA.

A
$294 {cm}^{2}$

B
$193 {cm}^{2}$

C
$201 {cm}^{2}$

D
$257 {cm}^{2}$

Solution
Question 9
1 / -0

Direction : If f (–x) = –f (x), then f (x) is an odd function & if f (–x) = f (x) then f (x) is an even function.

$Also \int_{- a}^{a} f (x) d x = \{\begin{array}{cl} 2 \int_{0}^{a} f (x), & if f (x) is even \\ 0 & if f (x) is odd \end{array}$

The value of $\int_{-\pi / 2}^{\pi / 2} \sin ^{2} x d x$ is equal to

A
$π / 4$

B
$π / 2$

C
0

D
$- π / 2$

Solution

$Let I = \int_{- π / 2}^{π / 2} \sin^{2} x d x$

$\begin{array}{l} Here, f (x) = \sin^{2} x \\ f (- x) = \sin^{2} (- x) = [\sin (- x)]^{2} = [- \sin x]^{2} = \sin^{2} x = f (x) \end{array}$

$f (x) i s a n e v e n f u n c t i o n .$

$\begin{array}{l} I = \int_{- π / 2}^{π / 2} \sin^{2} xdx = 2 \int_{0}^{π / 2} \sin^{2} xdx \\ = 2 \int_{0}^{π / 2} [\frac{1 - \cos 2 x}{2}] d x \\ = {[x - \frac{\sin 2 x}{2}]}_{0}^{π / 2} = \frac{π}{2} - 0 = \frac{π}{2} \end{array}$
Question 10
1 / -0

The curve given below represent a/an?

A
pie diagram

B
bar diagram

C
Ogive

D
histogram

Solution

The curverepresents anOgive.