Mathematics Test

Result

Mathematics Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

The coordinate of the point where the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}$ meets the plane $x+y+4 z=6$ is

A
$(1, - 1, - 1)$

B
(1,1,1)

C
$(- 1, - 1, - 1)$

D
(1,1,1)

Solution
Question 2
1 / -0

A
$x>\frac{1}{e}$

B
b

C
a

D
$x \in R$

Solution
Question 3
1 / -0

The focal distance of a point on the parabola $y^{2}=16 x$ whose ordinate is twice the abscissa, is

A
6

B
8

C
10

D
12

Solution

Undefined control sequence \therefore points are (0,0),(4,8)

Hence, focal distance are respectively

Undefined control sequence \because
Question 4
1 / -0

A bag contains 3 red and 3 white balls. Two balls are drawn one by one. The probability that they are of different colours is

A
$\frac{3}{10}$

B
$\frac{2}{5}$

C
$\frac{3}{5}$

D
$\frac{1}{5}$

Solution
Question 5
1 / -0

If $f (x) = x^{3} + {a x}^{2} + b x + c$ attains it's local minimum at certain negative real number then -

A
b

B

C

D

Solution
Question 6
1 / -0

$\mathrm{A} 4,3,5, \mathrm{B} 0,-2,2$ and $C(3,2,1)$ are three points. The coordinates of the point at which the bisector of $b$ meets the side is

A
$(\frac{15}{8}, \frac{4}{8}, \frac{11}{8})$

B
$(\frac{12}{7}, \frac{2}{7}, \frac{10}{7})$

C
$(\frac{9}{5}, \frac{2}{5}, \frac{7}{5})$

D
$(\frac{3}{2}, 0, \frac{3}{2})$

Solution

D divide BC is ratio AB : AC

$A B=\sqrt{16+25+9}=\sqrt{50}=5 \sqrt{2}$

$A C=\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2}$

$A B: A C=5: 3$

$D=\left[\frac{15+0}{8}, \frac{10-6}{8}, \frac{5+6}{8}\right]$

$=\left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)$
Question 7
1 / -0

If $(9 a, 6 a)$ is a point bounded in region formed by parabola $y^{2}=16 x$ and $x=9$, then

A
$a \in (0,1)$

B
b

C
d

D
c

Solution

Since, the point (9a,6a)is bounded in the region formed by the parabola y² =16and x = 9then

$y^{2}-16 x<0, x-9<0$

$\Rightarrow 36 a^{2}-16,9 a<0,9 a-9<0$

$36 a(a-4)<0,9 a-9<0$
Question 8
1 / -0

If $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three square matrices of the same order such that $\mathrm{B}=\mathrm{CAC}^{-1}$, then $\mathrm{CA}^{3} \mathrm{C}^{-1}$ is equal to -

A
$B$

B
$\mathrm{B}^{2}$

C
$\mathrm{B}^{3}$

D
$\mathrm{B}^{9}$

Solution

$\because B=\mathrm{CAC}^{-1}$

$\Rightarrow B C=C A$

$\Rightarrow C^{-1} B C=A$

$\therefore C A^{3} C^{-1}=C A A A C^{-1}$

$=\mathrm{C}\left(\mathrm{C}^{-1} \mathrm{BC}\right)\left(\mathrm{C}^{-1} \mathrm{BC}\right)\left(\mathrm{C}^{-1} \mathrm{BC}\right) \mathrm{C}^{-1}$

$=\mathrm{B}(\mathrm{I})(\mathrm{B}) \mathrm{I}(\mathrm{B}) \mathrm{I}=\mathrm{B}^{3}$
Question 9
1 / -0

The co-ordinates of the points $A$ and $B$ are (2,3,4),(-2,5,-4) respectively. If a point $P$ moves so that $P A^{2}-P B^{2}=k$ where $k$ is a constant, then the locus of $P$ is

A
A line

B
A plane

C
A sphere

D
None of these

Solution

$b P A^{2}-P B^{2}=k$

$\therefore\left[(x-2)^{2}+(y-3)^{2}+(z-4)^{2}\right]-\left[(x+2)^{2}+(y-5)^{2}+(z+4)^{2}\right]=k$

or $-8 x+4 y-16 z-16=k,$ which is the equation of a plane. Main Concept:

Distance Between Two Points in $3 D$ Distance between $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\mathrm{Q}\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by

$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$
Question 10
1 / -0

The sum of all the proper divisors of 9900 is

A
33851

B
23952

C
23951

D
None of these

Solution