Mathematics Test

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Mathematics Test - 3

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Weekly Quiz Competition

Question 1
3 / -1

The value of K, for which the equation (K–2)x² + 8x + K + 4 = 0 has both the roots real distinct and negative is:

A
0

B
2

C
3

D
-4

Solution
Question 2
3 / -1

Number of real solutions of the equation (tan x + 1)(tan x + 3)(tan x + 5)(tan x + 7) = 33

A
will be 2 in the interval [-π/2, π/2]

B
will be 4 in the interval [-π/2, π/2]

C
will be 3 in the interval [-π/2,π ]

D
will be 3 in the interval [-π/3,π]

Solution

\begin{equation}
\begin{array}{l}
(\tan x+1)(\tan x+3)(\tan x+5)(\tan x+7)=33 \\
\left(\tan ^{2} x+8 \tan x+15\right)\left(\tan ^{2} x+8 \tan x+7\right)=33 \\
\text { Let } \tan ^{2} x+8 \tan x=p \\
\therefore(p+15)(p+7)=33 \\
\Rightarrow p^{2}+22 p+72=0 \\
\Rightarrow p=-18 \text { (rejected) or } p=-4 \\
\Rightarrow \tan ^{2} x+8 \tan x+4=0 \\
\text { So, tan } x+4=\pm \sqrt{12} \Rightarrow \tan x=-4+\sqrt{12} \text { or }-4-\sqrt{12}
\end{array}
\end{equation}

Hence option A is correct answer.
Question 3
3 / -1

If A and B are two square matrices such that B = –A^–1 BA, then (A+B)² is equal to

A
0

B
A² + B²

C
A² + 2AB + B²

D
A+B

Solution
Question 4
3 / -1

A hyperbola has focus at origin, its eccentricity is √2 and corresponding directrix is x + y + 1 = 0. The equation(s) of its asymptotes is/are

A
x + 1 = 0

B
x - 1 = 0

C
y + 1 = 4

D
y - 1 = 0

Solution

Equation of this rectangular hyperbola is:
$\sqrt{x^{2}+y^{2}}=\sqrt{2} \frac{|x+y+1|}{\sqrt{2}}$
$\Rightarrow 2 x y+2 x+2 y+1=0$
Equation of the pair of asymptotes is:
$2 x y+2 x+2 y+1+c=0$
$\Delta=0 \Rightarrow c=1$
$\Rightarrow x y+x+y+1=0$
$\Rightarrow(x+1)(y+1)=0$ represents asymptotes

Hence option A is the correct answer.
Question 5
3 / -1

The number of distinct integral factor pairs of 10,500 whose product is 10,500 is

A
12

B
56

C
27

D
48

Solution

$10,500=100 \times 105$
$=2^{2} \times 5^{2} \times 5 \times 21=2^{2} \times 5^{2} \times 3 \times 7$
Number of integral factors $=(2+1)(3+1)(1+1)(1+1)=48$
Hence option D is correct answer.
Question 6
3 / -1

The coefficient of the term independent of x in the expansion of ${(\sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{x^{2}})}^{10}$ is-

A
1/3

B
19/54

C
5/3

D
1/4

Solution

$T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$
Applying to the above question, we get $T_{r+1}={ }^{10} C_{r}\left(\frac{x}{3}\right)^{\left(5-\frac{5}{2}\right)} x^{-2 r} 3^{\frac{5}{2}}$
$={ }^{10} C_{r} x^{\left(5-\frac{5}{2}\right)} 3^{r-5} \ldots(i)$
For term independent of $x$ $\Rightarrow 5-\frac{5 r}{2}=0$
$\Rightarrow r=2$
Substituting in (i). we get $T_{3}={ }^{10} C_{2} 3^{-3}=\frac{9.10}{2 ! .27}=\frac{5}{3}$

Hence option C is the correct answer.
Question 7
3 / -1

If a, b, are in GP and $\log (\frac{a}{2 b}), \log (\frac{3 b}{3 c}) \log (\frac{3 c}{a})$ are in AP, then a, b, c are the lengths of the sides of a triangle which is

A
acute angled

B
obtuse angled with ∠A being obtuse

C
right angled with ∠A being 90 °

D
equilateral

Solution

$a, b, c$ are in $G P b^{2}=a c$
$\log \left(\frac{a}{2 b}\right), \log \left(\frac{2 b}{3 c}\right), \log \left(\frac{3 c}{a}\right)$
are in $\mathrm{AP}$. $2(\log 2 b-\log 3 c)=\log a-\log 2 b+\log 3 c-\log a$
$\log 2 b=\log 3 c 2 b=3 c$
$\therefore b^{2}=a c \& 2 b=3 c$
$b^{2}=a\left(\frac{2 b}{3}\right)$
$b=2 a / 3$ and $c=4 a / 9$
since $(a+b)=5 a / 3>c,(b+c)=10 a / 9>a$ and
$(c+a)=13 a / 9>b,$ therefore, $a, b, c$ are the sides of a triangle. Also, as 'a' is the greatest side, let us find angle $A$ of $\Delta A B C:$
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{-29}{48}<0$
Hence $\Delta A B C$ is an obtuse angled triangle.

Hence option B is the correct answer.
Question 8
3 / -1

The altitude of a cone is 20cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of

A
π/27 cm/sec

B
8π/27 cm/sec

C
8π/9 cm/sec

D
none of these

Solution

$\frac{d \theta}{d t}=2^{\circ}$
per second
$=2 \times \frac{\pi}{180} \mathrm{rad} / \mathrm{sec}$
$=\frac{\pi}{90} r a d / s e c$
$\theta=30^{\circ}=\frac{\pi}{6}$ radian
Let $\theta$ be the semi-vertical angle and $r$ be the base radius of the cone at time t. Then, $r=20$ tan $\theta$
$\frac{d r}{d t}=20 \sec ^{2} \theta \frac{d \theta}{d t}$
$\left.\frac{d r}{d t}\right]_{\theta=\frac{r}{6}}=\left(20 \sec ^{2} \frac{\pi}{6}\right) \times \frac{\pi}{90}$
$\left.\frac{d r}{d t}\right]_{\theta=\frac{x}{6}}=20 \times \frac{4}{3} \times \frac{\pi}{90}=\frac{8 \pi}{27} \mathrm{cm} / \mathrm{sec}$
Hence option B is the correct answer.
Question 9
3 / -1

A

B

C

D
none of these

Solution
Question 10
3 / -1

Directions: The following question has four choices, out of which ONE or MORE are correct.

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ If the hyperbola passes through a focus of the ellipse, then

A
the equation of the hyperbola is x²/3-y²/2=1

B
the focus of the hyperbola is (2, 0)

C
the eccentricity of the hyperbola is √5/3

D
None of these

Solution

Hence option B is the correct answer.
Question 11
3 / -1

If the roots of ax² + 2bx + c = 0 (a ≠ 0) are non-real complex and a + c < 2b, then

A
c > 0

B
4a - c < 4b

C
4a + c < 4b

D
4a + c > 4b

Solution

since roots of $a x^{2}+2 b x+c=0$ are non-real complex, $\therefore f(x)=a x^{2}+2 b x+c>0$ or $<0$ for all $x$
But $f(-1)=a-2 b+c<0$
$\therefore f(0)$ and $f(-2)$ must be less than zero $f(0)<0$
$\Rightarrow c<0$ and $f(-2)<0$
Hence option C is correct answer.
Question 12
3 / -1

A

B

C

D
none of these

Solution
Question 13
3 / -1

Find the 5 th term form the end in the expansion of $\left(x-\frac{1}{x}\right)^{12}$ ?

A
$-\frac{495}{x^{4}}$

B
$\frac{495}{x^{4}}$

C
$\frac{495}{x^{3}}$

D
$-\frac{495}{x^{3}}$

Solution

Given:
$\left(x-\frac{1}{x}\right)^{12}$
As we know that, in the expansion of $(a+b)^{n}$, the rth term from the end is $[(n+1)-r+1]=(n-r+2)$ th term from the beginning.
Here, $\mathrm{n}=12$ and $\mathrm{r}=5$
So, the 5 th term from the $=[(12+1)-5+1]=9$ th term from the beginning.
In the expansion of $(a+b)^{n}$ the general term is given by:
$\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{b}^{\mathrm{r}}$
$\Rightarrow T_{9}=T_{(8+1)}={ }^{12} C_{8} \cdot(x)^{4} \cdot(\frac{-1 }{ x})^{8}$
$\Rightarrow \mathrm{T}_{9}=\mathrm{T}_{(8+1)}=(-1)^{8} \cdot{ }^{12} \mathrm{C}_{8} \cdot({x})^{4} \cdot(\frac{1 }{x})^{8}$
$\Rightarrow \mathrm{Tg}=\frac{495}{x^{4}}$
Hence the correct option is (B).
Question 14
3 / -1

A
$\frac{\ln 3}{128 \ln 4}$

B
$\frac{9 \ln 3}{128 \ln 4}$

C
$\frac{9 \ln 3}{\ln 4}$

D
$\frac{9 \ln 3}{64 \ln 4}$

Solution
Question 15
3 / -1

A
1

B
2

C
4

D
8

Solution