Mathematics Test

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Mathematics Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is

A
69760

B
30240

C
99784

D
none of these

Solution

The total number of words that can be formed is 10⁵ and number of these words in which no letters are repeated is ¹⁰P₅.
Hence the required number,
= 10⁵ − ¹⁰P₅
= 100000 − 10 × 9 × 8 × 7 × 6
= 69760
Hence option A is correct.
Question 2
1 / -0

The tangent to the circle x² + y² = 5 at (1, − 2) also touches the circle x² + y² − 8x + 6y + 20 = 0. Then the point of contact is

A
(-1, 3)

B
(3, -1)

C
(3, 1)

D
(-3, - 1)

Solution

Tangent at (1,-2) to $x^{2}+y^{2}=5$ is $x-2 y-5=0 \ldots(i)$
Centre and radius of $x^{2}+y^{2}-8 x+6 y+20=0$ are
$C (4,-3)$ and radius $r=\sqrt{5}$
Perpendicular distance from $C(4,-3)$ to $(i)$ is radius.
$\therefore$ (i) is also a tangent to the second circle.
Let $P ( h , k )$ be the foot of the drawn circle from $C (4,-3)$ on $( i )$ $\frac{h-4}{1}=\frac{k+3}{-2}=-\frac{[1 \cdot 4-2 \cdot(-3)-5]}{1^{2}+2^{2}}$
$\therefore(h, k)=(3,-1)$
Hence option B is correct.
Question 3
1 / -0

Events $A , B , C$ are mutually exclusive events such that $P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4},$ and $P(C)=\frac{1-2 x}{2}$
Then set of possible values of $x$ are in the interval:

A

\begin{equation}\left[\frac{1}{3}, \frac{2}{3}\right]\end{equation}

B

\begin{equation}\left[\frac{1}{3}, \frac{13}{3}\right]\end{equation}

C

[0,1]

D

\begin{equation}\left[\frac{1}{3}, \frac{1}{2}\right]\end{equation}

Solution

$0 \leq \frac{3 x+1}{3} \leq 1,0 \leq \frac{1-x}{4} \leq 1,0 \leq \frac{1-2 x}{2} \leq 1$
and $0 \leq \frac{3 x+1}{3}+\frac{1-x}{4}+\frac{1-2 x}{2} \leq 1$
Considering the four inequalities $-\frac{1}{3} \leq x \leq \frac{2}{3},-3 \leq x \leq 1,-\frac{1}{2} \leq x \leq \frac{1}{2}$
and $\frac{1}{3} \leq x \leq \frac{13}{3},$ hence $\frac{1}{3} \leq x \leq \frac{1}{2}$
Hence option D is correct.
Question 4
1 / -0

Given vectors $\overrightarrow{ x }=3 \widehat{ i }-6 \widehat{ j }-\widehat{ k }, \overrightarrow{ y }=\widehat{ i }+4 \widehat{ j }-3 \widehat{ k }$ and $\vec{z}=3 \widehat{ i }-4 \widehat{ j }-12 \widehat{ k }$ then the projection of $\overrightarrow{ x } \times \overrightarrow{ y }$ on $\overrightarrow{ z }$ is

A
14

B
-14

C
12

D
15

Solution

The projection of $\frac{(\overrightarrow{ x } \times \overrightarrow{ y }) \cdot \overrightarrow{ z }}{|\overrightarrow{ z }|}$
$=\frac{[\overrightarrow{ x } \overrightarrow{ y } \overrightarrow{ z }]}{|\overrightarrow{ z }|}$
$=\frac{1}{13}\left|\begin{array}{ccc}3 & -6 & -1 \\ 1 & 4 & -3 \\ 3 & -4 & -12\end{array}\right|$
\begin{equation}\begin{array}{l}
=\frac{1}{13}(3(-48-12)+6(-12+9)-(-4-12)) \\
=\frac{1}{13}(-180-18+16) \\
=\frac{1}{13}(-182) \\
=-14
\end{array}\end{equation}
Hence option B is correct.
Question 5
1 / -0

The plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $y$ -axis also passes through the point.

A

(-3,0,1)

B
(3,3,-1)

C
(3,2,1)

D
(-3,1,1)

Solution

Equation of plane
$(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$
$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z-1+4 \lambda=0$
dr's of normal of the plane are
$1+2 \lambda, 1+3 \lambda, 1-\lambda$
Since plane is parallel to y-axis,
$1+3 \lambda=0$
$\Rightarrow \lambda=\frac{-1}{3}$
So the equation of plane is
$x+4 z-y=0$
Point (3,2,1) satisfies this equation
Hence option C is correct.
Question 6
1 / -0

f $\int_{-1}^{-4} f(x) d x=4$ and $\int_{2}^{-4}(3-f(x)) d x=7,$ then the value of $\int_{-2}^{1} f(-x) d x$ is

A
2

B
-3

C
5

D
none of these

Solution

\begin{equation}\begin{array}{l}
\int_{2}^{-4}(3-f(x)) d x=7 \\
\Rightarrow \int_{2}^{-4} 3 d x-\int_{2}^{-4}(f(x)) d x=7 \\
\Rightarrow[3 x]_{2}^{-4}-\int_{2}^{-4}(f(x)) d x=7 \\
\Rightarrow \int_{2}^{-4}(f(x)) d x=-25 \ldots \ldots \ldots \ldots \ldots(i) \\
\text { and } \int_{-1}^{-4}(f(x)) d x=4 \\
\Rightarrow \int_{-4}^{-1}(f(x)) d x=-4 \ldots \ldots \ldots \ldots(i i)
\end{array}\end{equation}now we can write
$\int_{-4}^{2}(f(x)) d x=\int_{-4}^{-1}(f(x)) d x+\int_{-1}^{2}(f(x)) d x$
$\Rightarrow 25=-4+\int_{-1}^{2}(f(x)) d x$
$\Rightarrow \int_{-1}^{2}(f(x)) d x=29 \ldots \ldots \ldots \ldots \ldots(i i i)$
$N o w$
$I=\int_{-2}^{1}(f(-x)) d x$
let $x=-t$
$d x=-d t$
when $x=-2, t=2$
and when $x=1, t=-1$
so $I=\int_{-2}^{1}(f(-x)) d x=\int_{2}^{-1}-(f(t)) d t$
$\Rightarrow I=\int_{-1}^{2}(f(t)) d t$
$\Rightarrow I=\int_{-1}^{2}(f(x)) d x$
$\Rightarrow I=29$ ( from equation (iii))
Hence option D is correct.
Question 7
1 / -0

\begin{equation}\operatorname{lf} g[f(x)]=|\sin x| \text { and } f[g(x)]=(\sin \sqrt{x})^{2} \text { then, }\end{equation}

A

$f(x)=\sin ^{2} x, g(x)=\sqrt{x}$

B

$f(x)=\sin x, g(x)=|x|$

C

$f(x)=x^{2}, g(x)=\sin \sqrt{x}$

D

f and g cannot be determined

Solution

Given $(g \circ f)(x)=|\sin x| \ldots \ldots \ldots \ldots .(1)$
and, $f[g(x)]=(\sin \sqrt{x})^{2}$
$(f \circ g)(x)=(\sin \sqrt{x})^{2}$
For option $(A), f(x)=\sin ^{2} x, g(x)=\sqrt{x}$
$(f \circ g)(x)=f[g(x)]=f(\sqrt{x})$
$\Rightarrow(f \circ g)(x)=f(y)=\sin ^{2} y=\sin ^{2}(\sqrt{x})$
$y=\sqrt{x}$
$(g \circ f)(x)=g[f(x)]=g\left(\sin ^{2} x\right)=g(t) t=\sin ^{2} x$
$\Rightarrow \sqrt{t}=\sqrt{\sin ^{2} x}=|\sin x|$
Hence option A is correct.
Question 8
1 / -0

\begin{equation}\text { The distance between the line } \vec{r}=2 \vec{i}-2 \vec{j}+3 \vec{k}+\lambda(\vec{i}-\vec{j}+4 \vec{k}) \text { and the Plane } \vec{r} .(\vec{i}+5 \vec{j}+\vec{k})=5 \text { is }\end{equation}

A
$\frac{10}{3}$

B
$\frac{3}{10}$

C
$\frac{10}{3 \sqrt{3}}$

D
$\frac{10}{9}$

Solution

The line is parallel to plane because $(\hat{i}-\hat{j}+4 \hat{k}) \cdot(\hat{i}+5 \hat{j}+\hat{k})=0$
General point on the line is $(\lambda+2,-\lambda-2,4 \lambda+3)$ for $\lambda=0$ (because at $\lambda=0,$ the distance of the line is minimum from the plane, a point on this line is (2,-2,3) and distance from $\overrightarrow{ r } \cdot(\widehat{ i }+5 \widehat{ j }+\widehat{ k })=5$
or $x+5 y+z=5$ is
$d=\frac{|2+5(-2)+3-5|}{\sqrt{1+25+1}}$
$\Rightarrow d=\frac{|-10|}{3 \sqrt{3}}=\frac{10}{3 \sqrt{3}}$
Hence option C is correct.
Question 9
1 / -0

If $g ( x )$ satisfies the conditions of Rolle's theorem in [1,2] and $g ^{\prime}( x )= f ( x ),$ then $\int_{1}^{2} f(x) d x$ is equal to

A
1

B
3

C
0

D
2

Solution

As $g ( x )$ satisfies the condition of Rolle's theorem in $[1,2], g ( x )$ is continuous in the interval and $g(1)=g(2)$
$n o w$
$g^{\prime}(x)=f(x)$
$\left.\therefore \int_{1}^{2} f(x) d x=g(x)\right]_{1}^{2}$
$=g(2)-g(1)=0$
Hence option C is correct.
Question 10
1 / -0

Which of the following is logically equivalent to ∼ (∼p ⇒ q) ?

A
p∧q

B
p∧∼ q

C
∼p ∧q

D
∼p ∧ ∼q

Solution

p q ~p ~q ~p⇒q ~(~p⇒q) p ∧ q
p ∧∼ q
∼p ∧ q
∼p∧ ∼q
F F T T F T F F F T
F T T F T F F F T F
T F F T T F F T F F
T T F F T F T F F F

The values in column 6 and column 10 are same.
Hence option D is correct.

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Re-Attempt Weekly Quiz Competition

p	q	~p	~q	~p⇒q	~(~p⇒q)	p ∧ q	p ∧∼ q	∼p ∧ q	∼p∧ ∼q
F	F	T	T	F	T	F	F	F	T
F	T	T	F	T	F	F	F	T	F
T	F	F	T	T	F	F	T	F	F
T	T	F	F	T	F	T	F	F	F

Mathematics Test - 1

Result

Mathematics Test - 1

Score

-

Rank

-

SHARING IS CARING

Question 1

69760

30240

99784

none of these

Solution

Question 2

(-1, 3)

(3, -1)

(3, 1)

(-3, - 1)

Solution

Question 3

\begin{equation}\left[\frac{1}{3}, \frac{2}{3}\right]\end{equation}

\begin{equation}\left[\frac{1}{3}, \frac{13}{3}\right]\end{equation}

[0,1]

\begin{equation}\left[\frac{1}{3}, \frac{1}{2}\right]\end{equation}

Solution

Question 4

14

-14

12

15

Solution

Question 5

(-3,0,1)

(3,3,-1)

(3,2,1)

(-3,1,1)

Solution

Question 6

2

-3

5

none of these

Solution

Question 7

\(f(x)=\sin ^{2} x, g(x)=\sqrt{x}\)

\(f(x)=\sin x, g(x)=|x|\)

\(f(x)=x^{2}, g(x)=\sin \sqrt{x}\)

f and g cannot be determined

Solution

Question 8

103

310

1033

109

Solution

Question 9

1

3

0

2

Solution

Question 10

p∧q

p∧∼ q

∼p ∧q

∼p ∧ ∼q

Solution

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$\frac{10}{3}$

$\frac{3}{10}$

$\frac{10}{3 \sqrt{3}}$

$\frac{10}{9}$