Mathematics Test

Result

Mathematics Test - 55

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant would stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000 .

A
The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1250000.

B
The merchant should stock 200 desktop models and 100 portable models to get the maximum profit of Rs.1150000.

C
The merchant should stock 100 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

D
The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

Solution

Let the merchant stock $x$ desktop model and $y$ portable models. Therefore,

$x \geq 0$ and $y \geq 0$

The cost of a desktop model is Rs. 25000 and of a portable model is Rs. 4000 . However, the merchant can invest a maximum of Rs. 70 lakhs.

$\therefore 25000 \mathrm{x}+40000 \mathrm{y} \leq 7000000$

$5 \mathrm{x}+8 \mathrm{y} \leq 1400$

The monthly demand of computers will not exceed 250 units.

$\therefore x+y \leq 250$

The profit on a desktop model is Rs. 4500 and the profit on a profit model is Rs. 5000 .

Total profit, $Z=4500 \mathrm{x}+5000 \mathrm{y}$

Thus, the mathematical formulation of the given problem is,

Maximum $\mathrm{Z}=4500 \mathrm{x}+5000 \mathrm{y}$....(1)

Subject to the constraints,

$5 x+5 y \leq 1400 \ldots \ldots . .(2)$

$x+y \leq 250 \ldots \ldots . .(3)$

$x, y \geq 0 \ldots \ldots \ldots(4)$

The feasible region determined by the system of constraints is as shown.

The corner points are $\mathrm{A}(250,0), \mathrm{B}(200,50)$ and $\mathrm{C}(0,175)$

The values of $Z$ at these corner points are as follows

Corner point A(250,0),

$Z=4500 \mathrm{x}+5000 \mathrm{y}=1125000$

Corner point A(200,50),

$Z=4500 \mathrm{x}+5000 \mathrm{y}=1150000$

Corner point A(0,175),

$Z=4500 \mathrm{x}+5000 \mathrm{y}=875000$

The maximum value of Z is 1150000 at (200,50).

Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1150000.
Question 2
1 / -0

If the function $f:(1, \infty) \rightarrow[1, \infty]$ is defined by $f(x)=2^{x(x-1)}$ then $f^{-1}(x)$ is

A
$\frac{1}{2} x(x-1)$

B
$\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right)$

C
$\frac{1}{2}\left(1-\sqrt{1+4 \log _{2} x}\right)$

D
Not defined

Solution

Given :
A function $\Rightarrow 4 \log _{2} y \geq 0$ defined by $f(x)=2^{x(x-1)}$
We have to find the value of $f^{-1}(x)$
Let $y=2^{x(x-1)}$ where $y \geq 1$ as $x \geq 1$
since, $x \geqslant 1$
$\Rightarrow f(x) \geqslant 1$
$\Rightarrow y \geqslant 1$
Now, taking log with base 2 on both sides, we get
$\log _{2} y=\log _{2} 2 x(x-1)$
[Using properties of logarithmic function-3 ]
$\Rightarrow \log _{2} y=x(x-1) \log _{2} 2$
[Using properties of logarithmic function-5 ]
$\Rightarrow \log _{2} y=x(x-1)$
$\Rightarrow x^{2}-x-\log _{2} y=0$
Which is a quadratic equation in $x$.
Thus, using the quadratic formula, we get
$x=\frac{1 \pm \sqrt{1+4 \log _{2} y}}{2}$
For $ \log _{2} y,~y \geq 1 $
$y=f(x) \in[1, \infty]$
$\Rightarrow 4 \log _{2} y \geq 0 \quad\left[\right.$ Multiplying '$^{} 4^{}$' on both sides]
$\Rightarrow 1+4 \log _{2} y \geq 1 \quad$ [Adding '1' both sides]
Taking square root on both sides,
$\sqrt{1+4 \log _{2} y} \geq 1$
$\Rightarrow-\sqrt{1+4 \log _{2} y} \leq-1$ [Multiplying '$^{}-1^{}$' on both sides. If $a \geq b$ then $\left.-a \leq-b\right]$
$\Rightarrow 1-\sqrt{1+4 \log _{2} y} \leq 0$ [Adding '$^{} 1^{}$' on both sides $]$
But $x \geq 1,$ so $x=\frac{1-\sqrt{1+4 \log _{2} y}}{2} \leqslant 1$ is not possible.
Therefore, we take $x=\frac{1}{2}(1+\sqrt{1+4\log _{2} y})$
$\Rightarrow x-f^{-1}(y)=\frac{1}{2}(1+\sqrt{1+4\log _{2} y})$
$\Rightarrow f^{-1}(x)=\frac{1}{2}(1+\sqrt{1+ 4\log _{2} x})$ [Replacing $y$ by $\left.x\right]$
Question 3
1 / -0

Let $f(x)=\frac{1}{\sqrt{10-x^{2}}} $. What is the value of $\lim _{x \rightarrow 1} \frac{{f}({x})-{f}(1)}{{x}-1}$?

A

$-\frac{1}{27}$

B
$0$

C
$\frac{1}{27}$

D
$1$

Solution

The given function is:
${f}({x})=\frac{1}{\sqrt{10-{x}^{2}}}$
${f}({x})=\frac{1}{\sqrt{10-{x}^{2}}}=\left(10-{x}^{2}\right)^{-\frac{1}{2}}$
Differentiating above equation with respect to $x$, we get:
${f}^{\prime}({x})=\frac{-1}{2}\left(10-{x}^{2}\right)^{\left(\frac{-1}{2}-1\right)} \times(0-2 {x})$
${f}^{\prime}({x})=\frac{{x}}{\left(10-{x}^{2}\right)^{\frac{3}{2}}}$ ....(1)
Therefore,
$ f(1)=\frac{1}{\sqrt{10-1^{2}}}=\frac{1}{3}$
$\Rightarrow\underset{{{{x} \rightarrow 1}}}{\lim} \frac{{f}({x})-{f}(1)}{{x}-1}=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{\frac{1}{\sqrt{10-x^{2}}}-\frac{1}{3}}{{x}-1}$
Putting $x=1$ in above equation, gives $\frac{0}{0}$ form.
We know that:
L-Hospital Rule as:
$\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$
Applying L-Hospital rule,
$\lim _{x \rightarrow 1} \frac{{f}({x})-{f}(1)}{{x}-1}=\lim _{x \rightarrow 1} \frac{\frac{d}{dx}({f}({x})-{f}(1))}{\frac{d}{dx}({x}-1)}$
$=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{\frac{x}{\left(10-x^{2}\right)^{\frac{3}{2}}}}{1}$ $\quad(\because$from equation (1)$)$
$=\frac{1}{27}$
Question 4
1 / -0

$\mathrm{ABC}$ is a triangle, right angled at $\mathrm{A}$. The resultant of the forces acting along $\overline{\mathrm{AB}}, \overline{\mathrm{BC}}$ with magnitudes $\frac{1}{\mathrm{AB}}$ and $\frac{1}{\mathrm{AC}}$ respectively is the force along $\overline{\mathrm{AD}}$, where $\mathrm{D}$ is the foot of the perpendicular from $\mathrm{A}$ onto $\mathrm{BC}$. The magnitude of the resultant is:

A

AB2+AC2

(AB)2(AC)2

B

(AB)(AC)

AB+AC

C
$\frac{1}{\mathrm{AB}}+\frac{1}{\mathrm{AC}}$

D

1

AD

Solution

If we consider unit vectors $\hat{i}$ and $\hat{j}$ in the direction $A B$ and $A C$ respectively and its magnitude $\frac{1}{\mathrm{AB}}$ and $\frac{1}{\mathrm{AC}}$ respectively, then as per quesiton, forces along $\mathrm{AB}$ and $\mathrm{AC}$ respectively are $\left(\frac{1}{\mathrm{AB}}\right) \hat{\mathrm{i}}$ and $\left(\frac{1}{\mathrm{AC}}\right) \hat{\mathrm{j}}$

$\therefore$ Their resultant along $\mathrm{AD}=\left(\frac{1}{\mathrm{AB}}\right) \hat{\mathrm{i}}+\left(\frac{1}{\mathrm{AC}}\right) \hat{\mathrm{j}}$

$\therefore$ Magnitude of resultant is

$=\sqrt{\left(\frac{1}{\mathrm{AB}}\right)^2+\left(\frac{1}{\mathrm{AC}}\right)^2}=\sqrt{\frac{\mathrm{A} \mathrm{C}^2+\mathrm{AB}^2}{\mathrm{AB}^2+\mathrm{AC}^2}}\left[\because A \mathrm{AC}^2+\mathrm{AB}^2=\mathrm{BC}^2\right] \\$

$ =\frac{\mathrm{BC}}{\mathrm{AB} \cdot \mathrm{AC}}$

$ \because \triangle \mathrm{ABC} \sim \triangle \mathrm{DBA} \\$

$ \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{AC}}{\mathrm{AD}} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB} \times \mathrm{AC}}=\frac{1}{\mathrm{AD}}$

$\therefore$ The required magnitude of resultant becomes $\frac{1}{\mathrm{AD}}$
Question 5
1 / -0

Let $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\beta \mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ be two vectors, such that $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 \hat{k}$. Then the projection of $\vec{b}-2 \vec{a}$ on $\vec{b}+\vec{a}$ is equal to:

A
2

B
$\frac{39}{5}$

C
9

D
$\frac{46}{5}$

Solution

$\begin{aligned} & \vec{a}=\alpha_i+\hat{j}+\beta \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k} \\ & \vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k} \\ & \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4\end{array}\right|=-\hat{i}+9 \hat{j}+12 \hat{k} \\ & 4+5 \beta=-1 \Rightarrow \beta=-1 \\ & -5 \alpha-3=12 \Rightarrow \alpha=-3 \\ & \vec{b}-2 \vec{a}=3 \hat{i}-5 \hat{j}+4 \hat{k}-2(-3 \hat{i}+\hat{j}-\hat{k}) \\ & \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}=9 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \vec{b}+\vec{a}=(3 \hat{i}-5 \hat{j}+4 \hat{k})+(-3 \hat{i}+\hat{j}-\hat{k}) \\ & \vec{b}+\vec{a}=-4 \hat{j}+3 \hat{k} \\ & \text { Projection of } \vec{b}-2 \vec{a} \text { on } \vec{b}+\vec{a} \text { is }=\frac{(\vec{b}-2 \vec{a}) \cdot(\vec{b}+\vec{a})}{|\vec{b}+\vec{a}|} \\ & =\frac{28+18}{5}=\frac{46}{5} \\ & \end{aligned}$
Question 6
1 / -0

$\mathrm{P(n)=2 \times 7^{n}+3 \times 5^{n}}-5$ is divisible by:

A
$24, \forall \mathrm{n} \in \mathrm{N}$

B
$21, \forall\mathrm{n} \in \mathrm{N}$

C
$32, \forall\mathrm{n} \in \mathrm{N}$

D
$50, \forall\mathrm{n} \in \mathrm{N}$

Solution

Given,
$\mathrm{P(n)=2 \times 7^{n}+3 \times 5^{n}}-5$
By putting $\mathrm{n}=1$, we get
$\mathrm{P(1)}=2 \times 7^{1}+3 \times 5^{1}-5=24$
As we can say that $24$ is divisible by $24$.
Assume $\mathrm{P(k)}$ is true, then
$\mathrm{P(k)}=2 \times 7^{\mathrm{k}}+3 \times 5^{\mathrm{k}}-5=24 q$, where $\mathrm{q \in N}$$\cdots(i)$
By putting $\mathrm{n}=k+1$, we get
$\mathrm{P}(\mathrm{k}+1)=2 \times 7^{\mathrm{k}+1}+3 \times 5^{\mathrm{k}+1}-5$
$=2 \times 7^{\mathrm{k}} \times 7+3 \times 5^{\mathrm{k}} \times 5-5$
$= 7\left\{2 \times 7^{\mathrm{k}}+3 \times 5^{\mathrm{k}}-5-3 \times 5^{\mathrm{k}}+5\right\}+3 \times 5^{\mathrm{k}} \times 5-5$
$= 7\left\{24 \mathrm{q}-3 \times 5^{\mathrm{k}}+5\right\}+15 \times 5^{\mathrm{k}}-5$
$=(7 \times 24 \mathrm{q)-21 \times 5^{k}+35+15 \times 5^{k}}-5$
$=(7 \times 24 \mathrm{q)-6 \times 5^{k}+30}$
$=(7 \times 24 \mathrm{q)-6\left(5^{k}-5\right)}$
Assume $\mathrm{(5^k-5)=4p}$, then
$=(7 \times 24 \mathrm{q)-6\times(4 p)}$
As $\left(5^\mathrm{{k}}-5\right)$ is a multiple of $4$,
$=7 \times 24 \mathrm{q-24 p}$
$=\mathrm{24(7 q-p)}$
$\Rightarrow 24 \times \mathrm{r}$
where, $\mathrm{r=7 q-p}$, is some natural number.
Thus, by the Principle of Mathematical Induction $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.
Question 7
1 / -0

If the expected value of a random variable $X$ is 2 and its variance is 1, then what will be the variance of $3 X+4$?

A
9

B
6

C
7

D
11

Solution

In this we use the concept:
${V~ar}[a X+b]=a^{2} {Var}(X)$
$V(a x)=a^{2} V(x)$ where a is constant,
$V(a)=0$, i.e. variance of constant is zero ; where a is constant
Given,
${V~ar}(x)=1, {V~ar}(3 x+4)=?$
Let, $y=3 x+4$
$\Rightarrow V(y)=V(3 x+4)$
$\Rightarrow V(y)=3^{2}[V(x)]$
$\Rightarrow V(y)=9[1]$
$\Rightarrow V(y)=V(3 x+4)=9$
Question 8
1 / -0

What is the equation to the plane through (1, 2, 3) parallel to 3x + 4y - 5z = 0?

A
3x + 4y + 5z + 4 = 0

B
3x + 4y - 5z + 14 = 0

C
3x + 4y - 5z + 4 = 0

D
3x + 4y - 5z - 4 = 0

Solution

As we know,
Equation of a plane parallel to a given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0, where λ is a constant.
Equation of a plane parallel to a given plane 3x + 4y - 5z = 0 is 3x + 4y - 5z + λ = 0.
The plane is passing through the point (1, 2, 3).
So, point (1, 2, 3) satisfy the equation of plane 3x + 4y - 5z + λ = 0.
∴(3 × 1) + (4 × 2) - (5 × 3) + λ = 0
⇒ 3 + 8 - 15 + λ = 0
⇒ -4 + λ = 0
∴ λ = 4
So, Equation of plane is 3x + 4y - 5z + 4 = 0.
Question 9
1 / -0

Which of the following equals $1+\cot ^{2} \theta ?$

A
$\sec ^{2} \theta$

B
$\cos ^{2} \theta$

C
$\operatorname{cosec}^{2} \theta$

D
$\cot ^{2} \theta$

Solution

Given,
$1+\cot ^{2} \theta$
$=1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$$\quad\quad(\because\cot\theta =\frac{\cos\theta}{\\sin\theta})$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta}$ $\quad\quad(\because\sin ^{2} \theta+\cos ^{2} \theta=1)$
$=\frac{1}{\sin ^{2} \theta}$
$=\operatorname{cosec}^{2} \theta$
Question 10
1 / -0

Let $A=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2\end{array}\right]$, then the value of A'BA is:

A
1224

B
1042

C
540

D
539

Solution

$A^{\prime} B A=\left[\begin{array}{lll}1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2\end{array}\right] \mathrm{A}$
$=\left[\begin{array}{lll}9^2+12^2-15^2 & -10^2+13^2+16^2 & 11^2-14^2+17^2\end{array}\right]\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$\begin{aligned} & =\left[9^2+12^2-15^2-10^2+13^2+16^2+11^2-14^2+17^2\right] \\ & =\left[\left(9^2-10^2\right)+\left(11^2+12^2\right)+\left(13^2-14^2\right)+\left(16^2-15^2\right)+17^2\right] \\ & =[-19+265+(-27)+31+289] \\ & =[585-46]=[539]\end{aligned}$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 55

Result

Mathematics Test - 55

Score

-

Rank

-

SHARING IS CARING

Question 1

The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1250000.

The merchant should stock 200 desktop models and 100 portable models to get the maximum profit of Rs.1150000.

The merchant should stock 100 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

The merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs.1150000.

Solution

Question 2

\(\frac{1}{2} x(x-1)\)

\(\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right)\)

\(\frac{1}{2}\left(1-\sqrt{1+4 \log _{2} x}\right)\)

Not defined

Solution

Question 3

\(-\frac{1}{27}\)

\(0\)

\(\frac{1}{27}\)

\(1\)

Solution

Question 4

AB2+AC2 (AB)2(AC)2

(AB)(AC) AB+AC

\(\frac{1}{\mathrm{AB}}+\frac{1}{\mathrm{AC}}\)

1 AD

Solution

Question 5

2

395

9

465

Solution

Question 6

\(24, \forall \mathrm{n} \in \mathrm{N}\)

\(21, \forall\mathrm{n} \in \mathrm{N}\)

\(32, \forall\mathrm{n} \in \mathrm{N}\)

\(50, \forall\mathrm{n} \in \mathrm{N}\)

Solution

Question 7

9

6

7

11

Solution

Question 8

3x + 4y + 5z + 4 = 0

3x + 4y - 5z + 14 = 0

3x + 4y - 5z + 4 = 0

3x + 4y - 5z - 4 = 0

Solution

Question 9

\(\sec ^{2} \theta\)

\(\cos ^{2} \theta\)

\(\operatorname{cosec}^{2} \theta\)

\(\cot ^{2} \theta\)

Solution

Question 10

1224

1042

540

539

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

AB2+AC2

(AB)2(AC)2

(AB)(AC)

AB+AC

1

AD

$\frac{39}{5}$

$\frac{46}{5}$