Mathematics Test

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Mathematics Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The value of the sum of the series 3. ⁿC₀ - 8 . ⁿC₁ + 13 . ⁿC₂ - 18 . ⁿC₃ + ......upto (n + 1) terms where n>2, is

A
3ⁿ

B
5ⁿ

C
0

D
None of these

Solution

$3 .{ }^{n} C_{0}-8 .{ }^{n} C_{1}+13 .{ }^{n} C_{2}-18 .{ }^{n} C_{3}+\ldots \ldots . .$
$\begin{aligned}{=3\left({ }^{n} C_{0}-{ }^{n} C_{1}+{ }^{n} C_{2}-{ }^{n} C_{3}+\ldots \ldots .\right)}{ } &+5\left(-{ }^{n} C_{1}+2 \cdot{ }^{n} C_{2}-3 \cdot{ }^{n} C_{3}+\ldots . .\right) \end{aligned}$
$\because(1-x)^{n}={ }^{n} \mathrm{C}_{0}-{ }^{n} \mathrm{C}_{1} \mathrm{x}+{ }^{n} \mathrm{C}_{2} \mathrm{x}^{2}-{ }^{n} \mathrm{C}_{3} \mathrm{x}^{3}+\ldots \ldots \ldots$
and $-n(1-x)^{n-1}=-{ }^{n} C_{1}+2^{n} C_{2} x-3^{n} C_{3} x^{2}+\ldots \ldots$
Putting $x=1$ in Eqs. (i) $\&$ (ii), we get
$$
{ }^{\mathrm{n}} \mathrm{C}_{0}-{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}-{ }^{\mathrm{n}} \mathrm{C}_{3}+\ldots \ldots .=0
$$
and $\quad-{ }^{n} C_{1}+2 \cdot{ }^{n} C_{2}-3 \cdot{ }^{n} C_{3}+\ldots \ldots=0$
Hence, $3 .{ }^{n} C_{0}-8 .{ }^{n} C_{1}+13 .{ }^{n} C_{2}-18 .{ }^{n} C_{3}+\ldots . .=0$
Hence, the correct option is (C)
Question 2
1 / -0

The system x + 4y - 2z = 3, 3x + y + 5z = 7 and 2x + 3y + z = 5 has

A
Infinite number of solutions

B
Unique solution

C
Trivial solution

D
No solution

Solution

Given system of equations are
$$
x+4 y-2 z=3
$$
and $3 x+y+5 z=7$
$$
2 x+3 y+z=5
$$
$\therefore \Delta=\left|\begin{array}{ccc}1 & 4 & -2 \\ 3 & 1 & 5 \\ 2 & 3 & 1\end{array}\right|$
$=1(1-15)-4(3-10)-2(9-2)$
$=-14+28-14=0$
and $\Delta_{2}=\left|\begin{array}{ccc}1 & 3 & -2 \\ 3 & 7 & 5 \\ 2 & 5 & 1\end{array}\right|=1 \neq 0$
$\therefore$ No solution will exist.
Hence, the correct option is (D)
Question 3
1 / -0

Let $\int \mathrm{e}^{\mathrm{x}}\left\{f(\mathrm{x})-f^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}\right\}=\phi(\mathrm{x})+\mathrm{C}$ Then $\int \mathrm{e}^{\mathrm{x}} f(\mathrm{x}) \mathrm{d} \mathrm{x}$ is equal to

A
$\phi(\mathrm{x})+\mathrm{e}^{\mathrm{x}} f(\mathrm{x})$

B
$\phi(\mathrm{x})-\mathrm{e}^{\mathrm{x}} f(\mathrm{x})$

C
$\frac{1}{2}\left(\phi(x)+e^{x} f(x)\right)+k$

D
$\phi(\mathrm{x})-\mathrm{e}^{\mathrm{x}} f^{\prime}(\mathrm{x})$

Solution

Here, $\int \mathrm{e}^{\mathrm{x}}\left\{f(\mathrm{x})-\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{d} \mathrm{x}=\phi(\mathrm{x})+\mathrm{C}$
and $\int \mathrm{e}^{\mathrm{x}}\left\{f(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{d} \mathrm{x}=\mathrm{e}^{\mathrm{x}} f(\mathrm{x})+\mathrm{C}$
On adding, we get $2 \int \mathrm{e}^{\mathrm{x}} f(\mathrm{x}) \mathrm{dx}=\phi(\mathrm{x})+\mathrm{e}^{\mathrm{x}} f(\mathrm{x})+\frac{c+c^{\prime}}{k}$
$\int e^{x} f(x) d x=\frac{1}{2}\left(\phi(x)+e^{x} f(x)\right)+k$
Hence, the correct option is (C)
Question 4
1 / -0

$\operatorname{Letf}_{(x)}=x^{2}-4 x-3, x>2$and let g(x) be the inverse of f(x). Find the value of g'(2) where f (x) = 2

A
$\frac{1}{6}$

B
$\frac{1}{5}$

C
$\frac{1}{4}$

D
$\frac{1}{7}$

Solution

As $g(f(x))=x$
differentiating $\left(\because g(x)=f^{-1}(x)\right)$
$g^{\prime}(f(x)) f^{\prime}(x)=1$
$g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)}$
take $f(x)=2$
$x^{2}-4 x-3=2$
$x=5, x=-1$
$\because x>2$
Hence $\therefore \quad x=5$
$g^{\prime}(2)=\frac{1}{f^{\prime}(5)}$
$=\frac{1}{(2 x-4)_{x=5}}$
$=\frac{1}{6}$
Hence, the correct option is (A)
Question 5
1 / -0

$\int_{0}^{1} \frac{\tan ^{-1} x}{x} d x$.

A
$\frac{1}{2} \int_{0}^{\pi / 2} \frac{\mathrm{x}}{\sin \mathrm{x}} \mathrm{d} \mathrm{x}$

B
$\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos x} d x$

C
$\int_{0}^{\pi / 2} \frac{x}{\sin x} d x$

D
$\int_{0}^{\pi / 2} \frac{x}{\cos x} d x$

Solution

Put $\tan ^{-1} x=t$
$\Rightarrow x=\tan t \Rightarrow d x=\sec ^{2} d t$
$\therefore \int_{0}^{1} \frac{\tan ^{-1} x}{x} d x=\int_{0}^{\pi / 4} \frac{t}{\tan t} \sec ^{2} t d t$
$=\int_{0}^{\pi / 4} \frac{t}{\sin t} \cdot \cos t \frac{1}{\cos ^{2} t} d t$
$=\int_{0}^{\pi / 4} \frac{t}{\sin t \cos t} d t$
$=2 \int_{0}^{\pi / 4} \frac{t}{\sin 2 t} d t ;$ put $2 t=p$
$=\int_{0}^{\pi / 2} \frac{p}{\sin p}\left(\frac{1}{2} d p\right)$
$=\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\sin x} d x$
Hence, the correct option is (A)
Question 6
1 / -0

Let $f:[a, b] \rightarrow R$ be a function such that for $c \in(a, b), f^{\prime}+(c)=f(c)=f(c)=f^{i v}(c)=f^{v}(c)=0$, then :

A
f has local extremum at x = c

B
f has neither local maximum nor local minimum at x = c

C
f is necessarily a constant function

D
it is difficult to say whether (f has local extremum at x = c) or (f has neither local maximum nor local minimum at x = c)

Solution

If even derivative is + ve, then minima and if even derivative is -ve, then maxima.

But here, $\mathrm{f}^{\mathrm{vi}}(\mathrm{f}$ is necessarily a constant function) is not given,

Hence, it is difficult to say maximum or minimum, i.e. (f has local extremum at $\mathrm{x}=\mathrm{c}$ ) or $(\mathrm{f}$ has neither local maximum nor local minimum at $x=c$ ).
Hence, the correct option is (D)
Question 7
1 / -0

A car manufacturing company has two plants. Plant P manufactures 70% of cars and plant Q 30%. At plant P 80% of the cars are rated as of standard quality and at plant Q, 90% of the cars are rated as of standard quality. A car is chosen at random and is found to be of standard quality. The chance that it has come from plant P is.

A
$\frac{56}{83}$

B
52/83

C
$\frac{50}{83}$

D
$\frac{55}{83}$

Solution

Let $\mathrm{E}_{\mathrm{i}}(\mathrm{i}=1,2)$ be the event that $t^{\mathrm{th}}$ plant is selected $\mathrm{A}$ and the event when a standard quality car is selected.
$$
\begin{array}{l}
\therefore \quad \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{70}{100} \text { and } \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{30}{100} \\
\text { And } \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=\frac{80}{100} \text { and } \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=\frac{90}{100}
\end{array}
$$
By Bayes' Theorem,
$$
\begin{aligned}
\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=& \frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)} \\
=& \frac{\frac{70}{100} \times \frac{80}{100}}{\frac{70}{100} \times \frac{80}{100}+\frac{30}{100} \times \frac{90}{100}} \\
=& \frac{5600}{5600+2700}=\frac{56}{83}
\end{aligned}
$$
Hence, the correct option is (A)
Question 8
1 / -0

The p^thderivative of a q^thdegree monic polynomial, where p, q are positive integers and 2p⁴+ 3pq3⁄2= 3q^3⁄2+ 2qp³is given by?

A
Cannot be generally determined

B
(q – 1)!

C
(q)!

D
(q – 1)! * pq

Solution

First consider the equation
2p⁴+ 3pq^3⁄2= 3q^3⁄2+ 2qp³
After simplification, we get
(2p³+ 3q^1⁄2) (p – q) = 0
This gives us two possibilities
2p³= – 3q^1⁄2
OR
p = q
The first possibility can’t be true as we are dealing with positive integers
Hence, we get
p = q
Thus the pthderivative of any monic polynomial of degree p(p = q) is
p! =q!.

Hence, the correct option is (C)
Question 9
1 / -0

The solution of the differential equationis:dy/dx=1+x+y+xy { Where C is an arbitrary constant }

A
$\log _{e}|1+y|=\frac{x^{2}}{2}+x+c$

B
$(1+y)^{2}=x+\frac{x^{2}}{2}+c$

C
$\log _{e}|1+y|=\log _{e}|1+x|+c$

D
$\log _{e}|3+x|=\log _{e}|1+x|+c$

Solution

$\frac{d y}{d x}=(1+x)+(y+x y)$
$=(1+x)+y(1+x)$
$=(1+x)(1+y)$
$\therefore \frac{d y}{1+y}=d x(1+x)$
Integrating both sides
$\log _{e}|1+y|=\frac{x^{2}}{2}+x+c$
Hence, the correct option is (A)
Question 10
1 / -0

A purse contains 4 copper coins and 3 silver coins the second purse contains 6 copper coins and 2 silver coins. A coin is taken out from any purse. The probability that it is a copper coin is :

A
$\frac{4}{7}$

B
$\frac{37}{56}$

C
$\frac{3}{7}$

D
$\frac{3}{4}$

Solution

Required probability

12 × 47 + 12 × 68

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 34

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Mathematics Test - 34

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SHARING IS CARING

Question 1

3n

5n

0

None of these

Solution

Question 2

Infinite number of solutions

Unique solution

Trivial solution

No solution

Solution

Question 3

\(\phi(\mathrm{x})+\mathrm{e}^{\mathrm{x}} f(\mathrm{x})\)

\(\phi(\mathrm{x})-\mathrm{e}^{\mathrm{x}} f(\mathrm{x})\)

\(\frac{1}{2}\left(\phi(x)+e^{x} f(x)\right)+k\)

\(\phi(\mathrm{x})-\mathrm{e}^{\mathrm{x}} f^{\prime}(\mathrm{x})\)

Solution

Question 4

16

15

14

17

Solution

Question 5

\(\frac{1}{2} \int_{0}^{\pi / 2} \frac{\mathrm{x}}{\sin \mathrm{x}} \mathrm{d} \mathrm{x}\)

\(\frac{1}{2} \int_{0}^{\pi / 2} \frac{x}{\cos x} d x\)

\(\int_{0}^{\pi / 2} \frac{x}{\sin x} d x\)

\(\int_{0}^{\pi / 2} \frac{x}{\cos x} d x\)

Solution

Question 6

f has local extremum at x = c

f has neither local maximum nor local minimum at x = c

f is necessarily a constant function

it is difficult to say whether (f has local extremum at x = c) or (f has neither local maximum nor local minimum at x = c)

Solution

Question 7

5683

52/83

5083

5583

Solution

Question 8

Cannot be generally determined

(q – 1)!

(q)!

(q – 1)! * pq

Solution

Question 9

\(\log _{e}|1+y|=\frac{x^{2}}{2}+x+c\)

\((1+y)^{2}=x+\frac{x^{2}}{2}+c\)

\(\log _{e}|1+y|=\log _{e}|1+x|+c\)

\(\log _{e}|3+x|=\log _{e}|1+x|+c\)

Solution

Question 10

47

3756

37

34

Solution

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3ⁿ

5ⁿ

$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{4}$

$\frac{1}{7}$

$\frac{56}{83}$

$\frac{50}{83}$

$\frac{55}{83}$

$\frac{4}{7}$

$\frac{37}{56}$

$\frac{3}{7}$

$\frac{3}{4}$