Mathematics Test

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Mathematics Test - 29

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Weekly Quiz Competition

Question 1
4 / -1

If \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0,\) then

A
a = 1 and b = -2

B
a = 1 and b = -1

C
a = -1 and b = 1

D
a = 2 and b = -2

Solution

\(\begin{aligned} & \lim _{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-a x-b=0 \\ \Rightarrow & \lim _{x \rightarrow \infty} \frac{x^{2}+1-a x(x+1)-b(x+1)}{x+1}=0 \\ \Rightarrow & \lim _{x \rightarrow \infty} \frac{(1-a) x^{2}-(a+b) x+1-b}{x+1}=0 \end{aligned}\)
iff coefficient of \(x^{2}=0\) and coefficient of \(x=0\)
\(\Rightarrow(1-a)=0\) and \(-(a+b)=0\)
\(\Rightarrow a=1\) and \(b=-1\)
Question 2
4 / -1

If A and B are two matrices such that AB = B and BA = A, then A² + B² is equal to

A
2AB

B
2BA

C
A + B

D
AB

Solution

We have

\(A^{2}+B^{2}=(B A)^{2}+(A B)^{2}\)

\(=(B A)(B A)+(A B)(A B)\)

\(=\mathrm{B}(\mathrm{AB}) \mathrm{A}+\mathrm{A}(\mathrm{BA}) \mathrm{B}\)

\(=B(B A)+A(A B)=B A+A B=A+B\)
Question 3
4 / -1

The locus of the point of intersection of the lines x sinθ + (1 – cos θ) y = a sinθ and x sin – (1 + cosθ ) y + a sinθ = 0 is

A
x² – y²= a²

B
x²+ y²= a²

C
y² = ax

D
None of these

Solution

From the given equations we have
\(\frac{1-\cos \theta}{\sin \theta}=\frac{a-x}{y}\) and \(\frac{1+\cos \theta}{\sin \theta}=\frac{a+x}{y}\)
Multiplying we get \(\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=\frac{a^{2}-x^{2}}{y^{2}} \Rightarrow x^{2}+y^{2}=a^{2}\)
Question 4
4 / -1

If p^th, q^th and r^th terms of an AP, respectively are a, b and c, then a(q - r) + b(r - p) + c(p - q) is equal to

A
1

B
0

C
a + b + c

D
-1

Solution

Given:
\(\mathrm{p}^{\mathrm{th}}\) term \(=\mathrm{a} \Rightarrow \mathrm{t}_{1}+(\mathrm{p}-1) \mathrm{d}=\mathrm{a} \ldots(1)\)
\(\mathrm{q}^{\mathrm{th}}\) term \(=\mathrm{b} \Rightarrow \mathrm{t}_{1}+(\mathrm{q}-1) \mathrm{d}=\mathrm{b} \ldots(2)\)
\(r^{\text {th }}\) term \(=b \Rightarrow t_{1}+(r-1) d=c \ldots(3)\)
The given expression is: \(a(q-r)+b(r-p)+c(p-q)\)
This is equivalent to: \(q(a-c)+r(b-a)+p(c-b)\) Putting the values of \(a, b\) and \(c\) in terms of \(t_{1}\) and \(d\) in the given expression, we get
\(\mathrm{q}[(\mathrm{p}-1) \mathrm{d}-(\mathrm{r}-1) \mathrm{d}]+\mathrm{r}[(\mathrm{q}-1) \mathrm{d}-(\mathrm{p}-1) \mathrm{d}]+\mathrm{p}[(\mathrm{r}-1) \mathrm{d}-(\mathrm{q}-1) \mathrm{d}]\)
\(=q[p d-d-r d+d]+r[q d-d-p d+d]+p[r d-d-q d+d]\)
\(=\mathrm{qpd}-\mathrm{qrd}+\mathrm{rqd}-\mathrm{prd}+\mathrm{prd}-\mathrm{qpd}=0\)
\(a(q-r)+b(r-p)+c(p-q)=0\)
Thus, answer option (2) is correct.
Question 5
4 / -1

If \(A=\left[\begin{array}{lll}1 & 0 & 2 \\ 5 & 1 & x \\ 1 & 1 & 1\end{array}\right]\) is a singular matrix, then \(x\) is equal to

A
3

B
5

C
9

D
11

Solution

As \(A\) is a singular matrix.
\(\Rightarrow \quad \mathrm{L} \mid=0\)
\(\Rightarrow\left|\begin{array}{ccc}1 & 0 & 0 \\ 5 & 1 & x-10 \\ 1 & 1 & -1\end{array}\right|=0 \quad\left[\right.\) using \(\left.C_{3} \rightarrow C_{3}-2 C_{1}\right]\)
\(\Rightarrow\left|\begin{array}{cc}1 & x-10 \\ 1 & -1\end{array}\right|=0\)
\(\Rightarrow-1-x+10=0\)
\(\Rightarrow \quad x=9\)
Question 6
4 / -1

The sides of the rectangle of the greatest area, that can be inscribed in the ellipse x² + 2y²= 8, are given by

A
4√2,4

B
4,2√2

C
2,√2

D
2√2,2

Solution

Any point on the ellipse \(\frac{x^{2}}{8}+\frac{y^{2}}{4}=1\) is \((2 \sqrt{2} \cos \theta, 2 \sin \theta)\)

\[
\begin{aligned}
A &=\text { Area of the inscribed rectangle } \\
&=4(2 \sqrt{2} \cos \theta)(2 \sin \theta) \\
&=8 \sqrt{2} \sin 2 \theta \\
& \frac{d A}{d \theta}=16 \sqrt{2} \cos 2 \theta=0 \Rightarrow \theta=\pi / 4 \\
\text { Also } \frac{d^{2} A}{d \theta^{2}} &=-32 \sqrt{2} \sin 2 \theta<0 \text { for } \theta=\pi / 4
\end{aligned}
\]
Hence, the inscribed rectangle is of largest area if the sides are \(4 \sqrt{2} \cos \pi / 4\) and \(4 \sin (\pi / 4)\) i.e. 4 and \(2 \sqrt{2}\)
Question 7
4 / -1

Let \(f(x)=g(x) \frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}},\) where \(g\) is a continuous function. Then, \(\lim _{x \rightarrow 0} f(x)\) exists if

A
g(x) = x + 2

B
g(x) = x² + 4

C
g(x) = xh(x), where h(x) is a polynomial

D
g(x) is a constant function

Solution

\[
\begin{array}{c}
\lim _{x \rightarrow 0+e^{1 / x}+e^{-1 / x}}=\lim _{x \rightarrow 0+1+e^{-2 / x}} \frac{1-e^{-2 / x}}{1+1}=1 \\
\lim _{x \rightarrow 0-e^{1 / x}+e^{-1 / x}}=\lim _{x \rightarrow 0-e^{2 / x}+1}=-1
\end{array}
\]
Hence \(\lim _{x \rightarrow 0} f(x)\) exists if \(g(x)=x h(x)\) where \(h(x)\) is a continuous function. If \(g(x)=a(a \neq 0)\) then \(\lim _{x \rightarrow 0+} f(x)=a\)
and \(\lim _{x \rightarrow 0-} f(x)=-a .\) Thus \(\lim _{x \rightarrow 0} f(x)\) does not exist if \(f(x)\)
\(=x+2\) or \(x^{2}+4\) or is a constant function.
Question 8
4 / -1

If \(1+\omega+\omega^{2}=0\) and \(\omega^{3}=1,\) find the value of \(\left[\begin{array}{cc}\omega & \omega+1 \\ \omega+1 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -1 \\ 1+\omega & \omega-1\end{array}\right]\)

A
\(\left[\begin{array}{cc}0 & a^{2} \\ 0 & -2\end{array}\right]\)

B
\(\left[\begin{array}{cc}0 & 2 \omega \\ 0 & -2\end{array}\right]\)

C
\(\left[\begin{array}{cc}0 & 2 a^{2} \\ 0 & -2\end{array}\right]\)

D
\(\left[\begin{array}{cc}1 & 2 a^{2} \\ 0 & -2\end{array}\right]\)

Solution

Multiplying the matrices, we get \(\left[\begin{array}{cc}\omega & \omega+1 \\ \omega+1 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -1 \\ 1+\omega & \omega-1\end{array}\right]\)
\(=\left[\begin{array}{cc}-a+1+2 a+a^{2} & -a+a^{2}-1 \\ -a-1+1+a & -a-1+a-1\end{array}\right]\)
\(=\left[\begin{array}{cc}1+a+a^{2} & a^{2}-1-a \\ 0 & -2\end{array}\right]\)
\(=\left[\begin{array}{cc}0 & 2 a^{2} \\ 0 & -2\end{array}\right]\)
Question 9
4 / -1

\(\mathrm{n} \in \mathrm{N}\) satisfies the inequality \(\left.\mathrm{n}-1 \mathrm{C}_{3}+\mathrm{n}-1 \mathrm{C}_{4}\right\rangle \mathrm{n}_{\mathrm{C}_{3}}\) and \(\mathrm{E}=\left(\mathrm{x}^{2}-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{n}}\)

The terms independent of x in the expansion of E are in

A
A.P.

B
G.P.

C
A.G.P.

D
None of these

Solution

\(\mathrm{n}-\mathrm{l}_{3}+\mathrm{n}-1 \mathrm{C}_{4}>\mathrm{n} \mathrm{C}_{3}\)
\(\Rightarrow \mathrm{n}>7 \Rightarrow \mathrm{n} \in[8,9,10, \ldots .] \ldots \ldots .(1)\)
\(E=\left(x^{2}-\frac{1}{2 x}\right)^{n}\)
\(\Rightarrow \mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{\mathrm{n}-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}}\right)^{\mathrm{r}}=\frac{(-1)^{\mathrm{x} \mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{x}^{2 \mathrm{n}-3 \mathrm{r}}}{2^{\mathrm{r}}}\)
For the terms independent of \(\mathrm{x}\).
\(2 n-3 r=0 \Rightarrow 3 r=2 n \Rightarrow n=9,12,15,18 \ldots\) from (1) (as n should be a multiple of \(\Rightarrow r=6,8,10,12 \ldots\)
\(\Rightarrow\) The terms independent of \(x\) are \(\frac{{ }^{9} \mathrm{C}_{6}}{2^{6}}, \frac{12 \mathrm{C}_{8}}{2^{8}}, \frac{15 \mathrm{C}_{10}}{2^{10}}\)
Question 10
4 / -1

Let S = (x – 1)¹⁰ + (x – 1)⁹ (x + 1) + (x – 1)⁸ (x + 1)² + ……. + (x + 1)¹⁰

If O is the sum of coefficients of odd powers of x and E is the sum of the coefficients of even powers of x, which of the following statements is correct?

A
E = 2¹¹

B
E = O

C
O = 2¹¹

D
E = 2¹⁰

Solution

\(\begin{aligned} S &=(x-1)^{10}+(x-1)^{9}(x+1)+(x-1)^{8}(x+1)^{2}+\ldots \ldots+(x+1)^{10} \\ &=(x-1)^{10}\left\{\frac{\left(\frac{x+1}{x-1}\right)^{11}-1}{\frac{x+1}{x-1}-1}\right\} \\ &=\frac{1}{2}(x+1)^{11}-(x-1)^{11} \\ &=11 \mathrm{C}_{1} \mathrm{x}^{10}+{ }^{11} \mathrm{C}_{3} \mathrm{x}^{8}+{ }^{11} \mathrm{C}_{5} \mathrm{x}^{6}+{ }^{11} \mathrm{C}_{7} \mathrm{x}^{4}+{ }^{1} \mathrm{C}_{9} \mathrm{x}^{2}+1 \\ \Rightarrow & 0=0, \mathrm{E}=2^{\mathrm{n}-1}=2^{10} \end{aligned}\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 29

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Mathematics Test - 29

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SHARING IS CARING

Question 1

a = 1 and b = -2

a = 1 and b = -1

a = -1 and b = 1

a = 2 and b = -2

Solution

Question 2

2AB

2BA

A + B

AB

Solution

Question 3

x2 – y2 = a2

x2 + y2 = a2

y2 = ax

None of these

Solution

Question 4

1

0

a + b + c

-1

Solution

Question 5

3

5

9

11

Solution

Question 6

4√2,4

4,2√2

2,√2

2√2,2

Solution

Question 7

g(x) = x + 2

g(x) = x2 + 4

g(x) = xh(x), where h(x) is a polynomial

g(x) is a constant function

Solution

Question 8

\(\left[\begin{array}{cc}0 & a^{2} \\ 0 & -2\end{array}\right]\)

\(\left[\begin{array}{cc}0 & 2 \omega \\ 0 & -2\end{array}\right]\)

\(\left[\begin{array}{cc}0 & 2 a^{2} \\ 0 & -2\end{array}\right]\)

\(\left[\begin{array}{cc}1 & 2 a^{2} \\ 0 & -2\end{array}\right]\)

Solution

Question 9

A.P.

G.P.

A.G.P.

None of these

Solution

Question 10

E = 211

E = O

O = 211

E = 210

Solution

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x² – y²= a²

x²+ y²= a²

y² = ax

g(x) = x² + 4

E = 2¹¹

O = 2¹¹

E = 2¹⁰