Mathematics Test

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Mathematics Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

The solution of \(\frac{d y}{d x}=\frac{a x+h}{b y+k}\) represents a parabola, when :

A
a = 0, b = 0

B
a = 1, b = 2

C
a = 0, b≠ 0

D
a = 2, b = 1

Solution

The given differential equation is

\(\frac{d y}{d x}=\frac{a x+h}{b y+k}\)

On integrating both sides.

\(\int(b y+k) d y=\int(a x+h) d x\)

\(\Rightarrow \quad \frac{b y^{2}}{2}+k y=\frac{a x^{2}}{2}+h x+c\)

Thus, above equation represents a parabola, if

\(a=0\) and \(b \neq 0\)

Hence option C is the correct answer.
Question 2
1 / -0

If \(y=a^{x} b^{2 x-1},\) then \(\frac{d^{2} y}{d x^{2}}\) is equal to :

A
y².log ab²

B
y.log ab²

C
y.(log ab²)²

D
y.(log a²b)²

Solution

\(\because y=a^{x} b^{2 x-1}\)
Taking log on both sides, we get
\[\log y=x \log a+(2 x-1) \log b
\]On differentiating w.r.t. \(x,\) we get
\[\begin{aligned}
\frac{1}{y} \frac{d y}{d x} &=\log a+\log b^{2} \\
\frac{d y}{d x} &=y \log a b^{2}
\end{aligned}
\]Again differentiating, we get
\[\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x} \log a b^{2}=y\left(\log a b^{2}\right)^{2}
\]

Hence option C is the correct answer.
Question 3
1 / -0

A missile fired from the ground level rises x meters vertically upwards in t seconds, where x = 100t - 25/2 t². The maximum height reached is :

A
200 m

B
125 m

C
160 m

D
190 m

Solution

The given equation is

\(x=100 t-\frac{25}{2} t^{2}\)

On differentiating w.r.t. \(t,\) we get \(\frac{d x}{d t}=100-\frac{25}{2} \cdot(2 t)=100-25 t\) We know that the velocity of missile is zero at maximum height. \(\therefore\) On putting \(\frac{d x}{d t}=0,\) we get

\(\begin{aligned} 100-25 t &=0 \\ \Rightarrow \quad t=4 \\ \therefore x=100 \times 4-\frac{25 \times 16}{2} &=400-200 \\ &=200 \end{aligned}\)

Hence option A is the correct answer.
Question 4
1 / -0

A vector perpendicular to \(2 \hat{i}+\hat{j}+\hat{k}\) and coplanar with \(\hat{i}+2 \hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}+2 \hat{k}\) is :

A

\(5(\hat{j}-\hat{k})\)

B

\(\hat{i}+7 \hat{j}-\hat{k}\)

C

\(5(\hat{j}+\hat{k})\)

D

\(2 \hat{i}-7 \hat{j}-\hat{k}\)

Solution

Any vector \(\perp\) to \(\vec{a}\) and coplanar to \(\vec{b}\) and \(\vec{c}\) is given by \(\vec{a} \times(\vec{b} \times \vec{c})\) \(\therefore\) Required vector is

\((2 \hat{i}+\hat{j}+\hat{k}) \times[(\hat{i}+2 \hat{j}+\hat{k}) \times(\hat{i}+\hat{j}+2 \hat{k})]\)

\(=(2 \hat{i}+\hat{j}+\hat{k}) \times\left[\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right]\)

\(=(2 \hat{i}+\hat{j}+\hat{k}) \times(3 \hat{i}-\hat{j}-\hat{k})\)

\(=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & -1 & -1\end{array}\right|=5(\hat{j}-\hat{k})\)

Hence option A is the correct answer.
Question 5
1 / -0

If n(U) = 20, n(A) = 12, n(B) = 9, n(AB) = 4, where U is the universal set, A and B are subsets of U, then n | (A U B)^o| is equal to :

A
17

B
9

C
11

D
3

Solution

\(n(U)=20, n(A)=12, n(B)=9, n(A \cap B)=4\)

\(\therefore n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(=12+9-4=17\)

Hence, \(n\left[(A \cup B)^{c}\right]=n(U)-n(A \cup B)\)

\(=20-17=3\)

Hence option D is the correct answer.
Question 6
1 / -0

The derivative of y = (1 - x)(2 - x)….(n - x) at x = 1 is equal to :

A
0

B
(-1) (n - 1)!

C
n! - 1

D
(-1)n - 1 (n - 1)!

Solution

Derivative of y=(1-x)(2-x)…….(n-x)

dy/dx(uvw)=u’vw+uv’w+uvw’

dy/dx= -1(2-x)(3-x)…(n-x)+ (1-x)(-1)(3-x)..(n-x)+……+(1-x)(2-x)……(n-1-x)(-1)

at x=1,only one term will be left as other terms include (1-x) become zero.

dy/dx(at x=1)= -1(2-1)(3-1)….(n-1)+0+…..+….+0

=> -1*(1)(2)…..(n—1)

=>-1(n-1)!

Hence option B is the correct answer.
Question 7
1 / -0

The circumradius of the triangle whose sides are 13, 12 and 5, is :

A
15

B
13/2

C
15/2

D
6

Solution

Let sides are \(a=13, b=12, c=5\)
Now, \(a^{2}=b^{2}+c^{2}\)
\(\Rightarrow \quad(13)^{2}=(12)^{2}+5^{2}\)
\(\Rightarrow \quad 169=169\)
\(\Rightarrow \quad \angle A =90^{\circ}\)
We know, \(R=\frac{a}{2 \sin A}\)
\[R=\frac{13}{2 \cdot \sin 90^{\circ}}=\frac{13}{2}
\]

Hence option B is the correct answer.
Question 8
1 / -0

If ⁿC₁₂ = ⁿC₆, then ⁿC₂ is equal to :

A
72

B
153

C
306

D
2556

Solution

Given \({ }^{\mathrm{n}} \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_{6}\)

or \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-12}={ }^{\mathrm{n}} \mathrm{C}_{6}\)

\(\Rightarrow \mathrm{n}-12=6\)

\(\Rightarrow \mathrm{n}=18\)

\(\therefore \mathrm{C}_{2}={ }^{18} \mathrm{C}_{2}=\frac{18 \times 17}{2 \times 1}\)

\(=153\)

Hence option B is the correct answer.
Question 9
1 / -0

The sum of 24 terms of the series √2 + √8 + √18 + √32 + .... is:

A
300

B
200 √2

C
300 √2

D
250 √2

Solution

\begin{equation}\begin{array}{l}
\text { Now, } \sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots \\
\quad=1 \times \sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2}+\ldots \\
\quad=\sqrt{2}(1+2+3+4+\ldots \text { upto } 24 \text { terms }) \\
\quad=\sqrt{2} \times \frac{24 \times 25}{2}=300 \sqrt{2}\left[\because \Sigma n=\frac{n(n+1)}{2}\right]
\end{array}\end{equation}

Hence option C is the correct answer.
Question 10
1 / -0

The equation of the circle with centre (2, 1) and touching the line 3x + 4y = 5 is:

A
x² + y² - 4x - 2y + 5 = 0

B
x²+ y²- 4x - 2y - 5 = 0

C
x²+ y²- 4x - 2y + 4 = 0

D
x²+ y²- 4x - 2y - 4 = 0

Solution

Distance from centre (2,1) to the line
\(3 x+4 y-5=\) radius of circle
\(\Rightarrow \quad \frac{|3(2)+4(1)-5|}{\sqrt{3^{2}+4^{2}}}=r\)
\(\Rightarrow \quad \frac{5}{5}=r\)
\(\Rightarrow \quad r=1\)
\(\therefore\) Equation of circle is \((x-2)^{2}+(y-1)^{2}=1^{2}\)
\(\Rightarrow \quad x^{2}+y^{2}-4 x-2 y+4+1=1\)
\(\Rightarrow \quad x^{2}+y^{2}-4 x-2 y+4=0\)

Hence option C is the correct answer.

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Mathematics Test - 8

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Mathematics Test - 8

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Question 1

a = 0, b = 0

a = 1, b = 2

a = 0, b≠ 0

a = 2, b = 1

Solution

Question 2

y2.log ab2

y.log ab2

y.(log ab2)2

y.(log a2b)2

Solution

Question 3

200 m

125 m

160 m

190 m

Solution

Question 4

\(5(\hat{j}-\hat{k})\)

\(\hat{i}+7 \hat{j}-\hat{k}\)

\(5(\hat{j}+\hat{k})\)

\(2 \hat{i}-7 \hat{j}-\hat{k}\)

Solution

Question 5

17

9

11

3

Solution

Question 6

0

(-1) (n - 1)!

n! - 1

(-1)n - 1 (n - 1)!

Solution

Question 7

15

13/2

15/2

6

Solution

Question 8

72

153

306

2556

Solution

Question 9

300

200 √2

300 √2

250 √2

Solution

Question 10

x2 + y2 - 4x - 2y + 5 = 0

x2+ y2- 4x - 2y - 5 = 0

x2+ y2- 4x - 2y + 4 = 0

x2+ y2- 4x - 2y - 4 = 0

Solution

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y².log ab²

y.log ab²

y.(log ab²)²

y.(log a²b)²

x² + y² - 4x - 2y + 5 = 0

x²+ y²- 4x - 2y - 5 = 0

x²+ y²- 4x - 2y + 4 = 0

x²+ y²- 4x - 2y - 4 = 0