Mathematics Test

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Mathematics Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The equations of the sides AB,BC and CA of a triangle ABC are 2x+y=0,x+py=39 and x−y=3 respectively and P(2,3) is its circumcentre. Then which of the following is NOT true?

A
\((A C)^2=9 p\)

B
\((\mathrm{AC})^2+\mathrm{p}^2=136\)

C
\(32<\operatorname{area}(\triangle \mathrm{ABC})<36\)

D
\(34<\operatorname{area}(\triangle \mathrm{ABC})<38\)

Solution

Intersection of 2x+y=0 and x−y=3:A(1,−2)

Equation of perpendicular bisector of AB is

x−2y=−4

Equation of perpendicular bisector of AC is

x+y=5

Point B is the image of A in line x−2y+4=0 which is

obtained as B \(\left(\frac{-13}{5}, \frac{26}{5}\right)\)

Similarly vertex C:(7,4)

Equation of line BC:x+8y=39

So, p=8

\(\mathrm{AC}=\sqrt{(7-1)^2+(4+2)^2}=6 \sqrt{2}\)

Area of triangle ABC=32.4
Question 2
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If \(A B^{T}\) is defined as a square matrix then what is the order of the matrix \(B\), where matrix \(A\) has order \(2 \times 3\)?

A
\(3 \times 3\)

B
\(2 \times 3\)

C
\(4 \times 2\)

D
\(3 \times 2\)

Solution

Let the matrix \(B\) has an order \(p \times q\), i.e, \(p\) rows and \(q\) columns.
Transpose of \(B\) is \(B'\) will have order \(q \times p\).
For matrix, \(A B^{T}=\left[A_{(2 \times 3)} \times B_{(q \times p)}^{T }\right]\) is defined.
\(\therefore q=3\)
Given,
\(AB ^{T}\) is a square matrix.
\(AB ^{T}=\left[ A _{(2 \times 3)} \times B _{(3 \times p )}^{T}\right]\)
\(\therefore p=2\)
So, the order of \(B=p \times q=2 \times 3\)
Question 3
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What is the area bounded by the curves \(|y|=1-x^{2}\)?

A
\(\frac{4}{3}\) square units

B
\(\frac{8}{3}\) square units

C
4 square units

D
\(\frac{16}{3}\) square units

Solution

\(|\mathrm{y}|=\left\{\begin{array}{cc}-y, & y<0 \\ y, & y \geq 0\end{array}\right.\)

For \(y \geq 0\)

\(y=1-x^{2}\)

For \(y<0\)

\(-y=1-x^{2}\)

\(\Rightarrow y=x^{2}-1\)

This can be drawn as:

So, area under the curve \(=4 \times\) Area under the region OABO (symmetry)

\(=4 \times \int_{0}^{1}\left(1-\mathrm{x}^{2}\right) \mathrm{dx}\)

\(=4 \times\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{3}\right]_{0}^{1}\)

\(=4 \times\left(1-\frac{1}{3}\right)\)

\(=\frac{8}{3}\) square units
Question 4
1 / -0

The number of solutions, of the equation \(\mathrm{e}^{\sin x}-2 \mathrm{e}^{-\sin x}=2\) is

A
2

B
more than 2

C
1

D
0

Solution

\(\begin{aligned}& \text { Take } e^{\sin x}=t(t>0) \\& \Rightarrow t-\frac{2}{t}=2 \\& \Rightarrow \frac{t^2-2}{t}=2 \\& \Rightarrow t^2-2 t-2=0 \\& \Rightarrow t^2-2 t+1=3 \\& \Rightarrow(t-1)^2=3 \\& \Rightarrow t=1 \pm \sqrt{3} \\& \Rightarrow t=1 \pm 1.73 \\& \Rightarrow t=2.73 \text { or }-0.73 \text { (rejected as } t>0 \text { ) } \\& \Rightarrow e^{\sin x}=2.73 \\& \Rightarrow \log _e e^{\sin x}=\log _e 2.73 \\& \Rightarrow \sin x=\log _e 2.73>1\end{aligned}\)
So no solution.
Question 5
1 / -0

If an angle \(A\) of a \(\triangle A B C\) satisfies \(5 \cos A+3=0\), then the roots of the quadratic equation, \(9 x^2+27 x+20=0\) are.

A
\(\sin A, \sec A\)

B
\(\sec A, \tan A\)

C
\(\tan A, \cos A\)

D
\(\sec A, \cot A\)

Solution

Here, \(9 x^2+27 x+20=0\)
\(\begin{aligned}& \therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\& \Rightarrow x=\frac{-27 \pm \sqrt{27^2-4 \times 9 \times 20}}{2 \times 9} \\& \Rightarrow x=-\frac{4}{3},-\frac{5}{3}\end{aligned}\)
Given, \(\cos A=-\frac{3}{5}\)
\(\therefore \sec A=\frac{1}{\cos A}=-\frac{5}{3}\)
Here, \(A\) is an obtuse angle.
\(\therefore \tan A=-\sqrt{\sec ^2 A-1}=-\frac{4}{3}\)
Hence, roots of the equation are \(\sec A\) and \(\tan A\)
Question 6
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If \(z\) be a complex number satisfying \(|\operatorname{Re}(z)|+|\operatorname{Im}(Z)|=4\), then \(|Z|\) cannot be:

A
\(\sqrt{\frac{17}{2}}\)

B
\(\sqrt{10}\)

C
\(\sqrt{7}\)

D
\(\sqrt{8}\)

Solution

\(\begin{aligned}& z=x+i y|x|+|y|=4 \\& |z|=\sqrt{x^2+y^2}\end{aligned}\)

Minimum value of

\(|z|=2 \sqrt{2}\)

Maximum value of

\(|z|=4|z| \in[\sqrt{8}, \sqrt{16}]\)

So, \(|z|\) can't be \(\sqrt{7}\).
Question 7
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What is the value of \(\lambda\) for which the vectors \(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\) are perpendicular?

A
\(21\)

B
\(-18\)

C
\(-22\)

D
\(22\)

Solution

Given: \(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\) are perpendicular
Let \(\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\)
We know that,
If vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular then \(\vec{a} \cdot \vec{b}=0\)
\(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}})=0\)
\(\Rightarrow-2-20-\lambda=0\)
\(\Rightarrow-22-\lambda=0\)
\(\therefore \lambda=-22\)
Question 8
1 / -0

If \(Z=1+i\), where \(i=\sqrt{-1}\), then what is the modulus of \(Z+\frac{2}{Z} ?\)

A
\(1\)

B
\(2\)

C
\(3\)

D
\(4\)

Solution

Given,
\(Z=1+i\)
We have to find the modulus of \(Z+\frac{2}{Z}\)
\(\Rightarrow(1+i)+\frac{2}{1+i}\)
On rationalizing the second term, we get
\(\Rightarrow(1+i)+\frac{2}{1+i} \times \frac{1-i}{1-i}\)
\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{1- i ^{2}}\)
\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{1-(-1)}\)
\(\Rightarrow(1+ i )+\frac{2 \times(1- i )}{2}\)
\(\Rightarrow 1+ i +1- i\)
\(\Rightarrow 2\)
\(\therefore\) The modulus of \(Z +\frac{2}{ Z }=2\).

Question 9

1 / -0

The equations x = a cos θ - b sin θ , and y = a sin θ + b cos θ, 0 ≤ θ ≤ 2π represent:

An ellipse

A parabola

A hyperbola

None of these

Solution

As we know,

The general equation of a non-degenerate conic section is \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) where \(a, h\) and \(b\) are all not zero

The above-given equation represents a non-degenerate conics whose nature is given below in the table:

S.No	Condition	Nature of Conic
1	h = 0 and a = b	Circle
2	h = 0 and Either a = 0 or b = 0	Parabola
3	h = 0, a≠ b and ab > 0	Ellipse
4	h = 0, a≠ b and sign of a and b are opposite	Hyperbola

Given,

\(x=a \cos \theta-b \sin \theta \)...(i)

\(y=a \sin \theta+b \cos \theta\)...(ii)

Squaring both sides of (i) and (ii) and adding both the equation we get,

\(x^{2}+y^{2}=a^{2}+b^{2}\)

By comparing the given equation with \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\), we get

\( a =1, h =0, b =1\)

As we can see that \(h =0, a = b\)

So, the equation represents the locus of the circle.

Question 10
1 / -0

The equations of the sides \(\mathrm{AB}, \mathrm{BC}\) and \(\mathrm{CA}\) of a triangle \(\mathrm{ABC}\) are \(2 \mathrm{x}+\mathrm{y}=0, \mathrm{x}+\mathrm{py}=15 \mathrm{a}\) and \(\mathrm{x}-\mathrm{y}=3\) respectively. If its orthocentre is \((2, \mathrm{a}),-\frac{1}{2}<\mathrm{a}<2\), then \(\mathrm{p}\) is equal to.................

A
3

B
4

C
5

D
6

Solution

Slope of \(\mathrm{AH}=\frac{\mathrm{a}+2}{1}\)

Slope of \(B C=-\frac{1}{p}\)

\(\therefore \mathrm{p}=\mathrm{a}+2\)

Coordinate of \(\mathrm{C}=\left(\frac{18 \mathrm{p}-30}{\mathrm{p}+1}, \frac{15 \mathrm{p}-33}{\mathrm{p}+1}\right)\)

Slope of \(\mathrm{HC}=\frac{\frac{15 \mathrm{p}-33}{\mathrm{p}+3}-\mathrm{a}}{\frac{18 \mathrm{p}-33}{\mathrm{p}+1}-2}=\frac{15 \mathrm{p}-33-(\mathrm{p}-2)(\mathrm{p}+1)}{18 \mathrm{p}-30-2 \mathrm{p}-2}\)

\(\begin{aligned}& =\frac{16 p-p^2-31}{16 p-32} \\& \because \frac{16 p-p^2-31}{16 p-32} \times-2=-1 \\& \therefore p^2-8 \mathrm{p}+15=0 \\& \therefore \mathrm{p}=3 \text { or } 5\end{aligned}\)

But if \(\mathrm{p}=5\) then \(\mathrm{a}=3\) not acceptable

\(\therefore \mathrm{p}=3\)

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Mathematics Test - 19

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Mathematics Test - 19

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Question 1

\((A C)^2=9 p\)

\((\mathrm{AC})^2+\mathrm{p}^2=136\)

\(32<\operatorname{area}(\triangle \mathrm{ABC})<36\)

\(34<\operatorname{area}(\triangle \mathrm{ABC})<38\)

Solution

Question 2

\(3 \times 3\)

\(2 \times 3\)

\(4 \times 2\)

\(3 \times 2\)

Solution

Question 3

\(\frac{4}{3}\) square units

\(\frac{8}{3}\) square units

4 square units

\(\frac{16}{3}\) square units

Solution

Question 4

2

more than 2

1

0

Solution

Question 5

\(\sin A, \sec A\)

\(\sec A, \tan A\)

\(\tan A, \cos A\)

\(\sec A, \cot A\)

Solution

Question 6

\(\sqrt{\frac{17}{2}}\)

\(\sqrt{10}\)

\(\sqrt{7}\)

\(\sqrt{8}\)

Solution

Question 7

\(21\)

\(-18\)

\(-22\)

\(22\)

Solution

Question 8

\(1\)

\(2\)

\(3\)

\(4\)

Solution

Question 9

An ellipse

A parabola

A hyperbola

None of these

Solution

Question 10

3

4

5

6

Solution

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