Mathematics Test

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Mathematics Test - 15

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Weekly Quiz Competition

Question 1
4 / -1

The value of \(\int_{0}^{1} \mathrm{x}(1-\mathrm{x})^{99} \mathrm{dx}\) is

A
1/10010

B
1/10110

C
100/10110

D
1/10100

Solution

\(\begin{aligned} I &=\int_{0}^{1} x(1-x)^{99} d x \\ &=\int_{0}^{1}(1-x)\{1-(1-x)\}^{99} d x \\ &=\int_{0}^{1}(1-x) x^{99} d x \\ &=\int_{0}^{1}\left(x^{99}-x^{100}\right) d x \\ &=\left[\frac{x^{100}}{100}-\frac{x^{101}}{101}\right]_{0}^{1} \\ &=\frac{1}{100}-\frac{1}{101}=\frac{1}{10100} \end{aligned}\)
Question 2
4 / -1

The orthocenter of the triangle with vertices
\(\left(2, \frac{\sqrt{3}-1}{2}\right) \cdot\left(\frac{1}{2},-\frac{1}{2}\right)\) and \(\left(2-\frac{1}{2}\right)\) is

A
\(\left(\frac{3}{2}, \frac{\sqrt{3}-3}{6}\right)\)

B
\(\left(2-\frac{1}{2}\right)\)

C
\(\left(\frac{5}{4},-\frac{\sqrt{3}-2}{4}\right)\)

D
\(\left(\frac{1}{2}, \frac{-a)}{2}\right)\)

Solution

As we can see that the triangle so formed is a rightangle triangle, right angled at the point \(\left(2-\frac{1}{2}\right)\), hence

the orthocenter in a right-angle triangle is the point of right angle.
Question 3
4 / -1

If one root of the quadratic equation \(a x^{2}+b x+c=0\) is equal to the \(n^{\text {th }}\) power of the other root, then the value of \(\left(a c^{n}\right)^{\frac{1}{n+1}}+\left(a^{n} c\right)^{\frac{1}{n+1}}\)

A
b

B
-b

C
b(1/n+1)

D
-b(1/n+1)

Solution

Let \(\alpha, \alpha^{n}\) be two roots, Then \(\alpha+\alpha^{n}=-\frac{b}{a}, \alpha \alpha^{n}=\frac{c}{a}\)
Eliminating \(\alpha,\) we get \(\left(\frac{c}{a}\right)^{\frac{1}{n+1}}+\left(\frac{c}{a}\right)^{\frac{n}{n+1}}=-\frac{b}{a}\)
\(\Rightarrow a \cdot a^{-\frac{1}{n+1}} \cdot c^{\frac{1}{n+1}}+a \cdot a^{-\frac{n}{n+1}} \cdot c^{\frac{n}{n+1}}=-b\)
or \(\left(a^{n} c\right)^{\frac{1}{n+1}}+\left(a c^{n}\right)^{\frac{1}{n+1}}=-b\)
Question 4
4 / -1

In a ΔABC, a = 13 cm, b = 12 cm and c = 5 cm. The distance of A from BC is

A
25/13

B
60/13

C
65/12

D
144/13

Solution

ABC is a right angled triangle. Given, \(a=13, b=12\) and \(c=5\) \(s=\frac{13+12+5}{2}\)
Area \((\Delta)=\sqrt{s(s-a)(s-b)(s-c)}-\sqrt{15(2)(3)(10)}-\sqrt{900}-30\)
Also area \(=\frac{1}{2}\) base \(\times\) alt \(30=\frac{1}{2} x+3 x a t\)
Altitude \(=\frac{60}{13}\)
Question 5
4 / -1

If the sum of odd coefficients of (1+z)⁵ is x, find the number of solutions of a+b+c+d=x where -x

A
³⁴C₃

B
³³C₃

C
³⁵C₃

D
³⁶C₄

Solution

Sum of odd coefficients of \((1+z)^{5}\) is \(2^{5-1}=2^{4}=16=x\) \(a+b+c+d=16\) and \(-16Let \(a=e-16\) then \(e>0\) \(e-16+b+c+d=16\)
\(e+b+c+d=32\)
Number of positive integral solutions \(-(n+r-1) c_{r-1}\) \((32+4-1) c_{4-1}=(35) c_{3}\)
Question 6
4 / -1

The lengths of sides of a triangle are in the ratio 5 : 12 : 13 and its area is 270 cm². The respective lengths of sides of the triangle (in cm) are

A
5, 12, 13

B
10, 24, 26

C
15, 36, 39

D
20, 48, 52

Solution

Let the lengths of sides of the triangle be 5x, 12x, 13x; Obviously, the triangle is right-angled.

Hence, area of the Δ = ½ (12x) (5x) ⇒ 30x² = 270 ⇒ x = 3

Hence, the lengths of sides (in cm) are 15, 36 and 39.
Question 7
4 / -1

The value of the function \(f(n)\) is \(\frac{35}{16}\) which is represented \(\operatorname{as} f(n)=\sum_{n=0}^{\infty}(1+3 n) x^{n}\) Find the value of \(x\)

A
3/7

B
1/5

C
1/3

D
3/5

Solution

\(f(n)=1+4 x+7 x^{2}+10 x^{3}+\)
(2) \(x f(n)=x+4 x^{2}+7 x^{3}+10 x^{4}+\frac{x}{x}\)
Subtracting (1) and (2) , we get \((1-x) f(n)=1+3 x+3 x^{2}+3 x^{3}+\)
\(\Rightarrow(1-x) f(n)=1+\frac{3 x}{1-x}\)
\(\Rightarrow(1-x) \times \frac{35}{16}=\frac{1+2 x}{1-x}\)
\(\Rightarrow 35(1-x)^{2}=16+32 x\)
\(\Rightarrow 35 x^{2}-102 x+19=0\)
\(\Rightarrow(7 x-19)(5 x-1)=0\)
\(x \neq \frac{19}{7}\) (. For infinity series, common ratio \(\left.<1\right)\)
\(\therefore x=\frac{1}{5}\)
Question 8
4 / -1

A
x/√(1+x²)

B
x

C
x√(1+x²)

D
√(1+x²)

Solution
Question 9
4 / -1

If the lengths of the sides of a triangle are 3, 4, 5 units, then R (the circum-radius) is equal to

A
2.0

B
2.5

C
3.0

D
3.5

Solution

Sides of length 3.4 .5 form \(r t\) angled

Let B be \(90^{\circ}\) \(\mathrm{ABC} \rightarrow \Delta\)

So, \(R=\frac{B}{2 \sin B}=\frac{5}{2 a}=2.5\)
Question 10
4 / -1

In a triangle, the lengths of the two larger sides are 10 cm and 9 cm respectively. If the angles of the triangle are in A.P, then the length of the third side (in cm) can be

A
(5 - √6) only

B
(5 + √6) only

C
(5 - √6) or (5 + √6)

D
Neither (5 - √6) nor (6 + √6)

Solution

We know that in a triangle, larger the side larger the angle. since \(\angle \mathrm{A}, \angle \mathrm{B},\) and \(\angle \mathrm{C}\) are in \(\mathrm{AP}\)
\(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \Rightarrow \frac{1}{2}=\frac{100+a^{2}-81}{20 a}\)
\(\Rightarrow a^{2}+19=10 a\)
\(\Rightarrow a^{2}-10 a+19\)
\(a=\frac{10 \pm \sqrt{100-76}}{2}=5 \pm \sqrt{6}\)

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Mathematics Test - 15

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Mathematics Test - 15

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Question 1

1/10010

1/10110

100/10110

1/10100

Solution

Question 2

\(\left(\frac{3}{2}, \frac{\sqrt{3}-3}{6}\right)\)

\(\left(2-\frac{1}{2}\right)\)

\(\left(\frac{5}{4},-\frac{\sqrt{3}-2}{4}\right)\)

\(\left(\frac{1}{2}, \frac{-a)}{2}\right)\)

Solution

Question 3

b

-b

b(1/n+1)

-b(1/n+1)

Solution

Question 4

25/13

60/13

65/12

144/13

Solution

Question 5

34C3

33C3

35C3

36C4

Solution

Question 6

5, 12, 13

10, 24, 26

15, 36, 39

20, 48, 52

Solution

Question 7

3/7

1/5

1/3

3/5

Solution

Question 8

x/√(1+x2)

x

x√(1+x2)

√(1+x2)

Solution

Question 9

2.0

2.5

3.0

3.5

Solution

Question 10

(5 - √6) only

(5 + √6) only

(5 - √6) or (5 + √6)

Neither (5 - √6) nor (6 + √6)

Solution

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³⁴C₃

³³C₃

³⁵C₃

³⁶C₄

x/√(1+x²)

x√(1+x²)

√(1+x²)