Mathematics Test

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Mathematics Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

The value of the expression \(x^{2}+\frac{1}{x^{2}+1}\) for \(x \in R\) is

A
\([1, \infty)\)

B
\([2, \infty)\)

C
\([0, \infty)\)

D
\(\left[\frac{1}{2}, \infty\right)\)

Solution

The given expression can be written \(\frac{x^{4}+x^{2}+1}{x^{2}+1}=\frac{x^{4}}{x^{2}+1}+1\)

Now, for real values of \(x\), the quantity \(x^{4} / x^{2}+1\) is greater than or equal to zero. attaining this value at \(x=0\).

Therefore,

\(x^{2}+\frac{1}{x^{2}+1} \geq 1 \quad \forall x \in R\)

and its minimum value 1 is attained at x = 0.
Question 2
1 / -0

Let \(A=\left[\begin{array}{ccc}0 & \sin \alpha & \sin \alpha \sin \beta \\ -\sin \alpha & 0 & \cos \alpha \cos \beta \\ -\sin \alpha \sin \beta & -\cos \alpha \cos \beta & 0\end{array}\right]\), then

A
\(|A|\) is independent of \(\alpha \& \beta\)

B
\(A^{-1}\) depends only on \(\alpha\)

C
\(A^{-1}\) depends only \(\beta\)

D
IA I is dependent of \(\alpha\)

Solution

\(|A|=-\sin \alpha(\sin \alpha \sin \beta \cos \alpha \cos \beta)+\sin \alpha \sin \beta\)
\((\sin \alpha \cos \alpha \cos \beta)\)
\(=-\sin ^{2} \alpha \sin \beta \cos \alpha \cos \beta+\sin ^{2} \alpha \sin \beta \cos \alpha \cos \beta\)
\(=0\)
\(\because|A|=0 \Rightarrow A^{-1}\) does not exists.
Question 3
1 / -0

Six numbers are in AP such that their sum is 3. The first term is 4 times the third term. Then, the fifth term is

A
-15

B
-3

C
9

D
-4

Solution

Let the numbers be \(a-5 d, a-3 d, a-d, a+d, a+3 d, a+5 d\)
\(\therefore a-5 d+a-3 d+a-d+a+d+a+3 d+a+5 d=3(\because S u m=3)\)
\(\Rightarrow 6 a=3 \Rightarrow a=\frac{1}{2}\)
Also, given \(T_{1}=4 T_{3}\), where \(T_{1}, T_{3}\) are respectively, first and third terms of AP. \(\Rightarrow a-5 d=4(a-d)\)
\(\Rightarrow d=-3 a=-\frac{3}{2}\)
\(\therefore\) The fifth term \(a+3 d=\frac{1}{2}+3\left(-\frac{3}{2}\right)=\frac{1}{2}-\frac{9}{2}=-4\)
Question 4
1 / -0

If |a|,|b|,|c|<1 and a,b,cϵA.P. then (1+a+a²+...∞),(1+b+b²+...∞),(1+c+c²+...∞) are in

A
A.P

B
G.P

C
H.P

D
None of these

Solution

Consider the terms
\(\left(1+a+a^{2}+\ldots \infty\right),\left(1+b+b^{2}+\ldots \infty\right),\left(1+c+c^{2}+\ldots \infty\right)\)
Using the property for sum of an infinite G.p., we get \(1+a+a^{2}+a^{3}+\ldots \infty=\frac{1}{1-a}\)
\(1+b+b^{2}+b^{3}+\ldots \infty=\frac{1}{1-b}\)
\(\operatorname{and} 1+c+c^{2}+\ldots . \infty=\frac{1}{1-c}\)
Given that, \(a, b, c\) are in \(\mathrm{A}\). P. \(\Rightarrow 1-a, 1-b, 1-c\) are in \(\mathrm{A} . P\)
\(\Rightarrow \frac{1}{1-a}, \frac{1}{1-b}, \frac{1}{1-c}\) are in \(\mathrm{H.P}\)
Therefore, \(\left(1+a+a^{2}+\ldots \infty\right),\left(1+b+b^{2}+\ldots \infty\right),\left(1+c+c^{2}+\ldots \infty\right)\) are in
H.P.
Question 5
1 / -0

Suppose a, b, c are in AP and a²,b²,c₂ are in GP and a

A
1/ 2√2

B
1/ 2√3

C
1/ 2 - 1/√3

D
1/ 2 - 1/√2

Solution

Given a, b,c are in A.P then \(2 b=a+c\) but \(a+b+c=\frac{3}{2}\)
\(\Rightarrow 3 b=\frac{3}{2} \Rightarrow b=\frac{1}{2}\)
Let the common difference of the A.P. be \(d\). Now, \(a^{2}, b^{2}, c^{2}\) are in G.P \(\Rightarrow(b-d)^{2}, b^{2},(b+d)^{2}\) are in G.P
\(\Rightarrow b^{4}=(b-d)^{2}(b+d)^{2}\)
\(=\{(b-d)(b+d)\}^{2}=\left(b^{2}-d^{2}\right)^{2}\)
\(\Rightarrow\left(b^{2}-d^{2}\right)=\pm b^{2}\)
Now since +ve sign will give \(d=0\) so we will take -ve sign. hence \(\left(b^{2}-d^{2}\right)=-b^{2}\)
\(\Rightarrow 2 b^{2}=d^{2}\)
\(\Rightarrow d=\pm b \sqrt{2}\)
Putting the value of \(b\) \(d=\pm \frac{1}{\sqrt{2}}\)
since \(a, b\) and \(c\) are in increasing order according to question. So, we will take +ve sign in \(d\) Hence \(a=b-d=\frac{1}{2}-\frac{1}{\sqrt{2}}\)
hence the correct option is \(D\).
Question 6
1 / -0

If a₁,a₂,a₃... are positive numbers in GP then loga_n,loga_n+1,loga_n+2 are in

A
AP

B
GP

C
HP

D
none of these

Solution

since, \(a_{1}, a_{2}, a_{3}, \ldots .\) are in G.P. Let \(a\) and \(r\) be the first term and common ratio respectively \(\Rightarrow \log a_{n}=\log a+n \log r\)
\(\log a_{n+1}=\log a+(n+1) \log r \quad \ldots(2)\)
and \(\log a_{n+2}=\log a+(n+2) \log r \quad \ldots(3)\)
From ( 1\(),(2)\) and (3) \(2 \log a_{n+1}=\log a_{n}+\log a_{n+2}\)
Therefore, \(\log a_{n}, \log a_{n+1}, \log a_{n+2}\) are in A.P.
Question 7
1 / -0

The value of x for which log₃(2^1−x+3),log₉4 and log₂₇(2x−1)³ form an A.P. is

A
11/6

B
6/11

C
0

D
1

Solution

Condition for \(A . P .\) is: \(2 b=a+c\) \(\Rightarrow 2 \log _{9} 4=\log _{3}\left(2^{1-x}+3\right)+\log _{27}\left(2^{x}+1\right)^{3}\)
\(\Rightarrow 2 \frac{\log 4}{\log 9}=\log _{3}\left(2^{1-x}+3\right)+3 \frac{\log \left(2^{x}+1\right)}{\log 27}\)
\(\Rightarrow 2 \frac{\log _{3} 4}{2 \log _{3} 3}=\log _{3}\left(2^{1-x}+3\right)+3 \frac{\log _{3}\left(2^{x}+1\right)}{3 \log _{3} 3}\)
\(\Rightarrow \frac{2}{2} \log _{3} 4=\log _{3}\left(2^{1-x}+3\right)+\frac{3}{3} \log _{3}\left(2^{x}-1\right)\)
\(\Rightarrow \frac{2}{2} \log 4=\log \left(2^{1-x}+3\right)+\frac{3}{3} \log \left(2^{x}-1\right)\)
\(\Rightarrow \log 4=\log \left[\left(2^{1-x}+3\right)\left(2^{x}-1\right)\right]\)
\(\therefore 4=\left(\frac{2}{y}+3\right)(y-1),\) where \(y=2^{x}\)
\(\therefore 4 y=(3 y+2)(y-1)\) or \(3 y^{2}-5 y-2=0\)
\(\therefore y=2,-\frac{1}{3}=2^{x}\)
since an exponential function cannot be \(-\)ve, \(\therefore 2^{x}=-\frac{1}{3}\) is rejected
\(\therefore 2^{x}=2 \Rightarrow x=1\)
Question 8
1 / -0

Find the sum of the first 5 terms of an arithmetic progression in which the 6^th term is 14 and the 11^th term is 22.

A
2.2

B
6.0

C
12.4

D
46.0

Solution

Let ' \(a\) ' is the first term and 'd' is the common difference between the given AP According to the question \(6^{t h}\) term is 14 \(\Rightarrow a+(6-1) d=14\)
\(\Rightarrow a+5 d=14 \ldots \ldots e q(1)\)
and \(11^{t h}\) term is 22 .
\(\Rightarrow a+(11-1) d=22\)
\(\Rightarrow a+10 d=22 \ldots . . e q(2)\)
Subtract eq (i) from eq ( 2 ), we get \(\Rightarrow-5 d=-8\)
\(\Rightarrow d=\frac{8}{5}\)
Substitute the value of \(d\) in eq (1) \(\Rightarrow a+5 \times \frac{8}{5}=14\)
\(\Rightarrow a=14-8\)
\(\Rightarrow a=6\)
Sum of first \(n\) terms of \(A P=S_{n}=\frac{n}{2}[2 a+(n-1) d]\) \(\therefore S_{5}=\frac{5}{2}\left[2 \times 6+(5-1) \frac{8}{5}\right]\)
\(\Rightarrow S_{5}=\frac{5}{2}\left[2 \times 6+4 \times \frac{8}{5}\right]\)
\(\Rightarrow S_{5}=\frac{5}{2}\left[12+\frac{32}{5}\right]\)
\(\Rightarrow S_{5}=\frac{5}{2}\left[\frac{60+32}{5}\right]\)
\(\Rightarrow S_{5}=\frac{5}{2} \times \frac{92}{5}=46\)
Question 9
1 / -0

There is a sequence of 4 positive integer terms, the first three are in AP. the last three arein G. P. where the 4th term exceeds the first term by 30, then the sum of all the terms is

A
127

B
128

C
129

D
None of these

Solution

Let \(a-d, a, a+d, a-d+30\) be the sequence of 4 positive integer terms. Given that the last three terms are in G.P. That is, \(a, a+d, a-d+30\) are in G.P. \(\Rightarrow(a+d)^{2}=a(a-d+30)\)
\(\Rightarrow a^{2}+2 a d+d^{2}=a^{2}-a d+30 a\)
\(\Rightarrow 3 a d=30 a-d^{2}\)
\(\Rightarrow 3 a=\frac{d^{2}}{10-d}\)
Therefore, R.H.S should be a multiple of \(3 .\) Now, For \(d=3,\) R.H.S is \(\frac{9}{7} ;\) which is not a multiple of \(3 .\) Also, for \(d=6, \mathrm{R} . \mathrm{H.S}\) is \(9 .\) \(\Rightarrow a=3 ;\) but in this case \(a-d\) is negative, which is not possible. Now, for \(d=9,\) R.H.S is 81 . \(\Rightarrow a=27\)
since \(d=12\) and afterwards make \(\mathrm{R}\).H.S as negative. Therefore, \(a=27\) and \(d=9\). Thus, the numbers are 18,27,36,48 . Hence, the sum of 18,27,36,48 is \(129 .\)
Question 10
1 / -0

If a,b,c are in A.P as well as in G.P, then

A
a=b≠c

B
a≠b=c

C
a≠b≠c

D
a=b=c

Solution

Given \(a, b, c\) are in A.P. \(\Rightarrow 2 b=a+c \rightarrow(1)\)
\(\Rightarrow(2 b)^{2}=(a+c)^{2}\)
\(\Rightarrow 4 b^{2}=a^{2}+2 a c+c^{2} \rightarrow(2)\)
Also, given that \(a, b, c\) are in G.P. \(\Rightarrow b^{2}=a c \rightarrow(3)\)
Now, substitute ( 3 ) in (2) we get \(4 a c=a^{2}+2 a c+c^{2}\)
\(\Rightarrow a^{2}+c^{2}-2 a c=0\)
\(\Rightarrow(a-c)^{2}=0\)
\(\Rightarrow(a-c)=0\)
\(\Rightarrow a=c \quad \ldots(4)\)
Now, substitute (4) in (1)\(;\) we get \(2 b=a+a=c+c\)
\(\Rightarrow 2 b=2 a=2 c\)
\(\Rightarrow a=b=c\)

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Mathematics Test - 38

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Mathematics Test - 38

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SHARING IS CARING

Question 1

\([1, \infty)\)

\([2, \infty)\)

\([0, \infty)\)

\(\left[\frac{1}{2}, \infty\right)\)

Solution

Question 2

\(|A|\) is independent of \(\alpha \& \beta\)

\(A^{-1}\) depends only on \(\alpha\)

\(A^{-1}\) depends only \(\beta\)

IA I is dependent of \(\alpha\)

Solution

Question 3

-15

-3

9

-4

Solution

Question 4

A.P

G.P

H.P

None of these

Solution

Question 5

1/ 2√2

1/ 2√3

1/ 2 - 1/√3

1/ 2 - 1/√2

Solution

Question 6

AP

GP

HP

none of these

Solution

Question 7

11/6

6/11

0

1

Solution

Question 8

2.2

6.0

12.4

46.0

Solution

Question 9

127

128

129

None of these

Solution

Question 10

a=b≠c

a≠b=c

a≠b≠c

a=b=c

Solution

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