Mathematics Test

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Mathematics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Four dice (six faced) are rolled. The number of possible outcomes in which atleast one die shows 2 is

A
1296

B
625

C
671

D
None of these

Solution

Total number of possible outcomes \(=6^{4}\)
Number of possible outcomes in which 2 does not appear on any dice \(=5^{4}\)
Required number \(=6^{4}-5^{4}=671\)
Question 2
1 / -0

Let \(f(x)\) and \(g(x)\) be twice differentiable functions on [0,2] satisfying \(f^{\prime \prime}(x)=g^{\prime \prime}(x), f^{\prime}(1)=4, g^{\prime}(1)=\) \(6, f(2)=3\) and \(\mathrm{g}(2)=9 .\) Then what is \(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\) at \(\mathrm{x}=4\) equal to?

A
–10

B
–6

C
–4

D
2

Solution

Given \(f^{\prime \prime}(x)=g^{\prime \prime}(x)\)
On integrating both sides, we get
\(f^{\prime}(x)=g^{\prime}(x)+c\)
\(\Rightarrow f^{\prime}(1)=g^{\prime}(1)+c\)
\(\Rightarrow 4=6+c\)
\(\Rightarrow c=-2\)
\(\therefore f^{\prime}(x)=g^{\prime}(x)-2\)
Again, on integrating both sides, we get
\(f(x)=g(x)-2 x+c_{1}\)

\(\Rightarrow \mathrm{f}(2)=\mathrm{g}(2)-2 \times 2+c_{1}\)
\(\Rightarrow 3=9-4+c_{1}\)
\(\Rightarrow c_{1}=-2\)
\(\therefore f(x)-g(x)=-2 x-2\)
\(\mathrm{At} \mathrm{x}=4\)
\([\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})]=-8-2=-10\)
Question 3
1 / -0

The range of the function \(f(x)=\frac{1}{(2-\sin 3 x)}\) is

A

]\(\frac{1}{3}, [1\)

B

\(\left[\frac{1}{3}, 1\right]\)

C

\(\left[\frac{1}{3}, [1\right.\)

D

\(\left[1, \frac{1}{3}\right]\)

Solution

We have, \(2 y-y \sin 3 x=1 \quad[\text { Let } f(x)=y]\)
\(\Rightarrow \sin 3 x=\frac{(2 y-1)}{y}\)
since, \(-1 \leq \sin 3 x \leq 1\)
We have \(-1 \leq \frac{(2 y-1)}{y} \leq 1\)________(1)
since \(y>0\) multiplying the inequality eq (1) by \(y\), we obtain
\(-y \leq 2 y-1 \leq y\)
Or \(1 \leq 3 y\) and \(y \leq 1\)
\(\Rightarrow \frac{1}{3} \leq y \leq 1\)
Question 4
1 / -0

If \(\mathrm{A} \& \mathrm{B}\) are two symmetric matrices of same order, then \((A B A)^{r}\) is

A
Symmetric matrix

B
skew – symmetric matrix

C
diagonal matrix

D
None of these

Solution

If A and B are symmetric matrices,then (ABA)^ris also a symmetric matric.Symmetric matrix
Question 5
1 / -0

The arithmetic mean of 1, 8, 27, 64 …… upto n terms is given by

A

\(\frac{n(n+1)}{2}\)

B

\(\frac{n(n+1)^{2}}{2}\)

C

\(\frac{n(n+1)^{2}}{4}\)

D

\(\frac{n^{2}(n+1)^{2}}{4}\)

Solution

Given, \(1,8,27,64,-\dots-\) upto n terms \(=1^{3}, 2^{3}, 3^{3}, 4^{3}, \dots-\) upto \(n\) terms
\[
\begin{array}{l}
\therefore A M=\frac{1^{2}+2^{2}+3^{3}+4^{3}+---+n^{3}}{n}=\frac{\left[\frac{n(n+1)}{2}\right]^{2}}{n}=\frac{n^{2}(n+1)^{2}}{4 n} \\
=\frac{n(n+1)^{2}}{4}
\end{array}
\]
Question 6
1 / -0

If sum of the squares of zeros of the quadratic polynomial \(f(x)=x^{2}-8 x+k\) is \(40,\) find the value of \(k\)

A
12

B
10

C
13

D
11

Solution

Let \(\alpha, \beta\) be the zeros of the polynomial
\(f(x)=x^{2}-8 x+k\). Then
\(\alpha+\beta=-\left(\frac{-8}{1}\right)=8\) and \(\alpha \beta=\frac{k}{1}=k\)
It is given that
\(\alpha^{2}+\beta^{2}=40\)
\(\Rightarrow(\alpha+\beta)^{2}-2 \alpha \beta=40\)
\(\Rightarrow 8^{2}-2 k=40[\because \alpha+\beta=8 \text { and } \alpha \beta=k]\)
\(\Rightarrow 2 \mathrm{k}=64-40\)
\(\Rightarrow 2 \mathrm{k}=24\)
\(\Rightarrow \mathrm{k}=12\)
Question 7
1 / -0

If \(\frac{1}{\log _{a} x}+\frac{1}{\log _{c} x}=\frac{2}{\log _{b} x}\), then \(a, b\) and \(c\) are in

A
AP

B
GP

C
HP

D
None of these

Solution

\(\log _{x} a+\log _{x} c=2 \log _{x} b\)
\(\Rightarrow \mathrm{ac}=\mathrm{b}^{2}\) i.e. \(\mathrm{a}, \mathrm{b} \& \mathrm{c}\) are in GP.
Question 8
1 / -0

If \(A=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right],\) then what is \(A+3 A^{-1}\) equal to?

A
3I

B
5I

C
7I

D
None of these

Solution

We have, \(A=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]\)
\(\Rightarrow|A|=10-7=3\)
Now \(A^{-1}=\frac{1}{|A|}\) adj \((A)=\frac{1}{3}\left[\begin{array}{cr}5 & -7 \\ -1 & 2\end{array}\right]\)
\(\therefore A+3 A^{-1}=\left[\begin{array}{ll}2 & 7 \\ 1 & 5\end{array}\right]+3 \times \frac{1}{3}\left[\begin{array}{cc}5 & -7 \\ -1 & 2\end{array}\right]\)
\(=\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]=7 I\)
Question 9
1 / -0

If \(\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4},\) then \(x\) is equal to

A
–1

B
1/6

C
–1, 1/6

D
None of these

Solution

\(\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\)
\(\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}\)
\(\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\tan ^{-1}(1)\)
\(\frac{5 x}{1-6 x^{2}}=1\)
\(\Rightarrow 6 x^{2}+5 x-1=0\)
\(\Rightarrow x=-1, \frac{1}{6}\)
But, -1 does not hold.
Question 10
1 / -0

Find a quadratic polynomial, the sum & product of whose zeroes are –3 & 2, respectively.

A

\(P(x)=x^{2}+2 x+1\)

B

\(P(x)=x^{2}+3 x+2\)

C

\(P(x)=2 x^{2}+3 x+1\)

D

\(P(x)=x^{2}-3 x+2\)

Solution

We have \(\alpha+\beta=-3=\frac{-b}{a}\)
\(\alpha \beta=2=\frac{c}{a}\)
If \(a=1\) then \(b=3 \& c=2\) So, one quadratic polynomial which fits the given conditions is \(x^{2}+3 x+2\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 28

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Mathematics Test - 28

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SHARING IS CARING

Question 1

1296

625

671

None of these

Solution

Question 2

–10

–6

–4

2

Solution

Question 3

]\(\frac{1}{3}, [1\)

\(\left[\frac{1}{3}, 1\right]\)

\(\left[\frac{1}{3}, [1\right.\)

\(\left[1, \frac{1}{3}\right]\)

Solution

Question 4

Symmetric matrix

skew – symmetric matrix

diagonal matrix

None of these

Solution

Question 5

\(\frac{n(n+1)}{2}\)

\(\frac{n(n+1)^{2}}{2}\)

\(\frac{n(n+1)^{2}}{4}\)

\(\frac{n^{2}(n+1)^{2}}{4}\)

Solution

Question 6

12

10

13

11

Solution

Question 7

AP

GP

HP

None of these

Solution

Question 8

3I

5I

7I

None of these

Solution

Question 9

–1

1/6

–1, 1/6

None of these

Solution

Question 10

\(P(x)=x^{2}+2 x+1\)

\(P(x)=x^{2}+3 x+2\)

\(P(x)=2 x^{2}+3 x+1\)

\(P(x)=x^{2}-3 x+2\)

Solution

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