Mathematics Test - 6

Given:

\(f(2 a-x)=f(x)\) ......(1)

\(\int_{0}^{a} f(x) d x=\lambda\) .....(2)

Using the property (1)

\(\int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\)

From equation (1)

\(\Rightarrow \int_{0}^{2 a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(x) d x\)

From equation (2)

\(\Rightarrow \int_{0}^{2 a} f(x) d x=\lambda+\lambda\)

\(\Rightarrow \int_{0}^{2 a} f(x) d x=2 \lambda\)

Given determinant is \(\left|\begin{array}{ccc}\mathrm{i} & \mathrm{i}^{2} & \mathrm{i}^{3} \\ \mathrm{i}^{4} & \mathrm{i}^{6} & \mathrm{i}^{8} \\ \mathrm{i}^{9} & \mathrm{i}^{12} & \mathrm{i}^{15}\end{array}\right|\)

Since, we have,

\(\mathrm{i}=\sqrt{-1}\)

\(\mathrm{i}^{2}=-1, \mathrm{i}^{3}=-\mathrm{i}, \mathrm{i}^{4}=1, \mathrm{i}^{6}=-1, \mathrm{i}^{8}=1, \mathrm{i}^{9}=\mathrm{i}, \mathrm{i}^{12}=1\), and \(\mathrm{i}^{15}=-\mathrm{i}\)

\(=\left|\begin{array}{ccc}\mathrm{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 \\ \mathrm{i} & 1 & -\mathrm{i}\end{array}\right|\)

\(=\mathrm{i}(\mathrm{i}-1)+1(-\mathrm{i}-\mathrm{i})-\mathrm{i}(1+\mathrm{i})\)

\(=i^{2}-i-2 i-i-i^{2}\)

\(=-4 i\)

We know that:

Suppose a set of \(n\) objects has \(n_{1}\) of one kind of object, \(n_{2}\) of a second kind, \(n_{3}\) of a third kind, and, 

So, on with \(n=n_{1}+n_{2}+n_{3}+\ldots+n_{k}\) then the number of distinguishable permutations of the \(n\) objects is: 

\(=\frac{n !}{n_{1} ! \times n_{2} ! \times n_{3} ! \ldots \ldots . n_{k} !}\)

Given: Three a’s and two b’s

Given:

\(\Rightarrow\) Number of coins having tail on both sides \(=n\)

\(\Rightarrow\) Number of fair coins \(=n+1\)

According to question,

\(\Rightarrow\) Probability of getting a tail \(=\frac{31}{42}\)

\(\Rightarrow P\) (tail) \(=\frac{^nC_1}{^{2n+1}C_1}\times 1+\frac{^{n+1}C_1}{^{2n+1}C_1} \times \frac{1}{2}=\frac{31}{42}\)

\(\Rightarrow \frac{n}{2 n+1}+\frac{n+1}{2(2 n+1)}=\frac{31}{42}\)

\(\Rightarrow(3 n+1) \times 21=31(2 n+1)\)

\(\Rightarrow 63 n+21=62 n+31\)

\(\Rightarrow n=10\)

\(\therefore\) Total coins in bag \(=2 n+1=21\)

A bounded feasible region will have both a maximum value and a minimum value for the objective function. It is bounded if it can be enclosed in any shape.

A convex polygon is a simple not self-intersecting closed shape in which no line segment between two points on the boundary ever goes outside the polygon.

So, the answer is convex polygon.

Given:

In two throws of a dice, total chances \(n(S)=(6 \times 6)=36\)

Let \(E\) is the event of getting a sum 9

\(\therefore E=\{(3,6),(4,5),(5,4),(6,3)\}\)

\(\Rightarrow n(E)=4 \)

\(\therefore P(E)=\frac{n(E)}{n(S)}\)

\( =\frac{4}{36}\Rightarrow\frac{1}{9}\)

Given that: 

\(\lim _{\mathrm{x} \rightarrow 0} \frac{\log (1+\sin x)}{\mathrm{x}}=\mathrm{k}\)

Put x = 0, to check form

\(\lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{x}\)

\(=\frac{\log (1+0)}{0}\)

\(=\frac{0}{0}\)

Applying L’ hospital’s rule as,

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(x)}{g(x)}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(x)}{g^{\prime}(x)}\)

\(\lim _{x \rightarrow 0} \frac{\log (1+\sin x)}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\log (1+\sin x))}{\frac{d}{d x}(x)}\)

\(=\lim _{\mathbf{x} \rightarrow 0} \frac{\frac{1}{1+\sin x} \times \cos x}{1}\)

\(=\frac{1}{1+\sin 0} \times \cos 0\)

\(=1\)

\(\therefore \mathrm{k}=1\)

Given,
\(\begin{aligned} 2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{cc}
3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right] \end{aligned}\)
\(\begin{aligned}\Rightarrow\left[\begin{array}{cc}
2 x & 10 \\
14 & 2 y -6
\end{array}\right]+\left[\begin{array}{cc}
3 & -4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right] \end{aligned}\)
\(\begin{aligned} \Rightarrow\left[\begin{array}{cc}
2 x +3 & 6 \\
15 & 2 y -4
\end{array}\right]=\left[\begin{array}{cc}
7 & 6 \\
15 & 14
\end{array}\right]
\end{aligned}
\)
As we know that,
If two matrices \(A\) and \(B\) are equal then their corresponding elements are also equal.
\(\therefore 2 x+3=7 \)
\(\Rightarrow 2x=7-3\)
\(\Rightarrow 2x=4\)
\(\Rightarrow x=2\)
And \(2 y-4=14\)
\(\Rightarrow 2y=14+4\)
\(\Rightarrow 2y=18\)
\(\Rightarrow y=9\)
Now,
\(y-x=9-2=7\)
So, the value of \(y-x\) is \(7\).

Given,

\(6 \sin ^{2} x-2 \cos ^{2} x=4\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4 \times 1\)

As we know that,

\(\sin ^{2} x+\cos ^{2} x=1\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4\left(\sin ^{2} x+\cos ^{2} x\right)\)

\(\Rightarrow 6 \sin ^{2} x-2 \cos ^{2} x=4 \sin ^{2} x+4 \cos ^{2} x\)

\(\Rightarrow 6 \sin ^{2} x-4 \sin ^{2} x=4 \cos ^{2} x+2 \cos ^{2} x\)

\(\Rightarrow 2 \sin ^{2} x=6 \cos ^{2} x\)

\(\Rightarrow \tan ^{2} x=3\)

\(\therefore \tan x=\sqrt{3}\)

Given, 

Three planes are

x + y = 0...(i)

y + z = 0...(ii)

x + z = 0...(iii)

Adding these three planes, we get

2(x + y + z) = 0

⇒ x + y + z = 0...(iv)

Putting value of (x + y) in (iv), we get

0 + z = 0

⇒ z = 0

Putting value of (y + z) in (iv), we get

x + 0 = 0

⇒ x = 0

Putting value of (x + z) in (iv), we get

y + 0 = 0

⇒ y = 0

So, (x, y, z) = (0, 0, 0)

So, the three planes meet in a unique point.