Mathematics Test - 6

$f\left(x\right)=2\sin x,g\left(x\right)=\cos ²x$

$(f+g):2\sin x+\cos ²x$

$(f+g)\frac{\mathrm{\pi }}{3}=2\sin \left(\frac{\mathrm{\pi }}{3}\right)+{\cos }^2\left(\frac{\mathrm{\pi }}{3}\right)$

$=2\left(\sqrt{\frac{3}{2}}\right)+{\left(\frac{1}{2}\right)}^2$

$=\sqrt{3}+\frac{1}{4}$

If,

f (x) = $\left|x+1\right|$

Now,

$f\left(x^2\right)=\left|x^2+1\right|$

${\left|f\left(x\right)\right|}^2={\left|x+1\right|}^2={\left|x\right|}^2+1^2+2\left|x\right|$

$\Rightarrow f\left(x^2\right)\ne {\left|f\left(x\right)\right|}^2$

$f\left|x\right|=\left|\left|x\right|+1\right|$

$\left|f\left(x\right)\right|=\left|x+1\right|$

$\Rightarrow f\left|x\right|\ne \left|f\left(x\right)\right|$

$f\left(x\right)+f\left(y\right)=\left|x+1\right|+\left|y+1\right|$

$f\left(x+y\right)\ne f\left(x\right)+f\left(y\right)$

A zero matrix is a matrix with all entries are zero. It may be or may not a square matrix.

A matrix has a determinant not numerical value. It is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns.

A unit matrix is a matrix whose diagonal entries are 1 i.e. all diagonal elements are same and remaining entries are zero.

Hence, a unit matrix is a diagonal matrix.

Null Set: A set which does not contain any element is called a null set. It is denoted by ϕ or {}.

ϕ is always a subset of any set.

Statement I: If A is a set then ϕ is a subset of A.

As we know that ϕ is always a subset of any set.

So, statement I is true.

Statement II: If B = {x ∈ N : x < 1} then B is an empty set.

As we know that if x ∈ N then x can never be less than 1.

⇒ B = ϕ

So, statement II is also true.

Let \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\mathrm{ABC}\)

Order of matrix \(\mathrm{A}\) is \((1 \times 3)\).

Order of matrix \(B\) is \((3 \times 3)\).

Order of matrix \(C\) is \((3 \times 1)\).

Now, Order of \(A_{(1 \times 3)} B_{(3 \times 1)}\) is \((1 \times 3)\).

\(\therefore\) Order of \(A B C=\) Order of \((A B)_{(1 \times 3)} C_{(3 \times 1)}=(1 \times 1)\)

Let \(f(x)=\frac{x^{2}}{1+x^{2}}\)

Clearly domain (f) =R

Let \(y=f(x) \Rightarrow y=\frac{x^{2}}{1+x^{2}}\)

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

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

Clearly x will take real values,

If \(\frac{y}{1-y} \geq 0 \Rightarrow \frac{y-0}{y-1} \leq 0\)

⇒ 0 ≤ y < 1 ⇒ y ∈ [0, 1)

Objective function \(: \mathrm{Z}=-3 \mathrm{x}+4 \mathrm{y}\)

We have to minimize \(\mathrm{Z}\) on constraints

\(\mathrm{x}+2 \mathrm{y} \leq 8 \)

\(3 \mathrm{x}+2 \mathrm{y} \leq 12 \)

\(\mathrm{x} \geq 0, \mathrm{y} \geq 0\)

After plotting inequalities, we got feasible region as shown in the image.

Now, there are 3 corner points \((0,4),(2,3)\) and \((4,0)\).

At \((0,4)\), value of \(Z=-3 \times 0+4 \times 4=16\)

At \((2,3)\), value of \(\mathrm{Z}=-3 \times 2+4 \times 3=6\)

\(\operatorname{At}(4,0)\), value of \(\mathrm{Z}=-3 \times 4+4 \times 0=-12\)

At \((0,0)\), value at \(Z=-3 \times 0+4 \times 0=0\)

So, minimum value of \(\mathrm{Z}=-12\).

Given,

\(\sin ^{-1} \sin \left(\frac{33 \pi}{5}\right)\)

\(\sin ^{-1} \sin \left(\frac{33 \pi}{5}\right)=\sin ^{-1} \sin \left(6 \pi+\frac{3 \pi}{5}\right) \)

\(=\sin ^{-1} \sin \left(\frac{3 \pi}{5}\right) \quad(: \sin (2 n \pi+\theta)=\sin \theta)\)

\(=\sin ^{-1} \sin \left(\pi-\frac{2 \pi}{5}\right) \)

\(=\sin ^{-1} \sin \left(\frac{2 \pi}{5}\right) \quad(\because \sin (\pi-\theta)=\sin \theta)\)

Here \(\frac{2 \pi}{5}\) is lies between \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\)

\(\therefore \sin ^{-1} \sin \left(\frac{2 \pi}{5}\right)=\frac{2 \pi}{5}\)

Given:

\(\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x\)

\(\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x\)

\(\Rightarrow \int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x=\left[\left(\sin \frac{\pi}{4}-\sin 0\right)+\left(\cos \left(\frac{\pi}{4}\right)-\cos 0\right)\right]=\frac{2}{\sqrt{2}}-1=\sqrt{2}-1\)

\(\Rightarrow \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x=\left[-\left(\cos \frac{\pi}{2}-\cos \frac{\pi}{4}\right)-\left(\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\right)\right]=\frac{2}{\sqrt{2}}-1=\sqrt{2}-1\)

\(\Rightarrow \int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x=2(\sqrt{2}-1)\)