Mathematics Test - 7

We have,

\(e^{x}=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

By the expansion of e^x, we get,

\(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots+\frac{x^{n}}{n !}+\ldots\)

\(=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

Equating the coefficient of xⁿ on both sides, we get,

\(B_{n}-B_{n-1}=\frac{1}{n !}\)

Given:

\(\left|\begin{array}{ccc}8 & -5 & 1 \\ 5 & x & 1 \\ 6 & 3 & 1\end{array}\right|=2\)

\(\Rightarrow 8(x-3)-(-5)(5-6)+1(15-6 x)=2\)

\(\Rightarrow 8(x-3)+5(5-6)+1(15-6 x)=2\)

\(\Rightarrow 8 x-24-5+15-6 x=2\)

\(\Rightarrow 2 x=2+14=16\)

\(\Rightarrow x=8\)

Power Set: Let A be set, then the set of all the possible subsets of A is called the power set of A is called the power set of A and is denoted by P(A).

Given:

A = {a, b}

As we know that if A is a set then the set of all the possible subsets of A is called the power set of A is called the power set of A and is denoted by P(A)

First let's find out the subsets of A = {a, b}

So, the subsets of A are: ϕ, {a}, {b}, {a, b}

⇒ P(A) = {ϕ, {a}, {b}, {a, b}}

If A, B, C are matrices then A(BC) = (AB)C

A^-1 A = I

Given that, A has inverse B

AB = I = BA ....(1)

Also, A has inverse C.

AC = I = CA ....(2)

From (1) and (2),

AB = AC ....(3)

Pre multiply equation (3) by A^-1

A^-1 (AB) = A^-1 (AC)

⇒ A^-1 (AB) = A^-1 (AC)

⇒ (A^-1 A)B = (A^-1A)C

⇒ (IB) = (IC)

⇒ B = C

Thus, if a matrix A has inverses B and C then, B should be equal to C.

Given,

\(\lim _{x \rightarrow 2} \frac{x^{3}+x^{2}}{x^{2}+3 x+2}\)

\(=\lim _{x \rightarrow 2} \frac{x^{2}(x+1)}{(x+1)(x+2)}\)

\(=\lim _{x \rightarrow 2} \frac{x^{2}}{x+2}\)

\(=\frac{2^{2}}{2+2}\)

\(=\frac{4}{4}\)

\(=1\)

\(\underset{{x \rightarrow 2}}{\lim} \frac{\sqrt{3-x}-1}{2-x}\)

At \(x=2,\) the value is \(\frac{0}{ 0},\) so limit is of an indeterminate form \((\frac 0 0, \frac{\infty}{ \infty}, 0 \times \infty, 00,1 \infty, \infty 0)\).

To avoid indeterminate form, rationalizing the numerator

\(\frac{\sqrt{3-x}-1}{2-x}=\frac{\sqrt{3-x}-1}{2-x} \times \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\)

\(=\frac{(\sqrt{3-x})^{2}-1}{(2-x)(\sqrt{3-x}+1)}=\frac{3-x-1}{(2-x)(\sqrt{3-x}+1)}\)

\(=\frac{1}{\sqrt{3-x}+1}\)

\(\therefore \underset{{x \rightarrow 2}}{\lim} \frac{\sqrt{3-x}-1}{2-x}=\underset{{x \rightarrow 2}}{\lim} \frac{1}{\sqrt{3-x}+1}=\frac{1}{1+1}=\frac{1}{2}\)

Let \(\mathrm{I}=\int \sqrt{\mathrm{x}} \mathrm{e}^{\sqrt{\mathrm{x}}} \mathrm{dx}\)

Substituting \(\sqrt{\mathrm{x}}=\mathrm{t},\) we get:

\(\frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{d} \mathrm{t}\)

\(d x=2 t d t\)

\(\therefore \mathrm{I}=\int \mathrm{t} \times \mathrm{e}^{\mathrm{t}} \times 2 \mathrm{t} \mathrm{d} \mathrm{t}\)

Integrating by parts, taking \(\mathrm{t}^{2}\) as the first function and \(\mathrm{e}^{\mathrm{t}}\) as the second function, we get:

\(\mathrm{I}=2\left[\mathrm{t}^{2} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t}^{2} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}\right) \mathrm{dt}\right]+\mathrm{C}\)

\(\mathrm{I}=2 t^{2} e^{t}-4 \int t e^{t} d t+C\)

Integrating \(\int \mathrm{te}^{\mathrm{t}} \mathrm{dt}\) by parts, we get:

\(\mathrm{I}=2 \mathrm{t}^{2} \mathrm{e}^{\mathrm{t}}-4\left[\mathrm{t} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}\right) \mathrm{dt}\right]+\mathrm{C}\)

\(\mathrm{I}=2 \mathrm{t}^{2} \mathrm{e}^{\mathrm{t}}-4\left(\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)+\mathrm{C}\)

\(\mathrm{I}=\left(2 t^{2}-4 t+4\right) e^{t}+C\)

Back substituting \(\sqrt{\mathrm{x}}=\mathrm{t},\) we get:

\(\mathrm{I}=(2 \mathrm{x}-4 \sqrt{\mathrm{x}}+4) \mathrm{e}^{\sqrt{\mathrm{x}}}+\mathrm{C}\)

Given equation dy = (1 + y²) dx

On solving, we get-

\(\frac{d y}{\left(1-y^{2}\right)}=d x\)

Integrating both sides

\(\int \frac{d y}{\left(1+y^{2}\right)}=\int d x\)

\(\tan ^{-1}(y)=x+c\)

\(y=\tan (x+c)\)

\(\mathrm{I}=\int \frac{1}{\mathrm{x}}(\log \mathrm{x}) \mathrm{dx}\)

Let \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}\) and \(\mathrm{g}(\mathrm{x})=\frac{1}{\mathrm{x}}\)

\(\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}} \log \mathrm{x}=\frac{1}{\mathrm{x}}\) and \(\int \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}=\log \mathrm{x}\)

\(\mathrm{I}=\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}-\int \mathrm{f}^{\prime}(\mathrm{x})\left[\int \mathrm{g}(\mathrm{x}) \mathrm{dx}\right] \mathrm{dx}\)

\(\Rightarrow \mathrm{I}=\log \mathrm{x} \times \log \mathrm{x}-\int \frac{1}{\mathrm{x}}(\log \mathrm{x}) \mathrm{dx}\)

\(\Rightarrow I=(\log \mathrm{x})^{2}-1+\mathrm{C}\)

\(\Rightarrow 2 \mathrm{l}=(\log \mathrm{x})^{2}+\mathrm{C}\)

\(\Rightarrow \mathrm{I}=\frac{1}{2}(\log \mathrm{x})^{2}+\mathrm{C}\)

Given,

Equation of line is \(\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\)

Equation of plane is \(10 x+2 y-11 z-3=0\)

As we know,

The angle between the line \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) and the plane \(a_{2} x+b_{2} y+c_{2} z\) \(+d=0\) is given by:

\(\sin \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\left(\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\right)\left(\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right)}\right|\)

Here,

\(a_{1}=2, b_{1}=3, c_{1}=6, a_{2}=10, b_{2}=2\) and \(c_{2}=-11\)

So,

\(\sin \theta=\left|\frac{\left|2× 10+3× 2+6 ×-11\right|}{\left(\sqrt{2^{2}+3^{2}+6^{2}}\right)\left(\sqrt{10^{2}+2^{2}+(-11)^{2}}\right)}\right|\)

\(\Rightarrow \sin \theta=\left|\frac{\left|20+6-66\right|}{\left(\sqrt{4+9+36}\right)\left(\sqrt{100+4+121}\right)}\right|\)

\(\Rightarrow \sin \theta=\left|\frac{\left|26-66\right|}{\left(\sqrt{49}\right)\left(\sqrt{225}\right)}\right|\)

\(\Rightarrow \sin \theta=\left|\frac{\left|-40\right|}{\left(7\right)\left(15\right)}\right|\)

\(\Rightarrow \sin \theta=\frac{40}{7 \times 15}\)

\(\Rightarrow \sin \theta=\frac{8}{21}\)

\(\Rightarrow \theta=\sin ^{-1}\left(\frac{8}{21}\right)\)