Mathematics Test - 32

Given, 1 ≤ |x – 1| ≤ 3

⇒ -3 ≤ (x – 1) ≤ -1 or 1 ≤ (x – 1) ≤ 3

i.e. the distance covered is between 1 unit to 3 units

⇒ -2 ≤ x ≤ 0 or 2 ≤ x ≤ 4

Therefore, the solution set of the given inequality is

x ∈ [-2, 0] ∪ [2, 4]

As we know,

Using the properties of determinants, if a row/column of a given matrix is multiplied by a scalar \(k\), then the value of the determinant is also multiplied by \(k\).

\(\because\left|\begin{array}{lll}{a}_{1} & {~b}_{1} & {c}_{1} \\ {a}_{2} & {~b}_{2} & {c}_{2} \\ {a}_{3} & {~b}_{3} & {c}_{3}\end{array}\right|=\Delta\) \( ..... (1) \)

Then, multiplying the equation \( (1) \) by \(x\) in the first column, \(y\) in the second column and \(z\) in the third column, we get

\( \left|\begin{array}{lll}{xa}_{1} & {yb}_{1} & {zc}_{1} \\ {xa}_{2} & {yb}_{2} & {zc}_{2} \\ {xa}_{3} & {yb}_{3} & {zc}_{3}\end{array}\right|\)

Taking the common from equation \( (1) \), \(x\) from the first column, \(y\) from the second column and \(z\) from the third column, we get

\(= {xyz}\left|\begin{array}{lll}{a}_{1} & {~b}_{1} & {c}_{1} \\ {a}_{2} & {~b}_{2} & {c}_{2} \\ {a}_{3} & {~b}_{3} & {c}_{3}\end{array}\right|\)

\(= {xyz} \Delta\)

Given,

\(a_{n}=(2 n-1)^{2}\)

As we know that,

\(S_{n}=\sum a_{n}\)

\(\Rightarrow S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left[4 k^{2}-4 k+1\right]\)

\(\Rightarrow S_{n}=-4 \sum_{k=1}^{n} k+4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1\)

As we know,

\(\sum \mathrm{n}^{2}\) = \(\frac{n(n+1)(2 n+1)}{8}\)

\(\sum \mathrm{n}\) = \(\frac{n (n+1)}{2}\)

\(\sum 1\) =n

\(\Rightarrow S_{n}=\frac{n \cdot(2 n+1) \cdot(2 n-1)}{3}\)

As we know,

Equivalent set: When two sets, for example, A and B have the same cardinality if there exists an objective function from set A to B are called Equivalent set, i.e \(n(A)=n(B)\).

Equal set: Two sets for example \(A\) and \(B\) can be equal only if each element of set \(A\) is also the element of the set \(B\). If two sets are the subsets of each other, they are said to be equal. So, it can be written as,

\(A=B\)

\(\mathrm{A} \subset \mathrm{B}\) and \(\mathrm{B} \subset \mathrm{A} \Longleftrightarrow \mathrm{A}=\mathrm{B}\)

From the defination of equivalent and equal set we can see that;

Statement 1 is true because the number of elements in \(A=\) Number of elements in \(B=3\)

Statement 2 is true because \(\{1,5,9\} \in A\) and \(\{1,5,9\} \in B\)

\( y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\\)

\( \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}\)

Again,

\(\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\)

Again

\(y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\)

at \(\mathrm{x}=\frac{1}{2}\) '

\(\mathrm{y}^{\prime}-\mathrm{y}^{\prime \prime}=\frac{736}{225}\)

Thus \(225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736\)

\(\begin{aligned} & I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin ^2 x}{1+\pi^{\sin x}} d x \ldots \text { (i) } \\ & I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin ^2(-x)}{1+\pi^{\sin (-x)}}\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right] \\ & \frac{\pi}{2} \frac{1+\sin x}{-\frac{\pi}{2}} d x \\ & I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} x} \frac{\pi^{\sin x}\left(1+\sin ^2 x\right)}{1+\pi^{\sin x}} d x \ldots \text { (ii) }\end{aligned}\)

Adding Eqs. (i) and (ii), we get

\(\left.2 I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+\pi \sin x}{(1+\pi \sin x}\right)\left(1+\sin ^2 x\right) \\\)

\( \Rightarrow 2 I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1+\sin ^2 x\right) d x \\\)

\( \Rightarrow 2 I=[x]{ }_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x \\\)

\( {\left[\because \sin ^2 x \text { is an even function, so } \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 d x=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x\right]} \\\)

\( \Rightarrow I=\frac{1}{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)+\frac{1}{2} \int_0^{\frac{\pi}{2}}(1-\cos 2 x) d x \\\)

\( \Rightarrow I=\frac{\pi}{2}+\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}} \\\)

\( \frac{\pi}{2}+\frac{1}{2}\left[\left(\frac{\pi}{2}-0\right)-(0-0)\right] \\\)

\( I=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3 \pi}{4}\)

Let the given statement be \(P(n)\), i.e.,

\(\mathrm{P}(n):(2 n+7)<(n+3)^{2}\)

Put \(n=1\) in \(\mathrm{P}(n)\)

\(2.1+7=9<(1+3)^{2}=16\), which is true.

Let \(\mathrm{P}(k)\) be true for some positive integer \(k\),

\((2 k+7)<(k+3)^{2} \ldots(1)\)

We shall now prove that \(\mathrm{P}(k+1)\) is true whenever \(\mathrm{P}(k)\) is true.

\(\{2(k+1)+7\}=(2 k+7)+2\)

\(\Rightarrow \{2(k+1)+7\}=(2 k+7)+2<(k+3)^{2}+2\) \(\quad\) [using (i).]

\(\Rightarrow 2(k+1)+7

\(\Rightarrow 2(k+1)+7

Now, \(k^{2}+6 k+11

\(\therefore 2(k+1)+7<(k+4)^{2}\)

\(2(k+1)+7<\{(k+1)+3\}^{2}\)

Thus, \(\mathrm{P}(k+1)\) is true whenever \(\mathrm{P}(k)\) is true.

So, by the principle of mathematical induction, statement \(\mathrm{P}(n)\) is true for all natural numbers.

Let, \( I=\int_{2}^{4} \frac{x}{x^{2}+1} d x \)

\(=\frac{1}{2} \int_{2}^{4} \frac{2 x}{x^{2}+1} d x\)

Put \(x^2=t\)

Differentiating with respect to \(x\).

\(2x=\frac{dt}{dx}\)

\(2xdx=dt\)

So,

\(\int_{2}^{4} \frac{x}{x^{2}+1} d x=\frac{1}{2}\int_{2}^{4}\frac{dt}{t+1}\)

\(\left[\frac{1}{2}\log(t)\right]_{2}^{4}\)

\(=\left[\frac{1}{2} \log \left(1+x^{2}\right)\right]_{2}^{4}=F(x)\)

Put the limit,

\(I=F(4)-F(2) \)

\(=\frac{1}{2}\left[\log \left(1+4^{2}\right)-\log \left(1+2^{2}\right)\right] \)

\(=\frac{1}{2}[\log 17-\log 5] \)

\(=\frac{1}{2} \log \left(\frac{17}{5}\right)\)

Given,

\(\frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}\)

We know that,

\(\operatorname{cosec}^{2} x=1+\cot ^{2} x\)

Therefore,

\(\frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-\operatorname{cosec}^{2} x-\cot ^{2} x}{\cot x-\operatorname{cosec} x+\operatorname{cosec}^{2} x-\cot ^{2} x}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-\operatorname{cosec}^{2} x-\cot ^{2} x}{\cot x-\operatorname{cosec} x+1}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)}{\cot x-\operatorname{cosec} x+1}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{(\cot x+\operatorname{cosec} x)(\operatorname{cotx}-\operatorname{cosec} x+1)}{(\cot x-\operatorname{cosec} x+1)}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=(\cot x+\operatorname{cosec} x)\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cos x+1}{\sin x}\)