Mathematics Test - 14

Given:

2sin²A + 3cos²A = 2

2sin²A + 3cos²A = 2

⇒ 2sin²A + 3(1 – sin²A) = 2

⇒ sin²A = 1

⇒ sinA = 1

⇒ A = 90°

(tanA – cotA)² = (tan90° - cot90°)²

⇒ (tanA – cotA)² = (∞ - 0)²

⇒ (tanA – cotA)² = ∞

∴ Value of (tanA – cotA)² is ∞.

Given, \(\mathrm{A}=\left|\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right|=2\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|=2 \mathrm{I}\) and \(\mathrm{B}\) is a square matrix given as,

\(\mathrm{B}=\left|\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 2\end{array}\right|\)

So, applying the first concept we get, \(A \cdot B=21 . B \equiv\) Multiplying each term of \(B\) by \(2\).

\(\therefore \mathrm{A} . \mathrm{B}=\left|\begin{array}{ccc}2 & 4 & 6 \\ 0 & 2 & 6 \\ 0 & 0 & 4\end{array}\right|\)Now A.B is upper triangular matrix. So, applying the second concept we get,

\(\operatorname{det}\{A \cdot B\}=\) product of diagonal terms of \(A \cdot B=2 \times 2 \times 4=16\)

Since the given function is differentiable, we can write:

\(\lim _{x \rightarrow-1^{-}} f^{\prime}(x)=\lim _{x \rightarrow-1^{+}} f^{\prime}(x)\)

\(\Rightarrow \lim _{x \rightarrow-1^{-}}(2 b x)=\lim _{x \rightarrow-1^{+}}(2 a x-b)\)

\(\Rightarrow -2 b=-2 a-b\)

\(\Rightarrow -b=-2 a\)

\(\Rightarrow 2 a=b\quad....(i)\)

Since \(f(x)\) will also be continuous, we can write:

\(\lim _{x \rightarrow-1^{-}}\left(b x^{2}-a\right)=\lim _{x \rightarrow-1^{+}}\left(a x^{2}-b x-2\right)\)

\(\Rightarrow {b}-{a}={a}+{b}-2\)

\(\Rightarrow 2 {a}=2\)

\(\Rightarrow {a}=1\)

Using Equation \((i)\):

\(2 a=b\)

\(\Rightarrow 2 ×1=b\)

\(\Rightarrow b=2\)

These are the roots of the equation:

\(x^{2}-3 x+2=0\)

Given: \(A=\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)\)

\(A^{2}=\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right) \times\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)\)

\(\Rightarrow \mathrm{A}^{2}=\left(\begin{array}{ll}\{(-2) \times(-2)+2 \times 2\} & \{(-2) \times 2+2 \times(-2)\} \\ \{2 \times(-2)+(-2) \times 2\} & \{2 \times 2+(-2) \times(-2)\}\end{array}\right)\)

\(\Rightarrow A^{2}=\left(\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right)\)

\(\Rightarrow \mathrm{A}^{2}=-4 \times\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)\)

\(\Rightarrow A^{2}=-4 A\)

Maximizing \(z=x_{1}-x_{2}\)

Constraints \(x_{1}+x_{2} \leq 10\)

\(x_{1} \geq 0 \)

\(x_{2} \geq 0 \)

\(x_{2} \leq 5\)

\(\text { Corner points are: }(0,0)(5,0),(5,5),(0,10)\).

\(Z(0,0)=0-0=0 \)

\(Z(0,5)=0-5=-5 \)

\(Z(5,5)=5-5=0 \)

\(Z(10,0)=10-0=10\)

Maximum \(\mathrm{Z}=\mathrm{Z}(10,0)=10\)

Thus, problem has one optimum solution.

Negation of greater is ‘not greater’. ‘Not greater’ means either smaller than or equal to. So, “Sum of 2 and 3 is smaller than or equal to 4”, “Sum of 2 and 3 is not greater than 4”, “Sum of 2 and 3 is not greater than 4” are negation of the given statement. And “Sum of 2 and 3 is smaller than 4” is not the negation of given statement as it should include equal to also.

A box contains \(100\) round discs, \(50\) square-shaped discs and \(30\) triangular discs.

Probability \(=\frac{\text { Number of observation }}{\text { Total number of observation }}\)

Total number of discs \(=180\)

Total number of triangular discs \(=30\)

The probability that in first pass a triangular disc is attracted \(=\frac{30}{180}\) \(=\frac{1}{6}\)

The first disc is kept aside and one square disc is kept inside.

The probability that in second pass a triangular disc is attracted \(=\frac{29}{180}\)

Required Probability \(=\frac{1}{6} \times \frac{29}{180}\) \(=\frac{29}{1080}\)

The quadratic expression \(a x^{2}+b x+c, x \in R\) is always positive, if \(a>0\) and \(D<0\)

So, the quadratic expression

\((1+2 m) x^{2}-2(1+3 m) x+4(1+m), x \in R\) will be always positive,

if \(1+2 m>0\) .....(i)

and \(D=4(1+3 m)^{2}-4(2 m+1) 4(1+m)<0\) .....(ii)

From inequality Eq. (i), we get

\(m>-\frac{1}{2}\) .....(iii)

From in equality Eq. (ii), we get

\(1+9 m^{2}+6 m-4\left(2 m^{2}+3 m+1\right)<0\)

\(\Rightarrow m^{2}-6 m-3<0\)

\(\Rightarrow[\mathrm{m}-(3+\sqrt{12})][\mathrm{m}-(3-\sqrt{12})]<0\)

\({\left[\because m^{2}-6 m-3=0\right] }\)

\(\Rightarrow m=\frac{6 \pm \sqrt{36+12}}{2}=3 \pm \sqrt{12}\)

\(\Rightarrow 3-\sqrt{12}<\mathrm{m}<3+\sqrt{12}\) .....(iv)

From in equalities Eqs. (iii) and (iv), the integral values of \(m\) are \(0,1,2,3,4,5,6\)

So, the number of integral values of \(\mathrm{m}\) is \(7\).

Given,

\(-2<2 x-1<2\)

\(\Rightarrow-2+1<2 x<2+1 \)

\(\Rightarrow-1<2 x<3 \)

\(\Rightarrow \frac{-1}{2}

\(\Rightarrow x \in\left(\frac{-1}{2}, \frac{3}{2}\right)\)

Given:

Pearsaons measure of skewness \(=0.42\)

Mean \(=\bar{x}=86\)

Median \(=M_{d}=80\)

We know that:

Skewness \((Sk_{p})=\) \(\frac{3(\text{Mean}-\text{median})}{\text{standard deviation}}\)

\(0.42=\frac{3\left(\overline{ x }- M _{ d }\right)}{ \sigma}\)

\(\Rightarrow 0.42=\frac{3(86-80)}{\sigma}\)

\(\Rightarrow \sigma=\frac{18}{0.42}\)

\(\Rightarrow \sigma=\frac{300}{7}\)

Coefficient of variation \(=\) standard

CV \(=\left(\frac{\frac{300}{7}}{86}\right) \times 100\)

\(\Rightarrow \left(\frac{300}{7} \times 86\right) \times 100\)

\(\therefore\) Coefficient of variation is 50.