Mathematics Test - 21

Given, \(P ( A \cup B )\)= \(\frac{5}{6}\) 

 \(P ( A \cap B )\) = \(\frac{1} {3}\), \(P ( B )\) = \(\frac{1}2\)

We know that, \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

\(\frac{5}{6}=P(A)+\frac{1}{2}-\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5}{6}-\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5-3+2}{6}\)

\(\Rightarrow P(A)=\frac{2}{3}\)

We know that for independent events,

\(P(A) \cdot P(B)\) = \(P(A \cap B)\)

\(P(A \cap B)\) = \((\frac{2}{3}) \times(\frac{1}{2})\) =\(\frac{1}{3}\)

This is equal to \(P ( A \cap B )\).

Thus events \(A\) and \(B\) are independent events.

Given, the curve \(y=x^3-11 x+5---(1)\)

The slope of the tangent \(=\frac{d y}{d x}--(2)\)

Differentiate equation (1) w.r.t. x, we have

\(\Rightarrow \frac{d y}{d x}=3 x^2-11\)

The tangent to the curves is \(y=x-11--(3)\)

The slope of the tangent \(=1\)

From equation (2),

\(\therefore \frac{d y}{d x}=1\)

\(\Rightarrow 3 x^2-11=1\)

\(\Rightarrow 3 x^2=12\)

\(\Rightarrow x^2=4\)

\(\Rightarrow x= \pm 2\)

From equation (1), at \(x=2, y=2^3-11 \times 2+5=-9\) and at \(x=-2, y=(-2)^3-11 \times(-2)+5=19\)

The points are \((2,-9)\) and \((-2,19)\).

These points should satisfy the equation of tangent also.

Substitute \(x=2\) and \(y=-9\) in the equation (3), we have

\(\Rightarrow-9=2-11\)

\(\Rightarrow-9=-9\)

Therefore, the point \((2,-9)\) lies on the tangent.

Now, substitute \(x=-2\) and \(y=19\) in the equation (3), we have

\(\Rightarrow 19 \neq-2-11\)

\(\Rightarrow 19 \neq-13\)

The point \((-2,19)\) does not lie on the tangent.

Therefore, the point is \((2,-9)\).

Given series:

\(\left[\alpha x-\frac{1}{2}(\alpha x)^{2}+\frac{1}{3}(\alpha x)^{3}-\ldots\right]+\left[\beta x-\frac{1}{2}(\beta x)^{2}+\frac{1}{3}(\beta x)^{3}-\ldots\right]\)

\(=\log (1+\alpha x)+\log (1+\beta x)\)

\(=\log \left[1+(\alpha+\beta) x+\alpha \beta x^{2}\right]\)

Now, \(\alpha+\beta=p\) and \(\alpha \beta=q\)

Given series \(=\log \left(1-p x+q x^{2}\right)\)

Given that:

\({ }^{{n}} {P}_{{r}}=2760,{ }^{{n}} {C}_{{r}}=23\)

We know that,

\({ }^{{n}} {C}_{{r}}=\frac{{ }^{{n}} {P}_{{r}}}{{r} !}\)

\(\Rightarrow 23=\frac{2760}{r !}\)

\(\Rightarrow {r} !=\frac{2760}{23}=120\)

\(\Rightarrow {r} !=5 \times 24\)

\(\Rightarrow {r} !=5 \times 4 \times 6\)

\(\Rightarrow {r} !=5 \times 4 \times 3 \times 2 \times 1\)

\(\Rightarrow {r!}=5!\)

\(\therefore r=5\)

Let the missing number be \(x\).

Given:

Mean = \(25\), Total elements = \(8\)

Mean of n elements \(=\frac{\text { Sum of all n elements }}{\text { Total number of elements (n) }}\)

\(\Rightarrow \frac{30+24+27+22+18+26+32+x}{8}=25\)

\(\Rightarrow 179+x=200\)

\(x=21\)

Since \(\omega^{3}=1\) given determinant can be rewritten as:

\(A=\left|\begin{array}{ccc}1+2 \omega+\omega^{2} & \omega^{2} & 1 \\ 1 & 1+\omega^{2}+2 \omega & \omega \\ \omega & \omega^{2} & 2+\omega+2 \omega^{2}\end{array}\right|\)

\(\therefore |A|=\left|\begin{array}{ccc}\omega & \omega^{2} & 1 \\ 1 & \omega & \omega \\ \omega & \omega^{2} & 1+\omega^{2}\end{array}\right|\left(\because 1+\omega+\omega^{2}=0\right)\)

\(=\omega\left|\begin{array}{ccc}\omega & \omega & 1 \\ 1 & 1 & \omega \\ \omega & \omega & -\omega\end{array}\right|\left(\because 1+\omega+\omega^{2}=0\right)\)

\(\Rightarrow \omega(0)=0\)

Thus, A is singular.

Given,

\(|{x}|<-5\)

Now, LHS \(\geq 0\) and RHS \(<0\)

Since LHS is non-negative and RHS is negative So, \(|x|<-5\) does not posses any solution.

Given,

\(\frac{(1+\cos x)}{(1-\sin x)}(\sin x+\cos x-1)^{2}=M \sin ^{N} x\)

\(\Rightarrow \frac{(1+\cos x)}{(1-\sin x)}(\sin x+\cos x-1)^{2}=M \sin ^{N} x\)

\(\Rightarrow \frac{(1+\cos x)}{(1-\sin x)} \times 2(1-\sin x)(1-\cos x)=M \sin ^{N} x\)

\(\Rightarrow 2(1+\cos x)(1-\cos x)=\mathrm{M} \sin ^{N} x\)

\(\Rightarrow 2\left(1-\cos ^{2} x\right)=M \sin ^{N} x\)

\(\Rightarrow 2 \sin ^{2} x=M \sin ^{N} x\)

\(\text { Comparing both sides, we have- }\)

\(\Rightarrow M N=(2 \times 2)=4\)

The direction ratios of the line joining the points \(\left(x_1, y_1, z_1\right)\) and \(\left(x_2, y_2, z_2\right)\) is given by: \(a=x_2-x_1, b=y_2-y_1\) and \(c=z_2-z_1\)

The direction ratios of the line joining the points (k, 1, 3) and (1,-2, k+1) is: a=1-k, b=-3 and c=k-2

If a, b, and c are the direction ratio ratios of a line passing through the point \(\left(x_1, y_1, z_1\right)\), then the equation of a line is given by: \(\frac{x-x_1}{a}=\) \(\frac{y-y_1}{b}=\frac{z-z_1}{c}\)

The equation of the line with direction ratios a, b, c and passing through the point (k, 1, 3) is given by: \(\frac{z-k}{1-k}=\frac{y-1}{-3}=\frac{z-3}{k-2}\)

\(\because\) It is given that, the line represented by \(\frac{x-k}{1-k}=\frac{y-1}{-3}=\frac{z-3}{k-2}\) alsopasses through the point \((15,2,-4)\)

So, substitute \(x=15, y=2\) and \(z=-4\) in the equation \(\frac{x-k}{1-k}=\) \(\frac{y-1}{-3}=\frac{z-3}{k-2}\)

\(\Rightarrow \frac{15-k}{1-k}=\frac{2-1}{-3}=\frac{-4-3}{k-2}\)

\(\Rightarrow \frac{15-k}{1-k}=\frac{1}{-3}\)

\(\Rightarrow-45+3 \mathrm{k}=1-\mathrm{k}\)

\(\Rightarrow \mathrm{k}=\frac{23}{2}\)

\(\Rightarrow \frac{1}{-3}=\frac{-7}{k-2}\)

\(\Rightarrow \mathrm{k}-2=21\)

\(\Rightarrow \mathrm{k}=23\)

Therefore, there are two possible values of k.