Mathematics Test - 9

\(n(A)=40 \%\) of \(10000=4000\)

\(n(B)=20 \%\) of \(10000=2000\)

\(n(C)=10 \%\) of \(10000=1000\)

\(n(A \cap B)=5 \%\) of \(10000=500\)

\(n(B \cap C)=3 \%\) of \(10000=300\)

\(n(A \cap C)=4 \%\) of \(10000=400\)

\(n(A \cap B \cap C)=2 \%\) of \(10000=200\)

\(=n\left[A \cap(B \cup C)^{C}\right]\)

\(=n(A)-n[A \cap(B \cup C)]\)

\(=n(A)-[n(A \cap B) \cup n(A \cap C)]\)

\(=n(A)-[n(A \cap B)+n(A \cap C)-n(A \cap B \cap C)]\)

\(=4000-[500+400-200]=3300\)

We know that

If A, B and C are subsets of a set X. Then

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A∩C) ∪ (B∩C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

From the options,

⇒ A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B) = A --- (Using property I and II)

So, option (A) is not correct.

⇒ A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A --- (Using property I and II)

So, option (B) is correct.

⇒ (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C) --- (Using property III)

So, option (C) is correct.

⇒ (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) --- (Using property IV)

So, option (D) is correct.

Given \(2 \int_{0}^{1} \tan ^{-1} {x}\ {d} {x}=\int_{0}^{1} \cot ^{-1}\left(1-{x}+{x}^{2}\right) {dx}\)
Then, \(2 \int_{0}^{1} \tan ^{-1} {x} {dx}=\int_{0}^{1}\left[\frac{\pi}{2}-\tan ^{-1}\left(1-{x}+{x}^{2}\right)\right] {dx}\)
\(\mathrm{I}_{\mathrm{f}}=\int_{0}^{1} \tan ^{-1}\left(1-{x}+{x}^{2}\right) {d} {x}=\frac{\pi}{2}-2 \int_{0}^{1} \tan ^{-1} {x}{d} {x}\quad\quad\)....(1)
Let \(\mathrm{I}=\int_{0}^{1} \tan ^{-1} {x} {dx}\)
\(\tan ^{-1} x \int 1 d x-\int\left(\frac{1}{1+x^{2}} \int 1 d x\right) d x\quad\quad\)\([\)Using \(\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x]\)
\(=x \tan ^{-1} x-\int \frac{x}{1+x^{2}} d x\) \(=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x\) \(=x \tan ^{-1} x-\frac{1}{2} \log 1+x^{2}\)
\(\therefore \int_{0}^{1} \tan ^{-1} {x} {dx}=\left[{x} \tan ^{-1} {x}-\frac{1}{2} \log 1+{x}^{2}\right]_{0}^{1}\) \(=\left[\frac{\pi}{4}-\frac{1}{2} \log 2\right]-0\)
\(\mathrm{I}=\frac{\pi}{4}-\frac{1}{2} \log 2\)
Putting value of \(I\) in equation (1). \(\mathrm{I}_{\mathrm{f}}=\frac{\pi}{2}-2\left[\frac{\pi}{4}-\frac{1}{2} \log 2\right]\) \(=\log 2\)

It is given that,

Both p and q belong to the set {1, 2, 3, 4} and the quadratic equation of the form, px² + qx + 1 = 0 has real roots.

By comparing the quadratic equation of the form, px² + qx + 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = p, b = q and c = 1.

Now, we know that.

Discriminant (D) = b² - 4ac

Therefore, according to the question,

D = q² - 4p

∵ The roots are real.

⇒ D ≥ 0 ⇒ q² - 4p ≥ 0

⇒ q² ≥ 4p

∵ p and q belong to the set {1, 2, 3, 4}

Case -1:

If p = 1 then q² ≥ 4 ⇒ q ∈ {2, 3, 4}

Hence, for p = 1, q can take 3 values.

So, three quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 1.

Case -2:

If p = 2 then q² ≥ 8 ⇒ q ∈ {3, 4}

Hence, for p = 2, q can take 2 values

So, two quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 2.

Case -3:

If p = 3 then q² ≥ 12 ⇒ q ∈ {4}

Hence, for p =1, q can take 1 value

So, only one quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 3.

Case -4:

If p = 4 then q² ≥ 16 ⇒ q ∈ {4}

Hence, for p =4 , q can take 1 value

So, only one quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 4.

So, in total we get, 3 + 2 + 1 + 1 = 7 possible quadratic equations of the form px² + qx + 1 = 0 that can be formed such that both p and q belong to the set {1, 2, 3, 4} and have real roots.

Given: Equation of curve is \(x^{2}+y^{2}-2 x-4 y+1=0\) and tangent to the given curve is parallel to \(Y\) - axis.

Let point of the contact be \(\left(x_{1}, y_{1}\right)\).

As we know that, if a tangent of any curve is parallel to \(Y-\) axis then \(\frac{d x}{d y}=0\)

Now by differentiating equation of curve \(x^{2}+y^{2}-2 x-4 y+1=0\) with respect to y we get,

\(\Rightarrow 2 x \cdot \frac{d x}{d y}+2 y-2 \cdot \frac{d x}{d y}-4=0\)

\(\Rightarrow \frac{d x}{d y}=\frac{2-y}{x-1}\)

\(\Rightarrow\left[\frac{d x}{d y}\right]_{\left(x_{1}, y_{1}\right)}=\frac{2-y_{1}}{x_{1}-1}\)

\(\because\left[\frac{d x}{d y}\right]_{\left(x_{1}, y_{1}\right)}=0\)

\(\Rightarrow y_{1}=2\)

\(\because\left(x_{1}, 2\right)\) is the point of contact i.e \(x=x_{1}, y=2\) will satisfy the equation \(x^{2}+y^{2}-2 x-4 y+1=0\)

i.e., \(x_{1}^{2}+4-2 x_{1}-8+1=0\)

\(\Rightarrow {x}_{1}^{2}-2 {x}_{1}-3=0\)

\(\Rightarrow\left(x_{1}-3\right) \cdot\left(x_{1}+1\right)=0\)

\(\Rightarrow {x}_{1}=-1\) or 3

So, the required points are: \((-1,2)\) and \((3,2)\)

\(\vec{a} \cdot \vec{b}=0,\) If vectors a and \(\mathrm{b}\) are perpendicular.

Given:

\(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}\)

\(\vec{a}+t \vec{b}\) \(=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\mathrm{t}(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)

\(=(2-t) \hat{i}+(2+2 t) \hat{j}+(3+t) \hat{k}\)

Now \(\vec{a}+t \vec{b}\) and \(\vec{c}\), are perpendicular.

So, \((\vec{a}+t \vec{b}) \cdot \vec{c}=0\)

\(\Rightarrow((2-t) \hat{i}+(2+2 t) \hat{j}+(3+t) \hat{k}) \cdot(3 \hat{i}+\hat{j})=0\)

\(\Rightarrow 6-3 \mathrm{t}+2+2 \mathrm{t}=0\)

\(\Rightarrow \mathrm{t}=8\)

A box contains 3 white and 2 black balls.

Total balls = 3 + 2 = 5

So probability of first ball to be black \( = \frac{2}{5}\) and

Probability of second ball to be also black \(=\frac{1}{4}\) (∵second time we have 2 -1 = 1 black and 5 - 1 = 4 total balls)

∴ The probability that both the balls are black \(=\frac{ 2}{5} × \frac{1}{4}\)

\(= \frac{1}{10}\)

We know that, \(\sin 45^{\circ}=\frac{1}{\sqrt{2}}\)

\(\therefore \sin ^{2} 45^{\circ}=\frac{1}{2}\)

\(\cos ^{2} \theta=\frac{(1+\cos 2 \theta)}{2}\)

\(\Rightarrow \cos ^{2} 15^{\circ}=\frac{(1+\cos 2 \times 15)}{2}=\frac{(1+\cos 30)}{2}=\frac{2+\sqrt{3}}{4}\)

\(\Rightarrow \sin ^{2} 15^{\circ}=\frac{2+\sqrt{3}}{4}\)

Now,

\(\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=\frac{1}{2}-\left(\frac{2+\sqrt{3}}{4}\right)\)

\(\Rightarrow \cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=-\frac{\sqrt{3}}{4}\)

Given,

\(\left(\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right) \times\left(\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right)=\left(\begin{array}{cc}1 & -1 \\ 17 & \lambda\end{array}\right)\)

\(\mathrm{A}\) is in 2 nd row and 2 nd column, so, it will get by sum of multiplication of 2nd row of first matrix and 2 nd column of the second matrix

\(\therefore \lambda=4(-2)+1(1)\)

\(=-8+1=-7\)