Mathematics Test - 11

Given curve is \(x^2=4 y\) \(\Rightarrow x=2 \sqrt{y}\)

Area of ABCD \(=\int_2^4 x d y\)

\(=\int_2^4 2 \sqrt{y}~ d y \)

\(=\frac{32-8 \sqrt{2}}{3} \) sq. unit 

Given:

|x – 1| ≤ 5, |x| ≥ 2

⇒ (-5 ≤ (x – 1) ≤ 5), (x ≤ -2 or x ≥ 2)

⇒ (-4 ≤ x ≤ 6), (x ≤ -2 or x ≥ 2)

Now, required solution is:

x ∈ [-4, -2] ∪ [2, 6]

The given limit:

\(\lim _{\mathrm{x} \rightarrow 1} \frac{\sqrt{\mathrm{f}(\mathrm{x})}-1}{\mathrm{x}-1}\)

\(\Rightarrow \frac{\sqrt{\mathrm{f}(1)}-1}{1-1}\)

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

\(\Rightarrow \frac{0}{0}\), is an indeterminate form.

From L'Hospital's rule, we know that:

\(\lim _{\mathrm{x} \rightarrow \mathrm{c}} \frac{\mathrm{f(x)}}{\mathrm{g(x)}}=\lim _{\mathrm{x} \rightarrow \mathrm{c}} \frac{\mathrm{f^{\prime}(x)}}{\mathrm{g^{\prime}(x)}}\)

Using the L'Hospital's rule, we get:

\(\lim _{\mathrm{x} \rightarrow 1} \frac{\sqrt{\mathrm{f}(\mathrm{x})}-1}{\mathrm{x}-1}\)

\(= \lim _{\mathrm{x} \rightarrow 1} \frac{\frac{1}{2 \sqrt{\mathrm{f}(\mathrm{x})}} \mathrm{f}^{\prime}(\mathrm{x})-0}{1-0}\)

\(= \frac{\mathrm{f}^{\prime}(1)}{2 \sqrt{\mathrm{f}(1)}}\)

\(= \frac{3}{2}\)

Given, set \(\mathrm{A}=\left\{\mathrm{x} \in \mathrm{Z}: 2^{\mathrm{x}+2\left(\mathrm{x}^{2}-5 \mathrm{x}+6\right)}=1\right\}\)

Consider, \(2^{(x+2)\left(x^{2}-5 x+6\right)}=1=2^{\circ}\)

\(\Rightarrow(x+2)(x-3)(x-2)=0\)

\(\Rightarrow x=(-2,2,3)\)

\(\Rightarrow A=\{-2,2,3\}\)

Also, we have set \(B=\{x \in Z:-3<2 x-1<9\}\)

Consider, \(-3<2 x-1<9, x \in Z\)

\(\Rightarrow-2<2 x<10, x \in Z\)

\(\Rightarrow-1

\(\Rightarrow B=\{0,1,2,3,4\}\)

So, \(A \times B\) has 15 elements.

\(\therefore\) Number of subsets of \(A \times B=2^{15}\)

\(\left[\because\right.\) If \(n a=m\), the number of possible subsets \(\left.=2^{m}\right]\)

We know, the total number of terms in the expansion of \((x+y)^{n}\) are \((n+1)\)

\(\ln (\sqrt{3}+1)^{8}, \mathrm{n}=8\)

\(\therefore\) Number of terms \(=8+1=9\)

Given,

\(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\)

We know that,

Sum of squares of the direction cosines of a line is equal to unity.

So,

\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \)

\(\Rightarrow 1-\sin ^{2} \alpha+1-\sin ^{2} \beta+1-\sin ^{2} \gamma=1\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)\)

\(\Rightarrow 3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1\)

\(\Rightarrow 3-1=\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\)

\(\therefore \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2\)

As we know,

Equation of ellipse \(=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Foci \(=(\pm a e, 0)\)

Eccentricity \((e)=\sqrt{1-\frac{ b ^{2}}{ a ^{2}}}\)

Here, foci \(=(\pm 2,0)=(\pm a e, 0)\) and the eccentricity, \(e=\frac{1 }{ 2}\)

\(ae =2\)

\(\Rightarrow a \times \frac{1 }{ 2}=2 \)

\(\Rightarrow a =4 \)

\(\Rightarrow a ^{2}=16\)

Now, \(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

\(\Rightarrow b^{2}=a^{2}\left(1-e^{2}\right)\)

\(\Rightarrow b^{2}=16(1-\frac{1 }{ 4}) \)

\(\Rightarrow b^{2}=16 \times(\frac{3 }{ 4}) \)

\(\Rightarrow b^{2}=12\)

\(\therefore\) Equation of ellipse is \(\frac{ x ^{2}}{16}+\frac{ y ^{2}}{12}=1\)

The difference between the two Sets:

Let A and B be two sets. The difference between A and B is denoted as (A - B)

It is the set of all those elements of A which are not present in B i.e {x: x ∈ A and x ∉ B}

The Venn diagram representation of the difference between two sets is shown below

\(A-B=\) It is the set having all the elements of set \(A\) except the elements present in set B.

\(B-A=\) It is the set having all the elements of set \(B\) except the elements present in set A.

\(\therefore(A-B) \cap(B-A)\) is a null set \((\phi)\).

Given, the point \((2,3)\) and slope of the line is \(2\)

By, slope-intercept formula,

\(y-3=2(x-2)\)

\(\Rightarrow y-3=2 x-4\)

\(\Rightarrow 2 x-4-y+3=0\)

\(\Rightarrow 2 x-y-1=0\)