Mathematics Test-26

Let the line \(y=2 x\) cuts the parabola \(y=x^{2}\) at then coordinates of \(O\) and \(A\) are \((0,0)\) and \((2,4)\) respectively.

Then 

Area bounded \(=\) area of triangle \(\triangle OAB -\) Area of the region \(ODABO\)

\(=\frac{1}{2} \times OB \times AB -\int_{0}^{2} y~dx\)

\(=\frac{1}{2} \times 2 \times 4-\int_{0}^{2} x ^{2} dx\)

\(\left[\because\int x^{n} d x=\left(\frac{(x^{n+1})}{(n+1)}\right)\right]\)

\(=4-\left[\frac{ x ^{3}}{2}\right]_{0}^{2}\)

\(=4-\frac{8}{3}=\frac{12-8}{3}\)

\(=\frac{4}{3}\) Sq. units.

Given sequence,

-8, -5, -2, …, 7

Here,

a = -8

d = 3

n = 6

The sum of the sequence for n terms,

\(S_{n}=\frac{n}{2} \times[2 \times a+(n-1) \times d]\)

Substituting the values of a, n and d,

\(\Rightarrow S_{6}=\frac{6}{2} \times[2 \times(-8)+(6-1) \times 3]\)

\(\Rightarrow S_{6}=-3\)

\(f(x)=x^{3}+2 x^{2}+5 x+2 \cos x\)

\(f^{\prime}(x)=3 x^{2}+4 x+5-2 \cdot \sin x\)

\(=3\left(x+\frac{2}{5}\right)^{2}+\frac{11}{3}-2 \cdot \sin x\)

\(\Rightarrow f^{\prime}(X)>0, \forall x\)

\(f(x)\) is increasing for all \(x \in R\)

Also, \(f(0)=2 \Rightarrow f(x)=0\)

So, \(f(x)\) has no solution.

Given,

\(2 x+y-2 z+6=0\) and \(4 x+2 y-4 z-6=0\) are two planes.

Here, we can rewrite the equation of plane \(2 x+y-2 z+6=0\) as \(4 x+2 y-4 z+12=0\) by multiplying both the sides of \(2 x+y-2 z+6=0\) with \(2\).

As we can see that, the plane \(4 x+2 y-4 z+12=0\) and \(4 x+2 y-4 z-6=0\) are parallel planes.

As we know that, the distance between two parallel planes \(a x+b y+c z+d_1=0\) and ax \(+\) by \(+\mathrm{cz}+\mathrm{d}_2=0\) is given by: \(D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

Here, \(a=4, b=2, c=-4, d_1=12\) and \(d_2=-6\).

\(\Rightarrow D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

\(\Rightarrow D=\left|\frac{12-(-6)}{\sqrt{4^2+2^2+(-4)^2}}\right|\)

\(\Rightarrow D=\left|\frac{18}{\sqrt{16+4+16}}\right|\)

\(\Rightarrow D=\left|\frac{18}{6}\right|=3\)

Solving a \(3 \times 3\) determinant:

A \(3 \times 3\) determinant can be solved as follows:

\(\left|\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|=a(e i-f h)-b(d i-f g)+c(d h-g e)\)

Given

Simplify the given determinant as follows:

\(\left|\begin{array}{lll}a & b & 0 \\ 0 & a & b \\ b & 0 & a\end{array}\right|=0\)

\(a\left(a^2-0\right)-b\left(0-b^2\right)+0(0-a b)=0\)

\(\begin{aligned} a^3+b^3 &=0 \\ a^3 &=-b^3 \\ \frac{a^3}{b^3} &=-1 \end{aligned}\)

Therefore, the given condition holds only if \(\frac{a}{b}\) is one of the cube roots of \(-1\)

Given:

\(\left(1+3 x+3 x^{2}+x^{3}\right)^{6}\)

\(=\left[(1+\mathrm{x})^{3}\right]^{6} \)

\(=[1+\mathrm{x}]^{6(3)} \)

\(=(1+\mathrm{x})^{18}\)

So, total number of terms will be \(18+1=19\) terms.

Given,

\(\angle P A D=60^{\circ}\)

In \(\triangle A P D, \mathrm{AP}=\mathrm{DP}\) (Radii of circle)

In a triangle angles opposite to equal sides are equal.

\(\angle \mathrm{PAD}=\angle \mathrm{PDA}=60^{\circ}\)

\(\angle \mathrm{PAD}+\angle \mathrm{PDA}+\angle \mathrm{APD}=180^{\circ}\) (Angle sum property of \(\triangle\))

\(60^{\circ}+60^{\circ}+\angle \mathrm{APD}=180^{\circ}\)

\(\angle \mathrm{APD}=60^{\circ}\)

\(\angle \mathrm{APD}+\angle \mathrm{BPD}=180^{\circ}\) (Linear pair)

\(\angle \mathrm{BPD}=120^{\circ}\)