Mathematics Test-12

As we know that,

The equation of the line passing through two points \({P}\left(x_1, y_1, z_1\right)\) and \({Q}\left(x_2, y_2, z_2\right)\) are given by

\(\frac{x_2-x_1}{{PQ}}, \frac{y_2-y_1}{{PQ}}, \frac{z_2-z_1}{{PQ}}\)

Equation of the line is,

\(\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=r \)

\(\Rightarrow x=r+2, y=r-1, z=r+2\)

The point \(({r}+2, {r}-1, {r}+2)\) lies on the plane \(2 {x}+{y}+{z}=9\) is, 

\(2({r}+2)+{r}-1+{r}+2=9\) 

\( \Rightarrow r=1\)

So, the coordinate of \({Q}\) are \((3,0,3)\) and thus,

\(P Q=\sqrt{(3-2)^2+(0+1)^2+(3-2)^2}=\sqrt{3}\)

Given:

f(x) = |x - 1|

Since the minimum value of modulus functions is zero.

So, the minimum value of |x - 1|, where x ∈ R is 0.

Given,

\(\mathrm{T}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+4)\)

\(\mathrm{T}_{\mathrm{n}}=\mathrm{n}^{3}+5 \mathrm{n}^{2}+4 \mathrm{n}\)

Then, the sum of the series,

\(\mathrm{S}_{\mathrm{n}}=\sum \mathrm{n}^{3}+5 \sum \mathrm{n}^{2}+4 \sum \mathrm{n}\)

\(=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^{2}+\frac{5 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{8}+\frac{4 \mathrm{n}(\mathrm{n}+1)}{2}\)

\(=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{\mathrm{n}^{2}+\mathrm{n}}{2}+\frac{5}{3}(2 \mathrm{n}+1)+4\right]\)

\(=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{3 \mathrm{n}^{2}+23 \mathrm{n}+34}{6}\right]\)

Given,

\(y=a e^{3 x} \cos (x+b)\)

There are two constants a and b so differentiate two times Differentiating wr.t \(x\),

We get,

\(y^{\prime}=3 a e^{3 x} \cos (x+b)-a e^{3 x} \sin (x+b)\)

\(\Rightarrow y^{\prime}=3 y-a e^{3 x} \sin (x+b)\)

\(\Rightarrow a e^{3 x} \sin (x+b)=3 y-y^{\prime}\)

Differentiating again w.r.t \({x}\),

We get,

\( 3 {ae}^{3 x} \sin (x+b)+a e^{3 x} \cos (x+b)=3 y^{\prime}-y^{\prime \prime}\)

\(\Rightarrow 3\left(3 y-y^{\prime}\right)+y-3 y^{\prime}-y^{n}=0 \)

\(\Rightarrow 9 y-3 y^{\prime}+y-3 y^{\prime}-y^{n}=0 \)

\(\Rightarrow y^{\prime \prime}+6 y^{\prime}-10 y=0\)

This problem is solved in a graphical method.

From question, the equation and condition given is:

\(f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^{2}\right\}, & |x| \leq 2 \\ 8-2|x|, & 2<|x| \leq 4\end{array}\right.\)

Let us draw the graph of \(\mathrm{y}=\mathrm{f}(\mathrm{x})\),

From question, for \(|x| \leq 2\), then \(f(x)=\max \left\{|x|, x^{2}\right\}\),

Let us first draw the graph of \(y=|x|\) and \(y=x^{2}\) as shown in the following figure.

From graph, \(y=|x|\) and \(y=x^{2}\) intersect at \(x=-1,0\), and 1 .

Now, the graph of \(f(x)=\max \left\{|x|, x^{2}\right\}\), for \(|x| \leq 2\).

From question, for \(2<|x| \leq 4\), then \(f(x)=8-2|x|\),

Since, the \(x\) is in modulus, the value of \(x\) can positive or negative.

So,

\(\Rightarrow f(x)=8-2|x|=\left\{\begin{array}{lc}8-2 x, & x \in(2,4] \\ 8+2 x, & x \in[-4,-2)\end{array}\right.\)

The graph formed for the obtained equation is given below,

So, the graph of y = f(x) is,

From the graph it is clear that at x = -2, -1, 0, 1, 2 the curve has sharp edges and so at these points f is not differentiable.

As per the given question:

Permutation: Permutation is defined as an arrangement of \(r\) things that can be done out of total \(n\) things. This is denoted by \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}\).

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

\(720=\frac{n !}{(n-r) !}\)

\((n-r) !=\frac{n !}{720}\)

Combination: The number of selections of r objects from the given \(n\) objects is denoted by \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\).

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

\(120=\frac{n !}{r !(n-r) !}\)

Putting the value of \((\mathrm{n}-\mathrm{r}) !\) we get

\(120=\frac{n !}{r ! \frac{n !}{720}}\)

\(r !=\frac{720}{120}=6\)

\(r !=3 \times 2 \times 1\)

\(r=3!\)

As we know,

Identification of curves represented by the general equation of the second degree:

Let a general equation of second degree in \(x\) and \(y\) be \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\)...(1).

Then this equation represents,

1. A parabola if \(\Delta \neq 0\) and \(h^{2}=a b\)

2. An ellipse if \(\Delta \neq 0\) and \(h^{2}>a b\)

3. A hyperbola if \(Δ ≠ 0\) and \(h^2 > ab\)

4. A pair of straight line or empty set if \(\Delta=0\) and \(h^{2} \geq a b\)

5. A unique point if \(\Delta=0\) and \(h^2 < ab\)

6. A circle if \(a = b ≠ 0, h = 0\) and \(g^2 + f^2 -ac > 0\)

The given equation is \(2 x^{2}+2 y^{2}-5 x-7 y-3=0\).

Comparing it wth the equation (1) we get,

\(a=2, h=0, b=2, g=-(\frac{5 }{ 2}), f=-(\frac{7 }{ 2}), c=-3\)

As, \(a=b=2(>0), h=0\)

As we know,

\(g^2 + f^2 -ac > 0\)

\(\therefore (-\frac{5 }{ 2})^2+ (-\frac{7 }{ 2})^2-(2 \times -3)>0\)

\(\Rightarrow \frac{25 }{4}+ \frac{49}{4}+6>0\)

\(\Rightarrow \frac{25 +49+24}{4}>0\)

\(\Rightarrow \frac{98}{4}>0\)

\(\Rightarrow 24.5>0\)

So, the given equation represents a circle.

Given: The four sets have 150, 180, 210 and 240 elements respectively

n(A) = 150

n(B) = 180

n(C) = 210

n(D) = 240

Each pair of sets has 15 elements

n(A ∩ B) = 15

n(A ∩ C) = 15

n(A ∩ D) = 15

n(B ∩ C) = 15

n(B ∩ D) = 15

n(C ∩ D) = 15

Each triple of sets has 3 elements

n(A ∩ B ∩ C) = 3

n(A ∩ B ∩ D) = 3

n(A ∩ C ∩ D) = 3

n(B ∩ C ∩ D) = 3

A ∩ B ∩ C ∩ D = ϕ

n(A ∩ B ∩ C ∩ D) = 0

Now, number of elements in the union of 4 sets A, B, C and D

n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)

⇒ 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0

∴ The required number of elements is 702.

Given:

\(\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}\)

\(\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}\)

We know that:

\(\tan A=\frac{\sin A}{\cos A}\) and \(\cot A=\frac{\cos A}{\sin A}\)

Therefore inserting the value of \(\tan A\) and \(\cot A\) in the given equation, we get

\(\therefore \frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}\)

\(\Rightarrow \frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}\)

\(\Rightarrow \frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}\)

\(\Rightarrow \frac{\cos ^{2} A}{\cos A-\sin A}-\frac{\sin ^{2} A}{\cos A-\sin A}\)

\(\Rightarrow \frac{\cos ^{2}-\sin ^{2} A}{\cos A-\sin A}\)

\(\Rightarrow \frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}\)

\(\Rightarrow \cos A+\sin A\)

\(\Rightarrow \sin A+\cos A\)

Hence, \(\sin A +\cos A\) is the answer.