Mathematics Test - 5

\(f(x)=\frac{|x|}{1+|x|}=\left\{\begin{array}{ll}\frac{x}{1+x} & , x \geq 0 \\ \frac{-x}{1-x} & , x<0\end{array}\right.\)

Graph of f(x)

From graph we can see that, for two value of ‘x’, f(x) has same value so f(x) is not injective.

Also, range of f(x) is [0, 1)

So it is not surjective also.

Hence option D is the correct answer.

We have

\(\tan 60^{\circ}=\left|\frac{m+\sqrt{3}}{1-\sqrt{3} m}\right|\)

\(\sqrt{3}=\frac{m+\sqrt{3}}{1-\sqrt{3} m} \ldots(i)\) and \(\sqrt{3}=\frac{m+\sqrt{3}}{-1+\sqrt{3} m} \ldots(\) ii \()\)

From equation (i)

\(\sqrt{3}-3 m=m+\sqrt{3}\)

\(m=0\)

From equation (ii):

\(-\sqrt{3}+3 m=m+\sqrt{3}\)

\(2 m=2 \sqrt{3}\)

\(m=\sqrt{3}\)

The equations of the required lines are:

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

\(y+2=0\)

Hence option A is the correct answer.

Let the centre be at (0,0) Thus, the equation of the ellipse becomes \(\frac{(x)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) Differentiating w.r.t \(x,\) we get \(\Rightarrow \frac{2(x)}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \ldots \ldots \ldots\)

Thus, we get \(, \frac{d y}{d x}=-\left(\frac{x}{y}\right) \cdot \frac{b^{2}}{a^{2}}\)

Differentiating the equation 1 w. r.t \(x\)

\(\Rightarrow\)

\(\frac{2}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d^{2} y}{d x^{2}}+\frac{2}{b^{2}} \cdot\left(\frac{d y}{d x}\right)^{2}=0\)

\(\Rightarrow\)

\(\frac{b^{2}}{a^{2}}+y \cdot \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\)

\(-\frac{y}{x} \cdot \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\)

Hence option A is the correct answer.

At \(x=\frac{\pi}{6}\)

\(L H L=\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right]=0\)

\(R H L=\lim _{x \rightarrow \frac{x}{6}+}\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right]=1\)

\(f(x)\) is discontinuous at \(x=\frac{\pi}{6}\) At \(x=\frac{\pi}{3}\)

\(L H L=\lim _{x \rightarrow \frac{\pi}{3}}\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right]=1\)

\(R H L=\lim _{x \rightarrow \frac{\pi}{3}+\left[\frac{6}{x}\right] \cos \left[\frac{3 x}{x}\right]=2 \cos 1}\)

\(f(x)\) is discontinuous at \(x=\frac{\pi}{3}\)

Similarly, \(f(x)\) is discontinuous at \(\frac{\pi}{2}\)

Hence option D is the correct answer.

Let \(P\) be \((x, y, z),\) then \(x=r \sin \theta \cdot \cos \phi\)

\(y=r \sin q \sin f\)

\(z=r \cos \theta\)

\(\Rightarrow 1=r \sin \theta \cdot \cos \phi, 2=r \sin \theta \sin \phi, 3=r \cos \theta\)

\(\Rightarrow 1^{2}+2^{2}+3^{2}=r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \theta \sin ^{2} \phi+r^{2} \cos ^{2} \theta=r^{2}\)

\(\Rightarrow r=\sqrt{14}\)

Conclusions

1. \(\sin \theta \cos \phi=\frac{1}{\sqrt{14}}\)

\(2 \sin \theta \sin \phi=\frac{2}{\sqrt{14}}\)

3. \(\cos \theta=\frac{3}{\sqrt{14}}\)

4. \(\tan \theta=\frac{\sqrt{5}}{3}\)

\(\therefore\) Options 1,2 and 3 are correct. \(\tan \phi=2, \cos \theta=\frac{3}{\sqrt{14}}\)

\(\therefore \cos \theta \cos \phi=\frac{3}{\sqrt{14}} \cdot \frac{1}{\sqrt{5}}=\frac{3}{\sqrt{70}}\)

Sum of odd coefficients of \((1+z)^{5}\) is \(2^{5-1}=2^{4}=16\)

\(=\boldsymbol{x}\)

\(a+b+c+d=16\)

\(a=e-16\)

then \(e>0\) \(e-16+b+c+d=16\)

\(e+b+c+d=32\)

Number of positive integral solutions \(=(n+r-1) c_{r-1}\) \((32+4-1) c_{4-1}\)

\(=\)\({ }^{35} \mathrm{C}_{3}\)

Hence option C is the correct answer.

Let equation of the line through \(P(3,8,2)\) and parallel to the plane \(3 x+2 y-2 z+15=0\) be \(\frac{x-3}{A}=\frac{y-8}{B}=\frac{z-2}{C}\)

Then, \(3 A+2 B-2 C=0\)

Let line

(i) intersect

\(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\) at \(Q\)

then \(\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0\)

\(\Rightarrow\left|\begin{array}{ccc}3-1 & 8-3 & 2-2 \\ A & B & C \\ 2 & 4 & 3\end{array}\right|=0\)

\(\Rightarrow 15 A-6 B-2 C=0\)

From equation (ii) and equation \(\frac{A}{\left|\begin{array}{cc}2 & -2 \\ -6 & -2\end{array}\right|}=\frac{B}{\left|\begin{array}{cc}-2 & 3 \\ -2 & 15\end{array}\right|}=\frac{C}{\left|\begin{array}{cc}3 & 2 \\ 15 & -6\end{array}\right|}\)

\(\Rightarrow \frac{A}{2}=\frac{B}{3}=\frac{C}{6}\)

Substituting \(A, B\) and \(C\) in equation

\(\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}\)

This is the equation of line \(P Q\) parallel to the plane \(3 x+2 y-2 z+15=0\)

Now, 'Q' is the point where this line cuts the line \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\)

So, point 'Q' will satisfy this line also.

Any point 'Q' on line \(\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}=\lambda\)

is \(Q(2 \lambda+3,3 \lambda+8,6 \lambda+2)\)

When this point also lies on

\(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\)

\(\Rightarrow \frac{(2 \lambda+3)-1}{2}=\frac{3 \lambda+8-3}{4}=\frac{6 \lambda+2-2}{3}\)

\(\Rightarrow \lambda=1\)

Point \(Q=(5,11,8)\)

length of \(P Q=\sqrt{(5-3)^{2}+(11-8)^{2}+(8-2)^{2}}\)

\(=\sqrt{4+9+36}\)

\(=7\) unit

Hence option D is the correct answer.

For infinitely many solutions, the two equations become identical

\(\Rightarrow \frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1}\)

from first 2

\(\frac{k+1}{k}=\frac{8}{k+3}\)

\(\Rightarrow k=1,3\)

From last 2

\(\frac{8}{k+3}=\frac{4 k}{3 k-1}\)

\(\Rightarrow k=1,2\)

so value of \(k\) which satisfy both is 1 so \(k=1\) so number of values of \(\mathrm{k}\) is one.

Hence option B is the correct answer.

For planes to intersect in a straight line:

\(\left|\begin{array}{lll}k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1\end{array}\right|=0\)

\(k(k-4)-4(4-4)+1(8-2 k)=0\)

\(\Rightarrow k^{2}-4 k-2 k+8=0\)

\(\Rightarrow k^{2}-6 k+8=0\)

\(\Rightarrow k=2\)

Hence option B is the correct answer.

Hence option D is the correct answer.