Mathematics Test-35

Given,

\(\log _{\cos x} \sin x=1\)

\(\frac{\log \sin x}{\log \cos x}=1\)

\(\sin x=\cos x\)

The smallest positive value of \(x\) for which \(\sin x=\cos x\) is \(x=\frac{\pi}{4}\)

\( x^2 \frac{d y}{d x}=x^2+x y+y^2 \)

\(\Rightarrow \frac{d y}{d x}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2\)

Substituting \(y=v x\) and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

\( \Rightarrow v + x \frac{ dv }{ dx }=1+ v + v ^2 \)

\( \Rightarrow x \frac{ dv }{ dx }=1+ v ^2\)

Integrating both sides we get,

\(\int \frac{ dx }{ x }=\int \frac{ dv }{1+ v ^2}\)

\(\Rightarrow \log x=\tan ^{-1} v + c , c =\) constant of integration

Putting the value of \(v\) we get,

\(\therefore \log x =\tan ^{-1} \frac{ y }{ x }+ c\)

Let \(r^{\text {th }}\) term is independent of x.

\(T_r={ }^n C_r x^r y^{n-r}\)

\(={ }^{10} C_r(\sqrt{x})^r\left(\frac{1}{3 x^2}\right)^{10-r}\)

\(={ }^{10} C_r\left(\frac{1}{3}\right)^{10-r} \cdot(\sqrt{x})^r\left(\frac{1}{x^2}\right)^{10-r}\)

Equating the coefficient of \(x\) to zero.

\(\Rightarrow x^{\frac{r}{2}} \cdot x^{-2(10-r)}=x^0\)

\(\Rightarrow \frac{r}{2}-20+2 r=0\)

\(\Rightarrow \frac{5}{2} r=20 \Rightarrow r=8\)

Coefficient \(={ }^{10} C_r\left(\frac{1}{3}\right)^{10-r}\)

\(={ }^{10} C_8\left(\frac{1}{3}\right)^{10-8}=\frac{10 \times 9}{2} \times \frac{1}{9}=5\)

Given:

\(\frac{d x}{d y}+\frac{x}{y}=0\)

On solving, we get-

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

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

\(\Rightarrow \ln x=-\ln y+k\) or

\(\Rightarrow \ln x=\ln \frac{1}{y}+\ln c\)

\(\Rightarrow \ln x=\ln \frac{c}{y}\)

\(\Rightarrow x=\frac{c}{y}\)

\(\Rightarrow x y=c\)

Given:

x² + bx + c = 0

The roots of the equation are two consecutive integers.

Formula:

Quadratic equation = x² - (sum of roots)x + product of roots

Calculation:

x² + bx + c = 0 ---(1)

Let the roots are n, n + 1

So, a quadratic equation will be

x² - (n + n + 1)x + n(n + 1) = 0

⇒ x² - (2n + 1)x + n(n + 1) = 0

So,

⇒ b² - 4ac = (2n + 1)² - 4n(n+ 1)

∵ (a + b)² = a² + 2ab + b²

⇒ b² - 4ac = 4n² + 4n + 1 - 4n² - 4n

⇒ b² - 4ac = 1

Four-digit numbers divisible by10 using 1,5,0,6,7.

Divisibility by 10 \(\rightarrow\) no. should end with zero

After fixing 'zero' at unit place, we are left with 4 numbers.

After placing any number at tens place we are left with '3' numbers.

After placing any number at tens and hundreds place.

Total \(=4 \times 3 \times 2\)

\(=24\)

Vector equation of a plane at a distance ' \(d\) ' from the origin and unit vector to normal from origin \(\hat{n}\) is

\(\vec{r} \cdot \hat{n}= d\)

Unit vector of \(\vec{n}=\hat{n}=\frac{1}{|\vec{n}|}(\vec{n})\)

Given, equation of plane is

\(\vec{r} \cdot(6 \hat{\imath}-3 \hat{\jmath}-2 \hat{k})+1=0 \)

\(\vec{r} \cdot(6 \hat{\imath}-3 \hat{\jmath}-2 \hat{k})=-1\)

Multiplying with -1 on both sides,

\(-\vec{r} \cdot(6 \hat{\imath}-3 \hat{\jmath}-2 \hat{k})=-1 \times-1 \)

\(\vec{r} \cdot(-6 \hat{\imath}+3 \hat{\jmath}+2 \widehat{k})=1\)

Magnitude of \(\vec{n}=\sqrt{(-6)^2+3^2+2^2}\)

\(|\vec{n}|=\sqrt{36+9+4}=\sqrt{49}=7\)

Now,

\(\widehat{n} =\frac{1}{|\vec{n}|}(\vec{n}) \)

\(=\frac{1}{7}(-6 \hat{\imath}+3 \hat{\jmath}+2 \hat{k}) \)

\(=\frac{-6}{7} \hat{\imath}+\frac{3}{7} \hat{\jmath}+\frac{2}{7} \widehat{k}\)

\(\therefore\) Direction cosines of unit vector perpendicular to the given plane i.e. in are \(\frac{-6}{7}, \frac{3}{7}, \frac{2}{7}\).

Given:

\(I =\int \frac{(\cos x)^{1.5}-(\sin x)^{1.5}}{\sqrt{\sin x \cdot \cos x}} d x\)

\(\Rightarrow I =\int \frac{\cos x}{\sqrt{\sin x}}-\frac{\sin x}{\sqrt{\cos x}} d x\)

\(\Rightarrow I =\int \frac{\cos x}{\sqrt{\sin x}} d x-\int \frac{\sin x}{\sqrt{\cos x}} d x= I _1- I _2\) (let)

Computing \(I _1\),

\(I _1=\int \frac{\cos x}{\sqrt{\sin x}} d x\)

Put \(\sin x=t \rightarrow \cos x d x=d t\)

\(\Rightarrow I_1=\int \frac{d t}{\sqrt{t}}=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c_1\)

\(\Rightarrow I_1=2 \sqrt{t}+c_1\)

\(\Rightarrow I_1=2 \sqrt{\sin x}+c_1\)

Similarly,

\(I _2=\int \frac{\sin x}{\sqrt{\cos x}} d x\)

Put \(\cos x=t \rightarrow-\sin x d x=d t\)

\(\Rightarrow I_2=\int \frac{d t}{\sqrt{t}}=-\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c_2\)

\(\Rightarrow I_2=-2 \sqrt{t}+c_2\)

\(\Rightarrow I_2=-2 \sqrt{\cos x}+c_2\)

Putting these value in \(I\),

\(\Rightarrow I=I_1-I_2=2 \sqrt{\sin x}+c_1-(-2 \sqrt{\cos x})+c_2\)

\(\Rightarrow I=2 \sqrt{\sin x}+2 \sqrt{\cos x}+c\)

Given:

\(\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\)

\(\mathrm{A}^2=\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)

\(\therefore \mathrm{A}^2=\mathrm{I}\)

Given points \(A(-1,3,2), B(-4,2,-2)\) and \(C(5,5, \lambda)\)

Let the point \(A, B, C\) are collinear.

So, the point \(A, B, C\) lie on the same line.

We construct line \(A B\) and point \(C\) also lie on the line \(A B\).

Direction ratio of \({A B}=(-4+1,2-3,-2-2)=(-3,-1,-4)\)

Direction ratio of\({A C}=(5+1,5-3, \lambda-2)=(6,2, \lambda-2)\)

As we know that equation of a line if two points are given,

\(\frac{\left(x-x_1\right)}{a}=\frac{\left(y-y_1\right)}{b}=\frac{\left(z-z_1\right)}{c}\)

where a, b and c are direction ratio.

Equation of \(A B\),

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

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

If point \(C\) lie on this line because all three points are collinear. So, co-ordinate of point C will satisfy the equation (1).

Then, we get,

\(\frac{5+1}{3}=\frac{5-3}{1}=\frac{\lambda-2}{4}\)

Now, \(\frac{\lambda-2}{4}=\frac{5+1}{3}=2\)

\(\lambda-2=8\)

\(\lambda=10\)