Mathematics Test-2

Given here:

\(f(2)=2\) and \(f(x)=2\).

Let us check the form by putting \(x = 2\), we get:

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

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

\(=\frac{0}{0}\)

Apply L’ hospital’s rule,

\(\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(x)}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x}))}{\frac{d}{d x}(\mathrm{~g}(\mathrm{x}))}=\lim _{\mathrm{x} \rightarrow \mathrm{a}} \frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\)

\(\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}\)

Differentiating the numerator and the denominator in above, we get:

\(=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}(x f(2)-2 f(x))}{\frac{d}{d x}(x-2)}\)

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

Now, take limit at x = 2

\(\lim _{x \rightarrow 2}=f(2)-2 f^{\prime}(2)\)

\(=2-2(2)\)

\(=-2\)

Given,

\(\lim _{{x} \rightarrow 5} \frac{{x}^{2}-25}{{x}^{2}-2 {x}-10}\)

Putting the value \(x \rightarrow 5\) in above equation, we get:

\(=\frac{5^{2}-25}{5^{2}-2 \times 5-10}\)

\(=\frac{0}{5}\)

\(=0\)

Here, we have to find the equation of plane passing through the line of intersection of planes \(x+y+z=1\) and \(2 x+3 y+4 z=5\) which is perpendicular to plane \(x-y+z=0\)

As we know that, equation of a plane passing through the intersection of these planes is given by:\(\left(a_{1} x+b_{1} y+c_{1} z+d\right)+\lambda\left(a_{2} x+b_{2} y+c_{2} z+d\right)=0\), where \(\lambda\) is a scalar

\(\Rightarrow(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0\)

\(\Rightarrow(1+2 \lambda) x+(3 \lambda+1) y+(1+4 \lambda) z-1-5 \lambda=0\).....(1)

The direction ratios of the plane represented by (1) are: \((1+2 \lambda),(3 \lambda+1),(1+4 \lambda)\) The direction ratios of the plane \(x-y+z=0\) are: \(1,-1,1\)

\(\because\) plane represented by \((1)\) and is perpendicular to plane \(x-y+z=0\)

\(\Rightarrow 1+2 \lambda-3 \lambda-1+1+4 \lambda=0\)

\(\Rightarrow 3 \lambda+1=0\)

\(\Rightarrow \lambda=\frac{-1}{3}\)

By substituting \(\lambda=\frac{-1}{3}\) in equation (1), we get,

\(\Rightarrow \mathrm{x}-\mathrm{z}+2=0\)

In the expansion of \((a+b)^{n}\) the general term is given by: \(T_{r+1}={ }^{n} C_{r} . a^{n-r}.b^{r}\)

Given: \(\left(x-\frac{1}{x}\right)^{12}\)

As we know that, in the expansion of \((a+b)^{n}\), the rth term from the end is \([(n+1)-r+1]=(n-r+2)\) th term from the beginning.

Here, \(n=12\) and \(r=5\)

So, the 5 th term from the \(=[(12+1)-5+1]=9\) th term from the beginning.

\(T_{9}=T_{(8+1)}={}^{12} C_{8} .(x)^{4} .(\frac{-1}{x})^{8}\)

\(\Rightarrow T_{9}=T_{(8+1)}=(-1)^{8} .{}^{12} C_{8} .(x)^{4} .(\frac{1}{x})^{8}\)

\(=\frac{12 !}{8 ! 4 !} \times x^{4} \times \frac{1}{x^{8}}\)

\(=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times \frac{1}{x^{4}}\)

\(\Rightarrow T_{9}=\frac{495}{x^{4}}\)

The given AP series is \(5,12,19, \ldots, 215\)

First term \((a)=5\)

Common differenece \((d)=12-5=7\)

\(n^{\text {th }}\) term \(=a_{n}=215\)

As we know,

\(n^{\text {th }}\) term is given by,

\(a_{n}=a+(n-1) d\)

\(\therefore 215=5+(n-1) \times 7\)

\(\Rightarrow 210=(n-1) \times 7 \)

\(\Rightarrow n-1=30 \)

⇒ 31

Here, \(n\) is an odd number.

So, the middle term of the given AP will be \(\left(\frac{n+1}{2}\right)^{\text {th }}\) term.

\(\therefore\) Middle term \(=\left(\frac{ n +1}{2}\right)^{\text{th}}\) term

\(=\left(\frac{31+1}{2}\right)^{\text{th} }\) term

\(=16^{\text {th}}\) term

Now,

\(a_{16}=a+15 d\)

\(=5+15 \times 7\)

\(=5+105\)

\(=110\)

So, the middle of the given AP series is \(110\).

Given:

Number are\(15, 20, 18, 22 \mathrm {and} 25 \)

\(\text { Mean }=\frac{\sum x}{n} \)

\(=\frac{(15+20+18+22+25)}{5} \)

\(=\frac{100}{5}=20\)

Let, \(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\)

where \(x^{2}+y^{2}+z^{2}=1 \ldots(1)\)

\((\vec{d}\) being unit vector\() \quad \therefore \vec{a} \vec{d}=0\)

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

\([\vec{b} \vec{c} \vec{d}]=0 \Rightarrow\left|\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1=0 \\ x & y & z\end{array}\right|\)

\(\Rightarrow x+y+z=0\)

Using equation \((2)\)

\(\Rightarrow 2 x+z=0\)

\(\Rightarrow z=-2 x \ldots(3)\)

From \((1),(2)\) and \((3)\)

\(x^{2}+x^{2}+4 x^{2}=1 \Rightarrow x=\pm \frac{1}{\sqrt{6}}\)

\(\therefore d=\pm\left(\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}-\frac{2}{\sqrt{6}} \hat{k}\right)=\pm\left(\frac{\hat{i}+\hat{j}-2 \hat{k}}{\sqrt{6}}\right)\)

\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\)

\(=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \frac{2}{9}}\right)\)

\(=\tan ^{-1}\left(\frac{1}{2}\right)\)

\(=\frac{1}{2}\left(2 \tan ^{-1}\left(\frac{1}{2}\right)\right)\)

\(\Rightarrow \frac{1}{2} \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)\) \(\left\{\because 2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right\}\)

If \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) are a plane equations, then angle between planes can be found using the following formula:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{ a _{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

Here,

\(a_{1}=\sqrt{3}-1\)

\(a_{2}=-\sqrt{3}-1\)

\(b_{1}=-\sqrt{3}-1\)

\(b_{2}=\sqrt{3}-1\)

\(c_{1}=4\)

\(c_{2}=4\)

Therefore, required angle will be given by:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{a_{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

\(\cos \theta=\frac{-2-2+16}{\sqrt{24} \sqrt{24}}\)

\(=\frac{12}{24}\)

\(=\frac{1}{2}\)

\(\cos \theta = \cos \frac{\pi}{3}\)

\(\Rightarrow \theta=\frac{\pi}{3}\)

Equation of hyperbola can be written as, \(\frac{4x^{2}}{36}-\frac{y^{2}}{36}=1\) or \(\frac{x^{2}}{9}-\frac{y^{2}}{36}=1\)

Let the equation of tangent at any point \(P(x_{1}, y_{1})\) is \(\frac{xx^{2}}{a^{2}}-\frac{yy^{2}}{b^{2}}=1\)

As tangent is passing through the point \((0,3)\)

So, \(\frac{0}{a^{2}}-\frac{3y}{36}=1\)

\(y=-12\) and

Again, \(4 x^{2}-y^{2}=36\)

Using value of \(y\), we have \(x=\pm 3 \sqrt{5}\)

\(\frac{x^{2}}{9}-\frac{y^{2}}{36}=1\)

So, the coordinates of \(P\) and \(Q\) are \((3 \sqrt{5},-12)\) and \((-3 \sqrt{5},-12)\) respectively.

Now, area of \(\triangle TPQ\),

\(A=\frac{1}{2}\left|\begin{array}{ccc}3 \sqrt{5} & -12 & 1 \\ -3 \sqrt{5} & -12 & 1 \\ 0 & 3 & 1\end{array}\right|\)

\(=\frac{1}{2}|3 \sqrt{5}(-12-3)+12(-3 \sqrt{5}-0)+1(-9 \sqrt{5}-0)|\)

\(=\frac{1}{2}|-45 \sqrt{5}-36 \sqrt{5}-9 \sqrt{5}|\)

\(=\frac{1}{2}|90 \sqrt{5}|\)

\(=45 \sqrt{5}\)

Concept:

If \(x=f(t)\) defines a position vector of a particle then the first derivative represents the velocity of the particle and the second derivative represent the acceleration of the particle.

Now If, \(\frac{d^{2} x}{d t^{2}} \neq f(t)\) then it represents a constant acceleration.

Calculation:

Given:

Particle \(1: x(t)=3.5-2.7 t^{3}\)

Particle \(2: x(t)=3.5+2.7 t^{3}\)

Particle \(3: x(t)=3.5-2.7\) \(t ^{3}\)

Particle \(4: x ( t )=3.5-2.7 t ^{3}\)

The second derivative of Particle \(1\):

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d^{2}}{d t^{2}}\left(3.5-2.7 t^{3}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d}{d t}\left(0-2.7 \times 3 t^{2}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=-16.2 t\)

which is a function of time '\(t\)' hence acceleration is not constant.

The second derivative of Particle \(2\):

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d^{2}}{d t^{2}}\left(3.5+2.7 t^{3}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d}{d t}\left(0+2.7 \times 3 t^{2}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=16.2 t\)

Which is a function of time '\(t\)' hence acceleration is not constant.

The second derivative of Particle \(3\):

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d^{2}}{d t^{2}}\left(3.5-2.7 t^{2}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d}{d t}(0-2.7 \times 3 t)\)

\(\frac{d^{2}}{d t^{2}} x(t)=-8.1\)

Which is not a function of time '\(t\)' hence acceleration is constant.

The second derivative of Particle \(4\):

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d^{2}}{d t^{2}}\left(3.5-3.4 t-2.7 t^{2}\right)\)

\(\frac{d^{2}}{d t^{2}} x(t)=\frac{d}{d t}(0-3.4-2.7 \times 2 t)\)

\(\frac{d^{2}}{d t^{2}} x(t)=-5.4\)

Which is not a function of time '\(t\)' hence acceleration is constant.

So only particle c and particle d have Constant acceleration.

Let the common ratio of the qiven G.P. be \(r\).

Let \(D=\left|\begin{array}{lll}\ln a_{1} & \ln a_{2} & \ln a_{3} \\ \ln a_{4} & \ln a_{5} & \ln a_{6} \\ \ln a_{7} & \ln a_{8} & \ln a_{9}\end{array}\right|\)

Using \(C_{1} \rightarrow C_{1}-C_{2}\) and \(C_{2} \rightarrow C_{2}-C_{3}\), we get,

\(\Rightarrow D=\left|\begin{array}{lll}\ln a_{1}-\ln a_{2} & \ln a_{2}-\ln a_{3} & \ln a_{3} \\ \ln a_{4}-\ln a_{5} & \ln a_{5}-\ln a_{6} & \ln a_{6} \\ \ln a_{7}-\ln a_{8} & \ln a_{8}-\ln a_{9} & \ln a_{9}\end{array}\right|\)

\(\Rightarrow D=\left|\begin{array}{lll}\ln \frac{a_{1}}{a_{2}} & \ln \frac{a_{2}}{a_{3}} & \ln a_{3} \\ \ln \frac{a_{4}}{a_{5}} & \ln \frac{a_{5}}{a_{6}} & \ln a_{6} \\ \ln \frac{a_{7}}{a_{8}} & \ln \frac{a_{8}}{a_{9}} & \ln a_{9}\end{array}\right|\)

\(\Rightarrow D=\left|\begin{array}{lll}\ln r & \ln r & \ln a_{3} \\ \ln r & \ln r & \ln a_{6} \\ \ln r & \ln r & \ln a_{9}\end{array}\right|\)

Using \(C_{1} \rightarrow C_{1}-C_{2}\), we get,

\(\Rightarrow D=\left|\begin{array}{ccc}0 & \ln r & \ln a_{3} \\ 0 & \ln r & \ln a_{6} \\ 0 & \ln r & \ln a_{9}\end{array}\right|\)

Expanding along \(C_{1}\), we get,

\(\Rightarrow D=0\)

Given:

Curve \(1: y=2 x^{2}=f(x)\) (say)

Curve 2: \(y+6 x-8=0\)

\(\Rightarrow \mathrm{y}=8-6 \mathrm{x}=\mathrm{g}(\mathrm{x})\) (say)

To find the intersections (or limits of the area) putting value of y from curve 1

\(\Rightarrow 2 x^{2}=8-6 x\)

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

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

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

\(\Rightarrow x_{1}=-4, x_{2}=1\)

Now the required area (A) is:

\(A=\left|\int_{x_{1}}^{x_{2}}[f(x)-g(x)] d x\right|\)

\(\Rightarrow A=\left|\int_{-4}^{1}\left[2 x^{2}-(8-6 x)\right] d x\right|\)

\(\Rightarrow A=\left|\int_{4}^{1}\left[2 x^{2}-8+6 x\right] d x\right|\)

\(\Rightarrow A=\left|\left[\frac{2 x^{3}}{3}-8 x+\frac{6 x^{2}}{2}\right]_{-4}^{1}\right|\)

\(\Rightarrow A=\left|\frac{2}{3}-8+3-\left(\frac{2(-4)^{3}}{3}-8(-4)+3(-4)^{2}\right)\right|\)

\(\Rightarrow A=\left|\frac{-13}{3}-\frac{(-128)}{3}-32-48\right|\)

\(\Rightarrow A=\left|\frac{(-125)}{3}\right|\)

\(\Rightarrow A=\frac{125}{3}\)