Mathematics Test-12

The given series is,

\(\frac{1}{(\sqrt{1}+\sqrt{3})}+\frac{1}{(\sqrt{3}+\sqrt{5})}+\frac{1}{(\sqrt{5}+\sqrt{7})}+\ldots\)

Let the general term of the given series is \(T_r\).

\(\therefore T_{r}=\frac{1}{\sqrt{2 r-1}+\sqrt{2 r+1}}\)

We have to find the sum of \(n\) terms of given series. Therefore,

\(S_{n}=\sum T_{r}\)

\(\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{1}{\sqrt{2 r-1}+\sqrt{2 A+1}}\)

By rationalizing the denominator, we get,

\(S_{n}=\sum_{r=1}^{n} \frac{1}{\sqrt{2 r+1}+\sqrt{2 r-1}} \times\left[\frac{\sqrt{2 r+1}-\sqrt{2r-1}}{\sqrt{2 r+1}-\sqrt{2 r-1}}\right]\)

\(\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{\sqrt{2 r+1}-\sqrt{2r-1}}{(\sqrt{2 r+1})^2-(\sqrt{2 r-1})^2}\quad[\because (a+b)(a-b)=a^2-b^2] \)

\(\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{\sqrt{2 r+1}-\sqrt{2r-1}}{2 r+1-2 r+1} \)

\(\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{\sqrt{2 r+1}-\sqrt{2r-1}}{2} \)

\(\Rightarrow S_{n}=\frac{1}{2}\left[\sum_{r=1}^{n} \sqrt{2r+1}-\sqrt{2 r-1}\right]\)

\(\Rightarrow S_n=\frac{1}{2}\left[(\sqrt{3}-\sqrt{1})+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+\ldots+\sqrt{2 n+1}-\sqrt{2 n-1}\right]\)

\(\Rightarrow S_n=\frac{1}{2}\left[\sqrt{2 n+1}-1\right]\)

Let \(y=\frac{x^{2}+x+2}{x^{2}+x+1}\)

\(\Rightarrow x^{2}(y-1)+x(y-1)+(y-2)=0, \forall x \in R\)

Now, \(D \geq 0 \Rightarrow(y-1)^{2}-4(y-1)(y-2) \geq 0\)

⇒ \((y-1)\{(y-1)-4(y-2)\} \geq 0\)

⇒ \((y-1)(-3 y+7) \geq 0\)

\(\Rightarrow \leq \mathrm{y} \leq \frac{7}{3}\)

\(\therefore\) Range \(=\left[1, \frac{7}{3}\right]\)

Given,

\(G(x)=\sqrt{25-x^{2}}\)

As we know that, any a function \(f : I \rightarrow R\) is said to be differentiable at \(c\), if and only if \(\lim _{h \rightarrow 0}\left[\frac{f(c+h)-f(c)}{h}\right]=f^{\prime}(c)\).

\(\Rightarrow \lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}=G^{\prime}(1)\)

\(\because G(x)=\sqrt{25-x^{2}}\)

\(\Rightarrow G^{\prime}(x)=\frac{1}{2} \times \frac{1}{\sqrt{25-x^{2}}} \times(-2 x)=\frac{-x}{\sqrt{25-x^{2}}}\)

\(\Rightarrow G^{\prime}(1)=\frac{-x}{\sqrt{25-x^{2}}}=-\frac{1}{2 \sqrt{6}}\)

Given,

Equation of curve: \(\sqrt{ x }+\sqrt{ y }=1\)

\(\therefore \sqrt{ y }=1-\sqrt{ x }\)

\(\Rightarrow y =(1-\sqrt{ x })^{2}\)

\(=1-2 \sqrt{ x }+ x\)

Now let's find the limits,

\(\text { If } x=0\)

\(\sqrt{x}+\sqrt{y}=1\)

\(\Rightarrow \sqrt{y}=1\)

\(\Rightarrow y=1\)

And if, \(y=0, x=1\)

So the curve cut \(Y\)-axis at \((0,1)\) and \(X\)-axis at \((1,0)\)

\(\therefore x=0, x=1\)

Now,

\(\text { Area }= A =\int_{0}^{1} ydx\)

\(=\int_{0}^{1}(1-2 \sqrt{ x }+ x ) dx\)

\(=\left[ x -\frac{2( x )^{\frac{3}{2}}}{\frac{3}{2}}+\frac{ x ^{2}}{2}\right]_{0}^{1}\)

\(=\left[\left(1-\frac{4}{3}+\frac{1}{2}\right)-0\right]\)

\(=\left[-\frac{1}{3}+\frac{1}{2}\right]\)

\(=\frac{1}{6} \text { square unit }\)

Let, \(\tan ^{-1} \frac{ y }{2}=\theta\), where \(\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

\(\Rightarrow \tan \theta=\frac{y}{2}\)

\(\Rightarrow \tan ^{2} \theta=\frac{y^{2}}{4}\)

\(\Rightarrow 1+\tan ^{2} \theta=1+\frac{y^{2}}{4}=\frac{y^{2}+4}{4}\)

\(\Rightarrow \sec ^{2} \theta=\frac{y^{2}+4}{4}\)

\(\Rightarrow \sec \theta=\frac{\sqrt{y^{2}+4}}{2}\)

\(\Rightarrow \sec \left(\tan ^{-1} \frac{ y }{2}\right)=\frac{\sqrt{y^{2}+4}}{2}\)

Given, \(\sin ^{2} x+\sin x=1\)

\(\Rightarrow \sin x=1-\sin ^{2} x\)

\(\Rightarrow \sin x=\cos ^{2} x \quad \cdots \quad\) As we know, \((a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}\)

\(\Rightarrow \cos ^{12} x+3 \cos ^{10} x+3 \cos ^{8} x+\cos ^{6} x\)

\(=\left(\cos ^{4} x+\cos ^{2} x\right)^{3}[\) Substituting from eq. (1)]

\(=\left(\sin ^{2} x+\cos ^{2} x\right)^{3} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\)

\(=1^{3}\)

\(=1\)

Here, foci \(=(\pm 2,0)=(\pm ae , 0)\) and the eccentricity, \(e =\frac{1}{2}\)

\(a e=2\)

\(\Rightarrow a \times \frac{1}{2}=2\)

\(\Rightarrow a=4\)

\(\Rightarrow a^{2}=16\)

\(\text { Now, } e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

\(\Rightarrow b^{2}=a^{2}\left(1-e^{2}\right)\)

\(=16(1-\frac{1}{4})\)

\(=16 \times(\frac{3}{4})\)

\(=12\)

\(\therefore\) Equation of ellipse \(=\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\)

Let vector \(\vec{c}\) is perpendicular to both the vectors \(\hat{\imath}+\hat{\jmath}\) and \(\hat{\jmath}+\hat{k}\)

Let \(\vec{a}=\hat{\imath}+\hat{\jmath}\) and \(\vec{b}=\hat{\jmath}+\hat{k}\)

Therefore, \(\vec{c}=\vec{a} \times \vec{b}\)

\(=(\hat{\imath}+\hat{\jmath}) \times(\hat{\jmath}+\hat{k})\)

\(=\hat{\imath} \times \hat{\jmath}+\hat{\imath} \times \hat{k}+\hat{\jmath} \times\) \(\hat{\jmath}+\hat{\jmath} \times \hat{k}\)

\(=\hat{k}-\hat{\jmath}+0+\hat{\imath}\)

\(=\hat{\imath}-\hat{\jmath}+\hat{k}\)

Given:

Point \(C(1,2)\) divides the line passing through \(A(h, 0)\) and \(B(0, k)\) in the ratio \(2: 3\)

Here, \(m: n=2: 3,(x, y)=(1,2),\left(x_{1}, y_{1}\right)=(0, k),\left(x_{2}, y_{2}\right)=(h, 0)\)

Using sectional formula:

\(\therefore x=\frac{m x_{2}+nx_{1 }}{m+n}\)

\(\Rightarrow 1=\frac{3(\mathrm{~h})+2(0)}{2+3}\)

\(\Rightarrow 1=\frac{3 \mathrm{~h}}{5}\)

\(\Rightarrow 3 \mathrm{~h}=5\)

\(\Rightarrow \mathrm{h}=\frac{5}{3}\)

\(\mathrm{y}=\frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\)

\(\Rightarrow 2=\frac{2(k)+3(0)}{2+3}\)

\(\Rightarrow 2 \mathrm{k}=10\)

\(\Rightarrow \mathrm{k}=5\)

\(\therefore x=\frac{m x_{2}+nx_{1 }}{m+n}\)

Therefore, \(x\) and \(y\) intercept are \(\frac{5}{3}\) and \(5\).

Hence, required equation

\((\frac{x}{\frac{5}{3}})+(\frac{y}{ 5})=1 \)

\(\Rightarrow 3 x+y=5\)

\(\therefore\) Required equation is \(3 x+y=5\)

The given system of linear equations:

\(x+y+z=5 \)

\(x+2 y+2 z=6\)

\(x+3 y+\lambda z=\mu\) have infinite solution.

\(\therefore \Delta=0, \Delta \mathrm{x}=\Delta \mathrm{y}=\Delta \mathrm{z}=0\)

Now, forming determinant from the given equations,

\(\Delta=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda\end{array}\right|=0\)

\(\Rightarrow 1(2 \lambda-6)-1(\lambda-2)+1(3-2)=0\)

\(\Rightarrow 2 \lambda-6-\lambda+2+3-2=0\)

\(\Rightarrow \lambda-8+5=0\)

\(\Rightarrow \lambda-3=0\)

\(\therefore \lambda=3\)

Now, the determinant of \(y\) is:

\(\Delta \mathrm{y}=\left|\begin{array}{lll}1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & 3\end{array}\right|=0\)

\(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\)

\(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)

\(\Rightarrow\left|\begin{array}{ccc}1 & 5 & 1 \\ 1-1 & 6-5 & 2-1 \\ 1-1 & \mu-5 & 3-1\end{array}\right|=0\)

\(\Rightarrow\left|\begin{array}{ccc}1 & 5 & 1 \\ 0 & 1 & 1 \\ 0 & \mu-5 & 2\end{array}\right|=0\)

\(\Rightarrow 1(2-(\mu-5))-5(0-0)+1(0-0)=0 \)

\(\Rightarrow 1(2-(\mu-5))=0 \)

\(\Rightarrow 2-\mu+5=0 \)

\(\Rightarrow 7-\mu=0 \)

\(\therefore \mu=7\)

Now

\(\therefore \lambda+\mu=3+7=10\)

\(f(x)=\log x+3 x-10\) and \(g(x)=\tan x\)

\(fog(x)=f\{g(x)\}=\log (\tan x)+3 \tan x-10\)

\(\Rightarrow fog^{\prime}(x)=\frac{d[\log (\tan x)]}{d x}+3 \frac{d(\tan x)}{d x}-\frac{d(10)}{d x}\)

\(\Rightarrow fog^{\prime}(x)=\frac{1}{\tan x} \frac{d(\tan x)}{d x}+3 \sec ^{2} x\)

\(\Rightarrow fog^{\prime}(x)=\frac{1}{\tan x} \times \sec ^{2} x+3 \sec ^{2} x\)

\(\Rightarrow fog^{\prime}(x)=\frac{\cos x}{\sin x} \times \frac{1}{\cos ^{2}}+3 \sec ^{2} x\)

\(\Rightarrow fog^{\prime}(x)=\operatorname{cosec} x . \sec x+3 \sec ^{2}\)

\(\Rightarrow fog^{\prime}(x)=\sec x(\operatorname{cosec} x+3 \sec x)\)