Mix Test 2

\(sin[\frac{\pi}{3}-(\frac{-\pi}{6})]=sin(\frac{\pi}{2})=1\)

\(\underset{x\rightarrow0}{lim}\left(\frac{1-cos\,kx}{xsin\,x}\right)=\frac{1}{2}\)

\(\Rightarrow\underset{x\rightarrow0}{lim}\left(\frac{2sin^2\frac{kx}{2}}{xsin\,x}\right)=\frac{1}{2}\)

\(\Rightarrow\underset{x\rightarrow0}{lim}2\left(\frac{k}{2}\right)^2\left(\cfrac{sin\frac{kx}{2}}{\frac{kx}{2}}\right)^2\left(\frac{x}{sinx}\right)=\frac{1}{2}\)

\(\Rightarrow k^2=1\Rightarrow k=\pm1\) \(but\,k<0\)

\(\Rightarrow k=-1\)

\(A=\begin{bmatrix}0&1\\1&0\end{bmatrix}\)

\(A^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

As A is singular matrix

\(\Rightarrow|A|=0\)

\(\Rightarrow 2k^2-32=0\Rightarrow k=\pm4\)

Hence, the set of values of k for which the given matrix A is singular is {-4, 4}

\(as\,|adj\,A|=|A|^{n-1},\) where n is order of matrix A

Therefore, \(|adj A|=(-4)^2\) = 16

(1, 2)

\(\begin{cases}2a+b=4\\a-2b=-3\\5c-d=11\\4c+3d=24\end{cases}\) \(\Rightarrow{a=1\\b=2\\c=3\\d=4}\)

\(\therefore a+b-c+2d=8\)

\(f(x)=x+\frac{1}{x},x>0\) \(\Rightarrow f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2},x>0\)

As normal to \(f(x)\,is\perp\) to given line

\(\Rightarrow(\frac{x^2}{1-x^2})\times\frac{3}{4}=-1\,(m_1.m_2=-1)\)

\(\Rightarrow x^2=4\Rightarrow x=\pm2\)

But x > 0, \(\therefore x=2\)

Therefore point \(=(2,\frac{5}{2})\)

\(sin(tan^{-1}x)=sin\left[sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right]\) \(=\frac{x}{\sqrt{1+x^2}}\)

\(\because |1-1|=0\) which is a multiple of 4

Therefore, \((1, 1) \in R\)

and \(|1-5|=4\) which is a multiple of 4

Therefore, \((1, 5) \in R\)

Also \(|1-9|=8\) which is a multiple of 4

Therefore, \((1, 9) \in R\)

Since, \((1, 1), (1, 5), (1, 9) \in R\)

Therefore, the equivalence class of 1 is [1] = {1, 5, 9}

\(e^x+e^y=e^{x+y}\)

\(\Rightarrow e^{-y}+e^{-x}=1\)

Differentiating w.r.t. x:

\(\Rightarrow-e^{-y}\frac{dy}{dx}-e^{-x}=0\) \(\Rightarrow\frac{dy}{dx}=-e^{y-x}\)

Since, order of matrices A and B are \(3\times n\) and \(m\times5\), respectively.

Therefore, the order of matrices 5A and 3B are \(3\times n\) and \(m\times5\), respectively.

We know that matrix addition of two matrices is possible if both matrices have same order.

i.e., C is possible only if \(3\times n\) = \(m\times5\)

⇒ \(m=3\, \text{and}\, n=5\)

And order of matrix C is \(m\times n = 3\times 5\)

y = 5 cos x – 3 sin x

\(\Rightarrow\frac{dy}{dx}\) = −5 sin x − 3 cos x

\(\Rightarrow\frac{d^2y}{dx^2}\) = −5 cos x + 3 sin x = −y

\(adjA=\begin{bmatrix}7&-5\\11&2\end{bmatrix}\) \(\Rightarrow (adjA)'=\begin{bmatrix}7&11\\-5&2\end{bmatrix}\)

\(\frac{x^2}{9}+\frac{y^2}{16}=1\;\Rightarrow\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0\)

\(\Rightarrow\) slope of normal \(=\frac{-dx}{dy}=\frac{9y}{16x}\)

As curve’s tangent is parallel to y-axes

Therefore, the normal to the curve is parallel to x-axis

\(\Rightarrow\frac{9y}{16x}=0\Rightarrow y=0\;and\,x=\pm3\)

\(\therefore\,points=(\pm3,0)\)

\(|A|=-7\)

\(\therefore\sum_{i=1}^3a_{i2}A_{i2}\) \(=a_{12}A_{12}+a_{22}A_{22}+a_{32}A_{32}\) \(=|A|=-7\)

\(y=log(cos\,e^x)\)

Differentiating wrt x:

\(\frac{dy}{dx}=\frac{1}{cos(e^x)}.(-sin\,e^x).e^x\) (chain rule)

\(\Rightarrow\frac{dy}{dx}=-e^x\,tan\,e^x\)

\(f(x)=2cos\,x+x,x\in[0,\frac{\pi}{2}]\)

\(f'(x)=-2sin\,x+1\)

\(let\,f'(x)=0\Rightarrow x=\frac{\pi}{6}\in[0,\frac{\pi}{2}]\)

f(0) = 2

\(f(\frac{\pi}{6})=\frac{\pi}{6}+\sqrt3\)

\(f(\frac{\pi}{2})=\frac{\pi}{2}\)

\(\Rightarrow\) least value of f(x) is \(\frac{\pi}{2}\,at\,x=\frac{\pi}{2}\)

\(\text{let}\,f(x_1)=f(x_2)\forall x_1,x_2\in R\)

\(\Rightarrow x^3_1=x_2^3\)

\(\Rightarrow x^3_1-x_2^3=0\)

\(\Rightarrow (x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0\)

\(\Rightarrow x_1=x_2\)  \((\because x_1^2+x_1x_2+x_2^2 \neq0)\)

\(\Rightarrow\) f is one - one

\(\text{let}\,f(x)=x^3=y\;\forall y\in R\)

\(\Rightarrow x=y^{\frac{1}{3}}\)

Therefore, every image \(y\in R\) has a unique pre image \(y^{\frac{1}{3}}\) in \(R\).

\(\Rightarrow f\) is onto

\(\therefore\) f is one-one and onto.

\(x=a\,sec\,\theta\Rightarrow\frac{dx}{d\theta}=a\,tan\,\theta\,sec\,\theta\)

\(y=b\,tan\,\theta\Rightarrow\frac{dy}{d\theta}=b\,sec^2\theta\)

\(\therefore\frac{dy}{dx}=\frac{b}{a}coses\theta\)

\(\Rightarrow\frac{d^2y}{dx^2}=\frac{-b}{a}coses\theta.cot\theta.\frac{d\theta}{dx}\) \(=\frac{-b}{a^2}cot^3\theta\)

\(\therefore\frac{d^2y}{dx^2}]_{\theta=\frac{\pi}{6}}=\frac{-3\sqrt3b}{a^2}\)

\(let\,u=sin^{-1}\,(2x\sqrt{1-x^2})\)

\(and\,v=sin^{-1}\,x,\frac{1}{\sqrt2}<x<1\) \(\Rightarrow sin\,v=x...(1)\)

Using (1), we get

\(u=sin^{-1}(2\,sin\,v\,cos\,v)\)

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

\(\Rightarrow u=2v\)

Differentiating with respect to v, we get: \(\frac{du}{dv}=2\)

\(AB=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix},\)

\(\Rightarrow AB=6I\Rightarrow B^{-1}=\frac{1}{6}A\)