TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

The rank of the matrix

$\begin{bmatrix}3 & 5&-1&4 \\2 & 1&3&-2\\8&11&1&6\\-7&-14&6&-14 \end{bmatrix}$  is 

Answer:

Explanation:

We have,

$\begin{bmatrix}3 & -5&-1&4 \\2 & 1&3&-2\\8&11&1&6\\-7&-14&6&-14 \end{bmatrix}$

   [$R_{1}\rightarrow R_{1}-R_{2}$]

$\begin{bmatrix}1 & 4&-4&6 \\2 & 1&3&-2\\8&11&1&6\\-7&-14&6&-14 \end{bmatrix}$

  $[R_{2}\rightarrow R_{2}-2R_{1}, R_{3}\rightarrow R_{3}-8R_{1},R_{4}\rightarrow R_{4}+7R_{1}]$

 $\begin{bmatrix}1 & 4&-4&6 \\0 & -7&11&-14\\0&-21&33&-42\\0&14&-22&28 \end{bmatrix}$

$[R_{3}\rightarrow R_{3}-3R_{2}]$

$\begin{bmatrix}1 & 4&-4&6 \\0 & -7&11&-14\\0&0&0&0\\0&14&-22&28 \end{bmatrix}$

$[R_{3}\leftrightarrow R_{4}]$

$\begin{bmatrix}1 & 4&-4&6 \\0 & -7&11&-14\\0&14&-22&28\\0&0&0&0 \end{bmatrix}$

$[R_{3}\rightarrow R_{3}+2R_{2}]$

$\begin{bmatrix}1 & 4&-4&6 \\0 & -7&11&-14\\0&0&0&0\\0&0&0&0 \end{bmatrix}$

Number of non-zero rows =2

 $\therefore$ Rank of matrices=2

if the function $f:[a,b]\rightarrow \left[-\frac{\sqrt{3}}{4},\frac{1}{2}\right]$ defined by

$f(x)=\begin{bmatrix}1 & 1&1 \\1 & 1+\sin_{}x&1\\1+\cos x&1&1 \end{bmatrix}$

 is one-one and onto , then

Answer:

Explanation:

We have,  $f:[a,b]\rightarrow \left[-\frac{\sqrt{3}}{4},\frac{1}{2}\right]$ 

$f(x)=\begin{bmatrix}1 & 1&1 \\1 & 1+\sin_{}x&1\\1+\cos x&1&1 \end{bmatrix}$

$f(x)=\begin{bmatrix}1 & 1&0 \\0 & \sin_{}x&0\\1+\cos x&-\cos x&-\cos x \end{bmatrix}$

    $[C_{3}\rightarrow C_{3}-C_{1},C_{2}\rightarrow C_{2}-C_{1}]$

 f(x) =-sin x cos x

$f(x)=-\frac{1}{2} \sin 2x$

  $f(x)\in \left[\frac{\sqrt{3}}{4},\frac{1}{2}\right]$

$\therefore$    $-\frac{\sqrt{3}}{4}\leq \frac{-1}{2} \sin 2x \leq \frac{1}{2}$

$\Rightarrow$      $-1\leq  \sin 2x \leq \frac{\sqrt{3}}{2}$

 $\Rightarrow$    $-\frac{\pi}{4}\leq  x \leq \frac{\pi}{6}$

$\therefore$       $a=\frac{-\pi}{4},b=\frac{\pi}{6}$

If  $\alpha\in R, n\in N $  and n+2(n-1)+3(n-2)+....(n-1)2+n.1= $\alpha$ n(n+1)(n+2), then $\alpha$=  

Answer:

Explanation:

We have,

 $T_{r}=r[n-(r-1)]$

= r(n-r+1)

= $nr-r^{2}+r$

 $\therefore$     $\sum_{r=1}^{n}T_{r}=\sum_{r=1}^{n} (nr-r^{2}+r)$

$= \sum_{r=1}^{n}[(n+1)r-r^{2}]=(n+1)\sum_{r=1}^{n}r-\sum_{r=1}^{n}r^{2}$

  $= \frac{(n+1)n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$

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

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

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

$= \frac{n(n+1)(n+2)}{6}$

 $\therefore$     $\alpha$  = $\frac{1}{6}$

Let  $f:R\rightarrow R $ and $g:R\rightarrow R$  be the functions defined by  $f(x)= \frac{x}{1+x^{2}}$,

$x \in R, g(x)=\frac{x^{2}}{1+x^{2}},x\in R$  Then, the correct statement (s) among the following is/are

(a) both f.g are one-one

(b)  both f.g are onto

(c)  both f.g are not one-one  as well as onto

(d) f and g are onto but not one-one

Answer:

Explanation:

 We have,

 $f(x)= \frac{x}{1+x^{2}}$  $x \in R$

$g(x)=\frac{x^{2}}{1+x^{2}},x\in R$

$f'(x)= \frac{1+x^{2}-2x^{2}}{(1+x^{2})^{2}}$

$f'(x)= \frac{1-x^{2}}{(1+x^{2})^{2}}$

 Clearly  f'(x)  is not monotonic

$\therefore$    f(x)  is not one-one function range of f(x) , is    $\left[ -\frac{1}{2},\frac{1}{2}\right]$

 $\therefore$   f(x) is not onto

 Clearly g(x) is even function

 $\therefore$   g(x)  is not one-one function 

Range of g(x) is [0,1]

 $\therefore$   g(x) is also not onto . Here , f(x) and g(x) both are neither one-one not onto.

  The domain of the function 

$f(x)= \frac{1}{\sqrt{[x]^{2}-[x]^{}-2}}$  is 

 Here [x]  denotes the greatest integer not exceeding the value of [x]

Answer:

Explanation:

We have, 

$f(x)= \frac{1}{\sqrt{[x]^{2}-[x]^{}-2}}$ 

 f(x0 is defined when 

$[x]^{2}-[x]-2 >0$

$([x]-2)([x]+1)>0$

 [x] >2 and [x] <-1

 $x\geq 3 $and x<-1

 $\therefore$   Domain of f(x)  is  x $\in $$(-\infty,-1)\cup (3, \infty)$

• Mathematics - General questions