TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If   $f(x)= \begin{cases}\sin x & if x\leq0\\x^{2}+a^{2}, &if 0<x<1\\bx+2, & if 1\leq x \leq 2\\0&,ifx>2\end{cases}$ is continuous

On IR, then a+b+ab=

Answer:

Explanation:

Given,

$f(x)= \begin{cases}\sin x & if x\leq0\\x^{2}+a^{2}, &if 0<x<1\\bx+2, & if 1\leq x \leq 2\\0&,ifx>2\end{cases}$

is continuous on IR,

 $\therefore$   $\lim_{x \rightarrow {0^{-}}}f(x)=\lim_{x \rightarrow {0^{+}}}f(x) $ and

$\lim_{x \rightarrow {2^{-}}} f(x)=\lim_{x \rightarrow {2^{+}}}f(x)$  

 $\Rightarrow$   $\lim_{x \rightarrow {0^{-}}}\sin x=\lim_{x \rightarrow {0^{-}}}x^{2}+a^{2}  $

 and $\lim_{x \rightarrow {2^{-}}}bx+2=\lim_{x \rightarrow {2^{+}}}0  $

 $\Rightarrow$     $0=0+a^{2}$ and $2b+2=0$

$\Rightarrow$   a=0 and b=-1

Now, $a+b+ab=0+(-1)+0=-1$

In the expansion of $(1+x)^{n}$ , the coefficients of p th and (p+1) th terms are respectively p and q then p+q= 

Answer:

Explanation:

In the expansion of $(1+x)^{n}$ , the general term i.e, (r+1) th term is

  $T_{r+1}=^{n}C_{r}  (1)^{n-r} x^{r}=^{n}C_{r} x^{r}$

$\therefore$ Coefficient of (r+1) th term is $^{n}C_{r}$

Similarly , coefficent of pth term =$^{n}C_{p-1}$

$\therefore$           $p=^{n}C_{p-1}$(given)

            $p=\frac{n!}{(p-1)!(n-p+1)!}$   ........(i)

and coefficient of (p+1) th term= $^{n}C_{p}$

$\therefore$    $q=^{n}C_{p}$(given)     .........(ii)

On dividing Eq.(i) by Eq.(ii), we get

$\frac{p}{q}=\frac{\frac{n!}{(p-1)!(n-p+1)!}}{^{n}C_{p}}=\frac{\frac{n!}{(p-1)!(n-p+1)!}}{\frac{n!}{(p!)(n-p)!}}$

$=\frac{(p!)(n-p)!}{(n-p+1)!(p-1)!}=\frac{p(p-1)!(n-p)!}{(n-p+1)(n-p)!(p-1)!}$

$\Rightarrow$    $\frac{p}{q}=\frac{p}{n-p+1}$

$\Rightarrow$   $\frac{1}{q}=\frac{1}{n-p+1}$

$\Rightarrow$  $n-p+1=q$

$\Rightarrow$   $p+q=n+1$

The local maximum of $y=x^{3}-3x^{2}+5$ is attained at 

Answer:

Explanation:

Given , $y=x^{3}-3x^{2}+5$  .........(i)

 On differentiating  both sides w.r.t 'x' , we get

 $\frac{dy}{dx}=3x^{2}-6x$  ........(ii)

 For local maxima or local minima , put $\frac{dy}{dx}=0$

$\Rightarrow$   $3x^{2}-6x=0$

$\Rightarrow$   $3x(x-2)=0$

$\Rightarrow$      x=0 or x=2

Now, differentiating Eq.(ii)  w.r.t .'x' we get

    $\frac{d^{2}y}{dx^{2}}=6x-6$

$\Rightarrow$        $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0}=-6<0$

$\therefore$   x=0  is a point of local maxima

and   $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=2}=6\times2-6=12-6=6>0$

 $\therefore$    x=2 is a point of local minima

Let $a=2 \hat{i}+\hat{j}-3\hat{k}$ and $b= \hat{i}+3\hat{j}+2\hat{k}$ . Then the volume of the parallelopiped having coterminous edges as a,b and c, where c is the vector perpendicular to the plane of a, b and |c|=2 is 

Answer:

Explanation:

We have, 

$a=2 \hat{i}+\hat{j}-3\hat{k}$

and $b= \hat{i}+3\hat{j}+2\hat{k}$ 

Clearly , c is  parallel to $a \times b$ 

Here,  $a\times b=\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\2 & 1&-3\\1&3&2 \end{bmatrix}=\hat{i}(11)-\hat{j}(7)+\hat{k}(5)$

        $11 \hat{i}-7 \hat{j}+5\hat{k}=d$    (say)

Now, $\hat{d}=\frac{a\times b}{|a \times b|}=\frac{11\hat{i}-7\hat{j}+5\hat{k}}{\sqrt{195}}$

Thus,   $c=|c|\hat{d}=\frac{2}{\sqrt{195}}(11 \hat{i}-7\hat{j}+5\hat{k})$

Hence, volume of the parallelopiped= [a b c]

= $\begin{bmatrix}2 & 1&-3 \\1 & 3&2\\\frac{22}{\sqrt{195}}&\frac{-14}{\sqrt{195}}&\frac{10}{\sqrt{195}} \end{bmatrix}$

    =$\frac{2}{\sqrt{195}}\begin{bmatrix}2 & 1&-3 \\1 & 3&2\\11&-7&5 \end{bmatrix}$

   = $\frac{2}{\sqrt{195}}[2(29)-1(-17)-3(-40)]$

    =$\frac{2}{\sqrt{195}}[58+17+120]$

=  $\frac{2}{\sqrt{195}}\times195$

    =$2\sqrt{195}$

If $\tan \theta_{1}=k \cot \theta_{2}$ , then $\frac{\cos (\theta_{1}+\theta_{2})}{\cos (\theta_{1}-\theta_{2}) }=$

Answer:

Explanation:

 We have $\tan \theta_{1}=k \cot \theta_{2}$

$\Rightarrow$  $\tan \theta_{1} \tan \theta_{2}$=k

Consider, $\frac{\cos (\theta_{1}+\theta_{2})}{\cos (\theta_{1}-\theta_{2})}=\frac{\cos  \theta_{1}\cos \theta_{2}-\sin  \theta_{1}\sin \theta_{2}}{\cos \theta_{1}\cos\theta_{2}+\sin \theta_{1}\sin \theta_{2}}$

$=\frac{1-\tan\theta_{1} \tan \theta_{2}}{1+\tan\theta_{1} \tan \theta_{2}}$

$=\frac{1-k}{1+k}$

• Mathematics - General questions