Ask & Answer: India's Largest Education Community

Parle Tilak Vidyalaya Association's Sathaye College offers various courses at the UG, PG & Diploma level. The selection criteria are merit-based. The application process at PTVASC is conducted both online/offline. To secure a seat at Parle Tilak Vidyalaya Association's Sathaye College, students can check the following steps presented below:

1. Visit the official website and complete the online application process. 
2. Candidates can also apply offline for various courses.
3. Ensure eligibility and selection criteria for all programme.
4. The selection criteria is merit-based.
5. Upon selected, pay the specified fee to confirm and secure your seat.

91. Given, $x=a\left(cost+logtan\frac{t}{2}\right)y=asint$

Differentiating w r t we get,

$\frac{dx}{dt}=a\frac{d}{dt}\left[cost+log\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{dt}\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.se{c}^2\frac{t}{2}\frac{d}{dt}\left(\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{1}{co{s}^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{1}{2}\right]$

$=a\left[-sint+\frac{1}{2sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sin2\times \raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sint}\right]=a\left[\frac{1-si{n}^{2}t}{sint}\right]$

$=a\frac{co{s}^{2}t}{sint}\left\{?1=co{s}^{2}x+si{n}^{2}x\right\}$

$\frac{\mathrm{bd}y}{dt}=\frac{d}{dt}\left(asint\right)=acost$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{acost}{\raisebox{1ex}{$aco{s}^{2}t$}\!\left/ \!\raisebox{-1ex}{$sint$}\right.}=\frac{sint}{cost}=tant$

12. Let the given statement be P(n) i.e.,

P(n)=a+ar+ar²+ … +ar^n-1== $\frac{a\left(r^n-1\right)}{r-1}$

If n = 1, we get

P(1)=a= $\frac{a\left(r^1-1\right)}{r-1}$ =a

which is true.

Consider P(k) be true for some positive integer k

a+ar+ar²+ … +ar^k-1= $\frac{a\left(r^k-1\right)}{r-1}$ (1)

Now, let us prove that P(k+1) is true.

Here, {a+ar+ar²+ … +ar^k-1}+ar^{(k+1) –1}

By using (1),

= $\frac{a\left(r^k-1\right)}{r-1}+a{r}^k$

= $\frac{a\left(r^k-1\right)+a{r}^k(r-1)}{r-1}$

= $\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}$

= $\frac{a{r}^{k+1}-a}{r-1}$

= $\frac{a\left(r^{k+1}-1\right)}{r-1}$

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,

90. Given, x = 4t and y = $\frac{4}{t}$ Differentiating w r t. ‘t’ we get,

$\frac{dx}{dt}=4\frac{dy}{dt}=\frac{4d\left(t^{-1}\right)}{dt}$

$=-4t^{-2}$

$\frac{-4}{t^2}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$t^2$}\right.}{4}=-\frac{1}{t^2}.$

Given,  $x+y=ta{n}^{-1}y$

Differentiate with ‘x’ we get

$\begin{array}{l}1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}\\ =1+y^{|}=\frac{1}{1+y^2}{y}^{|}\\ =\left(1+y^{|}\right)\left(1+y^2\right)=y^{|}\\ =1+y^2{y}^{|}+y^{|}+y^2=y^{|}\\ =y^2{y}^{|}+y^2+1=0\end{array}$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E

11. we can write the given statement as

$P(n)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$ = $\frac{n(n+3)}{4(n+1)(n+2)}$

If n=1,

P(1)= $\frac{1}{1.2.3}$ = $\frac{1}{6}$ = $\frac{1(1+3)}{4(1+1)(1+2)}$ = $\frac{4}{4\times 2\times 3}$ = $\frac{1}{6}$

which is true.

Consider P(k) be true for some positive integer k

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}$ = $\frac{k(k+3)}{4(k+1)\left(k+2\right)}$

Let us prove that P(k+1) is true,

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ .

By equation (1), we get

= $\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

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

= $\frac{1}{(k+1)(k+2)}\left[\frac{k{(k+3)}^2+4}{4(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left[\frac{k\left(k^2+6k+9\right)+4}{4(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k\left(k^2+2k+1\right)+4\left(k^2+2k+1\right)}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k{(k+1)}^2+4{(k+1)}^2}{4(k+3)}\right\}$

= $\frac{{(k+1)}^2(k+4)}{4(k+1)(k+2)(k+3)}$

= $\frac{{(k+1)}^2\{(k+1)+3\}}{4(k+1)(k+2)(k+3)}$ = $\frac{(k+1)\{(k+1)+3\}}{4\{(k+1)+1\}\{(k+1)+2\}}$

P(k+1) is true whenever P(k) is true.

Hence, By the principle of mathematical induction, the P(n) is true for all natural number n.

Given $y-cosy=x$

Differentiate w.r.t ‘x’ we get

$\begin{array}{l}\frac{dy}{dx}-\left(-siny\right)\frac{dy}{dx}=\frac{dx}{dx}\\ \Rightarrow \frac{dy}{dx}\left[1+siny\right]=1\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+siny}=y^{|}\end{array}$

So, L.H.S of given D.E $=\left(ysiny+cosy+x\right)y^{|}$

$\begin{array}{l}=\left(ysiny+cosy+y-cosx\right)\left[\frac{1}{1+siny}\right]\\ =\frac{y\left(1+siny\right)}{\left(1+siny\right)}=y=R.H.S\end{array}$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given, $xy=logy+c$

Differentiate w.r.t. x we have

$\begin{array}{l}x\frac{dy}{dx}+y\frac{dx}{dx}=\frac{d}{dx}logy+\frac{d}{dx}C\\ \Rightarrow x\frac{dy}{dx}+y=\frac{1}{y}\frac{dy}{dx}+0\\ \Rightarrow x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}\left[x-\frac{1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}\left[\frac{xy-1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{xy-1}=\frac{\left(-1\right)\times -y^2}{\left(-1\right)\times \left(xy-1\right)}=\frac{y^2}{1-xy}\\ \therefore y^{|}=\frac{y^2}{1-xy}\end{array}$

Hence, y is a Solution of the given D.E

89. Given, x = sin t and y = cos2t. differentiation w r t. ‘t’ we get,

$\begin{array}{l}\frac{dx}{dt}=cost\frac{dy}{dt}=\left(-sin2t\right)\frac{d2t}{dt}\\ -t\\ -2\left(2sintcost\right)\\ -tt\end{array}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}$

$=\frac{-4sintcost}{cost}$

= -4 sin t

Parle Tilak Vidyalaya Association's Sathaye College admissions are open for various courses. Candidates are selected for courses based on merit. Aspirants can apply online on the official website of the institute. Interested candidates meeting the eligibility criteria can fill out the form.

Given,  $y=xsinx$

So,  $y^{|}=x\frac{d}{dx}sinx+sinx\frac{dx}{dx}=xcosx+sinx$

Now, L.H.S of the given D.E $=x{y}^{|}$

$=x\left(xcosx+sinx\right)$

$=x^{2}cosx+xsinx$

There are nearly 544 MBA colleges in Delhi/ NCR. Delhi Pharmaceutical Sciences and Research University is one of the university in Delhi/ NCR that provides MBA in three specialisations. The total tuition fee for the PG-level course is INR 1.6 lakh. Mentioned below are a few colleges/ universities that offers an MBA in Delhi/ NCR:

• IIM Rohtak
• GNIOT Institute of Management Studies
• Indian Institute of Foreign Trade
• Management Development Institute, etc.

10. Let the given statement be P(n) i.e.,

$P(n)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+?+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

For n=1,

P(1)= $\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}$

which is true.

Assume that P(k) is true for some positive integer k.

i.e.,P(k)= $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+?+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)}$ (1)

Now, let us prove P(k+1) is true,

Here, $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + … + $\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)[3(k+1)+2]}$

By using eqn.(1),

= $\frac{k}{6k+4}+\frac{1}{(3k+3-1)(3k+3+2)}$

= $\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}$

Taking 2 as common,

= $\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}$

= $\frac{1}{(3k+2)}\left\{\frac{k}{2}+\frac{1}{(3k+5)}\right\}$

= $\frac{1}{(3k+2)}\left\{\frac{k(3k+5)+2}{2(3k+5)}\right\}$

= $\frac{1}{(3k+2)}\left\{\frac{3k^2+5k+2}{\begin{array}{l}\bcancel{6k}+\bcancel{10}\\ 2(3k+5)\end{array}}\right\}$

= $\frac{1}{(3k+2)}\left\{\frac{3k^2+3k+2k+2}{2(3k+5)}\right\}$ = $\frac{1}{(3k+2)}\left\{\frac{3k(k+1)+2(k+1)}{2(3k+5)}\right\}$

= $\frac{1}{(3k+2)}\left\{\frac{(3k+2)(k+1)}{2(3k+5)}\right\}$

= $\frac{(k+1)}{6k+10}$ , so we get

$\frac{(k+1)}{6(k+1)+4}$

P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural number.

The salary after MBA in Quality Management can vary based on various factors such as experience, location, industry and job roles. Freshers may start with salaries ranging from INR 4 LPA to INR 7 LPA. However, professionals with certifications like Six Sigma Green/Black Belt, PMP or ISO Auditor and experience in quality roles can earn significantly higher.
Average Salary by Roles is mentioned below:

Note: This information is sourced from external website and may vary.

88. Kindly go through the solution