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The cost of an MBBS course at the Northern State Medical University (NSMU) is RUB 340,000 (INR 3.7 L) per year as per data curated from certain unofficial sources. NSMU has not officially released the fees for its MBBS courses.

However it has rolled out the fees for some of its other popular courses for the academic session 2025/26, the same is given below:

The course fees for three years BBA in Uttaranchal College of Science and Technology is INR 1.5 lakh. The course fee consists of multiple fee components, including registration fee, library, sports, and moreFee payment is the last step in the admission process. Candidates must pay the fee within the stipulated time period. 

NOTE: This information is sourced from the official website/ sanctioning body and is subject to change.

The highest packages offered to MMS graduates at TIMSR Mumbai for the academic batch of 2025 have not been released by the institution yet. However, as per the official placement brochure of TIMSR, the highest package offered during MMS placements 2024 stood at INR 15 LPA.

67. Given, f (x) = (ax² + sin x) (p +q cos x).

So, f? (x) = (ax² + sin x) $\frac{d}{dx}$ $(p+qcosx)+(p+qcosx)\frac{d}{dx}(a{x}^2+sinx)$

$=\left(a{x}^2+sinx\right)\left(0+q\frac{d}{dx}cosx\right)+(p+qcosx)\left(a\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}sinx\right)$

$=\left(a{x}^2+sinx\right)(q(-sinx)+(p+qcosx)(a·2x+cosx)$

= q sin x (ax² + sin x) + (p + q cos x) (2ax + cos x)

66. Given, f (x) = (x² + 1) cos x

f? (x) = (x² + 1) $\frac{d}{dx}cosx+cosx\frac{d}{dx}\left(x^2+1\right)$

$=\left(x^2+1\right)(-sinx)+cosx(2x+0)\left[?\frac{d}{dx}cosx=-sinx\right]$

= x² sin x sin x + 2x cos x.

No, you need to take an entrance test for admission  at Panjab University for BA (Hon).

65. Given, f (x) = x⁴. (5 sin x 3 cos x)

$f^{\prime }(x)=x^4\frac{d}{dx}(5sinx-3cosx)+(5sinx-3cosx)\frac{d{x}^4}{dx}.$

$=x^4\left[5\frac{d}{dx}sinx-3·\frac{dcosx}{dx}\right]+[5sinx-3cosx]·4x^3$

As $\frac{d}{dx}sinx=cosx$

and $\frac{d}{dx}cosx=-sinx$

Thus,

$f^{\prime }(x)=x^4[5cosx+3sinx]+[5sinx-3cosx]·4x^3$

$=x^3[5cosx+3xsinx+20sinx-12cosx]$

$=x^3[5xcosx+3xsinx+20sinx-12cosx].$

Uttaranchal College of Science and Technology Dehradun BBA admission is merit-based for various programmes. Selection is based on merit followed by a Personal Interview which is held at the university campus. Candidates must have Class 12 with 45% marks (SC/ST 40%) aggregate from a recognised board. 

64.  Given, f (x) = $\frac{sin(x+a)}{cosx}$

$f^{\prime }(x)=\frac{cosx\frac{d}{dx}sin(x+a)-sin(x+a)\frac{d}{dx}cosx}{co{s}^{2}x}$

Let g?(x) = sin (x + a)

So, g?(x) = $\underset{}{}$$\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

= cos (x + a)

And P(x) = cos x

So, P?(x) = $\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}$

Thus, f?(x) = $\frac{}{}$$\frac{cosx·cos(x+a)-sin(x+a)(-sinx)}{co{s}^{2}x}$

$=\frac{cosx·cos(x+a)+sin(x+a)sinx}{co{s}^{2}x}$

$=\frac{cos(x+a-x)}{co{s}^{2}x}$

$=\frac{cosa}{co{s}^{2}x}$

63. Given, f (x) = $\frac{a+bsinx}{c+dcosx}$

f?(x) = $\frac{(c+dcosx)\frac{d}{dx}(bsinx)-(a+bsinx)\frac{d}{dx}(c+dcosx)}{{(c+dcosx)}^2}$

$f^{\prime }(x)=\frac{(c+dcosx)·b·\frac{d}{dx}(sinx)-(a+bsinx)·d·\frac{d}{dx}(cosx)}{{(c+dcosx)}^2}\_\_\_\_\_(1)$

{Copy (A)}

So, g?(x) = $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}[cos(x+h)-cosx]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2·sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2sin\left(\frac{2x+h}{2}\right)sin\left(\frac{h}{2}\right)\right]$

$=-sin\left(\frac{2x+0}{2}\right)\times 1$

= sin x ______ (2)

And p?(x) = $\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}$

$=\underset{h\to 0}{lim}\frac{sin(x+h)-sinx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\cdot 2cos\left(\frac{2x+h}{2}\right)sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\frac{sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

= cos x _____ (3)

So, put (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(c+dcosx)(b·cosx)-(a+bsinx)(-d·sinx)}{{(c+dcosx)}^2}$

$=\frac{beccosx+bdco{s}^{2}x+adsinx+bdsi{n}^{2}x}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd\left(co{s}^{2}x+si{n}^{2}x\right)}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd}{{(c+dcosx)}^2}$

62. Given, f (x) =sinⁿx

By chain rule,

f? (x) = n (sin x)^n-1 $\frac{d}{dh}$ sin x

Let (gx) = sinx

So, g? (x) $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{sin(x+h)-sinx}{h}$

$=\underset{h\to 0}{lim}\frac{2}{h}cos\left(\frac{2a+h}{2}\right)sin\left(\frac{h}{2}\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\frac{sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

= $cos\frac{(2x+0)}{2}\times 1$

= cos x.

So, f? (x) = n (sin x)^n-1 cos x.

If you are in general category then it is possible to get SSN but don't be sure on this because this rank is on the edge of final cutoff but if you are from categories like BC then the cutoffs are more high than the general category one. If we talk about previous cutoffs then it seems possible to get admission with this cut off but if you are in other category then it's not possible

61. Given, f (x) = $\frac{secx-1}{secx+1}$

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

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

So, f?(x) = $\frac{(1+cosx)\frac{d}{dx}(1-cosx)-(1-cosx)\frac{d}{dx}(1+cosx)}{{(1+cosx)}^2}$

$=\frac{(1+cosx)(-1)\frac{d}{dx}(-cosx)-(1-cosx)\frac{d}{dx}(cosx)}{{(1+cosx)}^2}$

Let g(x) = cos x.

So, g?(x) $=\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{cos(x+h)-cosx}{h}$

$=\underset{h\to 0}{lim}-\frac{2}{h}sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)$

$=\underset{h\to 0}{lim}\frac{-2}{h}sin\left(\frac{2x+h}{2}\right)sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$=\underset{h\to 0}{lim}-sin\left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\frac{sin\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=-sin\left(\frac{2x+0}{2}\right)\times 1$

= -sin x.

So, f?(x) $\frac{(1+cosx)(-1)(-sinx)-(1-cosx)(-sinx)}{{(1+cosx)}^2}$

$=\frac{sinx+cosxsinx+sinx-sinxcosx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{\left(1+\frac{1}{secx}\right)}^2}=\frac{2sinx\times se{c}^{2}x}{{(secx+1)}^2}=\frac{2secx·sinx}{{(sinx+1)}^2·cosx}.$

$=\frac{2secx·tanx}{{(sinx+1)}^2}$

60. Given, f (x) = $\frac{sinx+cosx}{sinx-cosx}$

So, f?(x) = $\frac{sinx-cosx\frac{d}{dx}\left(sinx+cosx\right)-\left(sinx+cos2\right)\frac{d}{dx}\left(sinx-cosx\right)}{{\left(sinx-cosx\right)}^2}$

Let g(x) = cos x and p(x) = sin x.

{from so g’(x) A ) (upto equation 3)

Let g(x) = cos2 and p(x) = sin x.

So, g?(x) = $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}[cos(x+h)-cosx]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2·sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2sin\left(\frac{2x+h}{2}\right)sin\left(\frac{h}{2}\right)\right]$

$=-sin\left(\frac{2x+0}{2}\right)\times 1$

= -sin x ______ (2)

And p?(x) = $\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}$

$=\underset{h\to 0}{lim}\frac{sin(x+h)-sinx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\cdot 2cos\left(\frac{2x+h}{2}\right)sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\frac{sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

= cos x _____ (3)

Putting (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(sinx-cosx)[cosx-sinx]-(csinx+cosx)[cosx+sinx]}{{(sinx-cosx)}^2}$

$=\frac{-{(sinx-cosx)}^2-{(sinx+cosx)}^2}{{(sinx-cosx)}^2}$

$=\frac{-\left(si{n}^{2}x+co{s}^{2}x\right)+2sinxcosx-\left(si{n}^{2}x+co{s}^{2}x\right)-2sinxcosx}{{(sinx-cosx)}^2}$

$=\frac{-1-1}{{(sinx-cosx)}^2}=\frac{-2}{{(sinx-cosx)}^2}$