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115.

Solution :
The given function is $f\left(x\right)=x^2-4x-3$ f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

$\begin{array}{l}f\left(1\right)=1^2-4\times 1-3,f\left(4\right)=4^2-4\times 4-3\\ \therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{-3-\left(-6\right)}{3}=\frac{3}{3}=1\end{array}$

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f' (c) = 1

$\begin{array}{l}f\text{'}\left(c\right)=1\\ \Rightarrow 2c-4=1\\ \Rightarrow c=\frac{5}{2},where,c=\frac{5}{2}\in \left(1,4\right)\end{array}$

Hence, Mean Value Theorem is verified for the given function.

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Integrating both sides,

114. Solution :
It is given that f: [-5,5]? R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [?5, 5].

(b) f is differentiable on (?5, 5).

Therefore, by the Mean Value Theorem, there exists c? (?5, 5) such that

$\begin{array}{l}{f}^{\text{'}}\left(c\right)\frac{f\left(5\right)?f\left(?5\right)}{5?\left(?5\right)}\\ ?10f^{\text{'}}\left(c\right)=f\left(5\right)?f\left(?5\right)\end{array}$

It is also given that f' (x) does not vanish anywhere.

$\begin{array}{l}?f^{\text{'}}\left(c\right)?0\\ ?10f^{\text{'}}\left(c\right)?0\\ ?f\left(5\right)?f\left(?5\right)?0\\ ?f\left(5\right)?f\left(?5\right)\end{array}$

Hence, proved.

The given D.E. is

$\begin{array}{l}{x}^2\frac{dy}{dx}=x^2-2y^2++xy\\ \Rightarrow \frac{dy}{dx}=\frac{x^2-2y^2++xy}{x^2}=1-\frac{2y^2}{x^2}+\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=1-2{\left(\frac{y}{x}\right)}^2+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the D.E. is homogenous $f{x}^n$

Let, $y=vx.=\frac{y}{x}=v$ so that, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Thus, $v+x\frac{dv}{dx}=1-2v^2+v$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=1-2v^2\\ \Rightarrow \frac{dv}{1-2v^2}=\frac{dx}{x}\end{array}$

Integrating both sides,

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The Given D.E. is

$\begin{array}{l}\left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow 2\times ydy=-\left(x^2-y^2\right)dx\\ =\frac{dy}{dx}=-\frac{\left(x^2-y^2\right)}{2xy}=\frac{\left(y^2-x^2\right)}{2xy}\\ =\frac{1}{2}\frac{\left(\frac{y^2-x^2}{x^2}\right)}{\left(\frac{xy}{x^2}\right)}\\ =\frac{1}{2}\frac{\left[{\left(\frac{y}{x}\right)}^2-1\right]}{\left(\frac{y}{x}\right)}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx\Rightarrow \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

$\begin{array}{l}\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}\left[\frac{v^2-1}{v}\right]\\ \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=\frac{-1-v^2}{2v}\\ \Rightarrow \left(\frac{2v}{1+v^2}\right)dv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

Putting back $v=\frac{y}{x}$ we get,

$\begin{array}{l}\Rightarrow x\left[\frac{y^2}{x^2}+1\right]=c\\ \Rightarrow \frac{y^2+x^2}{x}=c\end{array}$

$\Rightarrow y^2+x^2=xc$ is the required solution.

2.

By binomial theorem,

(a + b)ⁿ = ⁿC₀ (a)ⁿ (b)⁰ + ⁿC₁ (a)^n–1 (b)¹ + …………. + ⁿC_r (a)^n–r (b)^r + …………… + ⁿC_n (a)^n–n (b)ⁿ

Where, b⁰ = 1 = a^n–n

So, (a + b)ⁿ = ⁿC_r (a)^n–r (b)^r

Putting a = 1 and b = 3 such that a + b = 4, we can rewrite the above equation as

(1 + 3)ⁿ = ⁿC_r (1)^n–r.3^r

=>4ⁿ = $\underset{r=0}{{\displaystyle ?}}.3r$ .ⁿC_r

Hence proved.

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The Given D.E. is $\left(x-y\right)dy-\left(x+y\right)dx=0$

$\begin{array}{l}\Rightarrow \left(x-y\right)dy=\left(x+y\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}=\frac{x\left(1+\frac{y}{x}\right)}{x\left(1-\frac{y}{x}\right)}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx=\frac{y}{x}=v,so,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then, $\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-1+v^2}{1-v}=\frac{1+v^2}{1-v}\\ \Rightarrow \left[\frac{1-v}{1+v^2}\right]dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1-v}{1+v^2}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1}{1+v^2}dv-\frac{1}{2}{\displaystyle \int \frac{2v}{1+v^2}dv=logx+c}}\\ \Rightarrow ta{n}^{-1}v-\frac{1}{2}log\left(1+v^2\right)=logx+c\end{array}$

Putting back $v=\frac{y}{x},weget,$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|1+\frac{y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|\frac{x^2+y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}\left[log\left(x^2+y^2\right)-log{x}^2\right]=logx+c\end{array}$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+\frac{1}{2}log{x}^2=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+log{\left(x^2\right)}^{\frac{1}{2}}=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+logx=log+c\\ =ta{n}^{-1}\frac{y}{x}=\frac{1}{2}log\left(x^2+y^2\right)+c\end{array}$

The fee range for the BA programme at Parle Tilak Vidyalaya Association's Sathaye College ranges between INR 2,400 to 30,000. This information is sourced from official website/ sanctioning body and is subject to change. The fee cover various academic expenses like tuition fees, examination fees, library fees, and other relevant charges. It is important to note that the fee structure is subject to change, and candidates are advised to refer to the official website of the university for the latest and most accurate fee structure.