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

P(n)= 1.3 + 2.3² + 3.3³ + … + n.3ⁿ = $\frac{(2n-1)3^{n+1}+3}{4}$

If n=1, we get

P(1) = 1.3=3= $\frac{(2.1-1)3^{1+1}+3}{4}$ = $\frac{1·3^2+3}{4}$ = $\frac{12}{4}$ =3

which is true.

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

1.3 + 2.3² + 3.3³ + … + k^3k = $\frac{(2k-1)3^{k+1}+3}{4}$ ------------------(1)

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

Here,

1.3 + 2.3² + 3.3³ + … + k^3k + (k + 1)3^{k + 1}

By using eqn. (1)

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

L.C.M

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

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

= $\frac{3^{k+1}\{2k-1+4k+4\}+3}{4}$

= $\frac{3^{k+1}\{6k+3\}+3}{4}$

= $\frac{3^{k+1}·3(2k+1)+3}{4}$ = $\frac{3^{k+1+1}\{2k+1\}+3}{4}$ ?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., n.

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NOTE: The mentioned fee is taken from the official website/ sanctioning body. It is still subject to change and, hence, is indicative.

85. Let y = (x2- 5x + 8) (x3 + 7x + 9) ____ (1)

(i) by product rule

$\frac{dy}{dx}=\left(x^2-5x+8\right)\frac{d}{dx}\left(x^3+7x+9\right)+\left(x^3+7x+9\right)\frac{d}{dx}\left(x^2-5x+8\right)$

$=\left(x^2-5x+8\right)\left(3x^2+7\right)+\left(x^3+7x+9\right)(2x-5).$

3x4 + 7x2- 15x3- 35x + 24x2 + 56 + 2x4- 5x3 + 14x2- 35x 18x-45

= 5x4- 20x3 + 45x2- 52x + 11

(ii) $y=\left(x^2-5x+8\right)\left(x^3+7x+9\right)$

$y=x^5+7x^2+9x-5x^4-35x^2-45x+8x^3+56x+72$

$y=x^5-5x^4+15x^3-26x^2+11x+72.$

$\therefore \frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11.$

Taking log in eqn (1)

$logy=log\left(x^2-5x+8\right)+log\left(x^3+7x+9\right)$

Now, Differe(iii) ntiating w r t ‘x’ we get,

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2-5x+8}\frac{d}{dx}\left(x^2-5x+8\right)+\frac{1}{x^3+7x+9}\frac{d}{dx}\left(x^3+7x+9\right).$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{(2x-5)\left(x^3+7x+9\right)+\left(3x^2+7\right)\left(x^2-5x+8\right)}{\left(x^2-5x+8\right)\left(x^3+7x+9\right).}\right]$

$=\frac{y}{y}$ 2x4 + 14x4 + 18x- 35x- 45 + 3x1- 15x3 + 24x2 + 7x2- 35x + 56]

$\left\{?e{q}^n(1)\right\}.$

$\frac{dy}{dx}$ = 5x4- 20x3 + 45x2- 52x + 11

We observed that all the methods give the same result.

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The highest order present in the D.E. is $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

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Combining all, the total fee to pursue this programme ranges from INR 6,800 to INR 3.05 lakh. Interested students can refer to the below table to know the breakdown of the total fees:

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

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The highest order derivative present in the D.E. is $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivation, its degree is 2.

84. Given, f(x) = (1 + x)(1 + x 4)(1 + x 8)

Taking log,

logf(x) = log (1 + x) + log (1 + x) + log (1 + x 4) + log (1 + x 8)

Now, Differentiating w r t ‘x’ we get,

$\frac{1}{f(x)}{f}^{\prime }(x)=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}\frac{d}{dx}\left(1+x^2\right)+\frac{1}{1+x^4}\frac{d}{dx}\left(1+x^4\right)+\frac{1}{1+x^8}\frac{d\left(1+x^8\right)}{dx}$

$\Rightarrow f^{\prime }(x)=f(x)\left\{\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right\}.$

$\Rightarrow f^{\prime }(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left\{\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right\}$

Putting x = 1

f'(x) = (1 +1)(1 + 14)(1 +18) $\left\{\frac{1}{1+1}+\frac{2\times 1}{1+1^2}+\frac{4\times 1^3}{1+1^4}+\frac{8\times 1^7}{1+1^8}\right\}$

$=2\times 2\times 2+2\left\{\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right\}$

$=16\times \left\{\frac{15}{2}\right\}=8\times 15=120.$

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

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

If n=1, we get

P(1): 1.2.3 = 6 = $\frac{1(1+1)(1+2)(1+3)}{4}$ = $\frac{1.2.3.4}{4}=6$

which is true.

considerP(k) is true for some positive integer k

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

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

Here,1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

By eqn (1), we get,

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

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

= $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$

By further Simplification,

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

? 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 numbers n.

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

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$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.