Maths

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

P (n)=1.2+2.3+3.4+ … +2 (n+1)= $\left[\frac{n(n+1)(n+2)}{3}\right]$

For n=1,

P (1)=1.2=2= $\frac{1(1+1)(1+2)}{3}$ = $\frac{2*3}{3}$ =2.

Which is true.

considerP (k) be true for some positive integer k

1.2 + 2.3 + 3.4 + … + k (k + 1) = $\left[\frac{k(k+1)(k+2)}{3}\right]$ - (1)

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

Here, 1.2 + 2.3 + 3.4 + … + k (k + 1) + (k+1) (k+2)

By using (1), we get

= $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

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

= $\frac{(k+1)(k+2)(k+3)}{3}$

By further simplification; $\frac{(k+1)(k+1+1)(k+1+2)}{3}$

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 no. i.e., n.

In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

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.

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.

The given order derivative present in the D.E. is $y^{|}$ so its order is 1.

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

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.

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) $\left\{\frac{1}{1+1}+\frac{2*1}{1+1^2}+\frac{4*1^3}{1+1^4}+\frac{8*1^7}{1+1^8}\right\}$

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

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

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.