if $y = m\log x + n{x^2} + x$ has its extreme values at $x = 2$ and $x = 1$, then $2m + 10n$ is equal to$  {\text{a}}{\text{. }} - 1 \\  {\text{b}}{\text{. }} - 4 \\  {\text{c}}{\text{. }} - 2 \\  {\text{d}}{\text{. 1}} \\  {\text{e}}{\text{. }} - 3 \\  $
hint- here, we will be proceeding by differentiating the given function with respect to x and then putting $\dfrac{{dy}}{{dx}} = 0$ and afterwards substituting the given extreme values of x to obtain the two equation in terms of m and n only and then we will solve them.given, $y = m\log x + n{x^2} + x{\text{ }} \to {\text{(1)}}$ which is a function in terms of variable x.it is also given that the function represented in equation (1) has its extreme values at $x = 2$ and $x = 1$.as we know that extreme values of any function y are obtained by finding $\dfrac{{dy}}{{dx}}$ and then putting $\dfrac{{dy}}{{dx}} = 0$.by differentiating the function given by equation (1) both sides with respect to x, we get$\dfrac{{dy}}{{dx}} = \dfrac{{d\left[ {m\log x + n{x^2} + x} \right]}}{{dx}} = \dfrac{{d\left( {m\log x} \right)}}{{dx}} + \dfrac{{d\left( {n{x^2}} \right)}}{{dx}} + \dfrac{{dx}}{{dx}}$since, m and n are constants (i.e., independent of variable x) so we can take them out of the differentiation.\[   \rightarrow \dfrac{{dy}}{{dx}} = m\left[ {\dfrac{{d\left( {\log x} \right)}}{{dx}}} \right] + n\left[ {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right] + 1 = m\left[ {\dfrac{1}{x}} \right] + n\left[ {2x} \right] + 1 \\   \rightarrow \dfrac{{dy}}{{dx}} = \dfrac{m}{x} + 2nx + 1 \\  \]put $\dfrac{{dy}}{{dx}} = 0$ for the extreme values of the variable x, we get\[   \rightarrow 0 = \dfrac{m}{x} + 2nx + 1 \\   \rightarrow 0 = \dfrac{{m + 2n{x^2} + x}}{x} \\   \rightarrow 0 = m + 2n{x^2} + x \\   \rightarrow 2n{x^2} + x + m = 0{\text{ }} \to {\text{(2)}} \\  \]so, we are having a quadratic equation in variable x which means there will be two extreme values of x. these extreme values are $x = 2$ and $x = 1$. these two given extreme values should satisfy the quadratic equation given by equation (2).put x=2 in equation (2), we get\[   \rightarrow 2n{\left( 2 \right)^2} + 2 + m = 0 \\   \rightarrow 8n + 2 + m = 0{\text{ }} \to {\text{(3)}} \\  \]put x=1 in equation (2), we get\[   \rightarrow 2n{\left( 1 \right)^2} + 1 + m = 0 \\   \rightarrow 2n + 1 + m = 0{\text{ }} \to {\text{(4)}} \\  \]in order to obtain the values of constants m and n we will subtract equation (4) from equation (3), we have\[   \rightarrow 8n + 2 + m - \left( {2n + 1 + m} \right) = 0 - 0 \\   \rightarrow 8n + 2 + m - 2n - 1 - m = 0 \\   \rightarrow 6n + 1 = 0 \\   \rightarrow 6n = - 1 \\   \rightarrow n = \dfrac{{ - 1}}{6} \\  \]put \[n = \dfrac{{ - 1}}{6}\] in equation (4) to obtain the value of m, we have\[   \rightarrow 2\left[ {\dfrac{{ - 1}}{6}} \right] + 1 + m = 0 \\   \rightarrow \dfrac{{ - 1}}{3} + 1 + m = 0 \\   \rightarrow m = \dfrac{1}{3} - 1 = \dfrac{{1 - 3}}{3} \\   \rightarrow m = \dfrac{{ - 2}}{3} \\  \]put \[m = \dfrac{{ - 2}}{3}\] and \[n = \dfrac{{ - 1}}{6}\] to find the value for the expression $2m + 10n$, we get$  2m + 10n = 2\left[ {\dfrac{{ - 2}}{3}} \right] + 10\left[ {\dfrac{{ - 1}}{6}} \right] = \dfrac{{ - 4}}{3} - \dfrac{5}{3} = \dfrac{{ - 4 - 5}}{3} = \dfrac{{ - 9}}{3} \\   \rightarrow 2m + 10n = - 3 \\  $hence, option e is correct.note- at the extreme values for any function y, $\dfrac{{dy}}{{dx}} = 0$. also, if $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ that means at this extreme value, minima occurs (i.e., we will be getting the minimum value of y corresponding to this extreme value of x) and if $\dfrac{{{d^2}y}}{{d{x^2}}}
in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] we know that if a number ends with n zeros then its square will have 2n zeroes conversely, if square of a number have 2n zeros at their ends then the number will have n zeroesthe square of a number ends 10 zeroes, then the number will have `(10)/(2)` = 5 zeroes given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> the sides of the triangle are 6 cm, 8 cm and 10 cm. the longest side is 10 cm. (10)2 = 100 ….(i) now, the sum of the squares of the other two sides will be, (6)2 + (8)2 = 36 + 64 = 100 ….(ii) (10)2 = (6)2 + (8)2 ….. from (i) and (ii) by the converse of pythagoras theorem: the given sides form a right angled triangle. log 80 can be written as; log (8 ✕ 10) log (4 ✕ 2 ✕ 10) here we can use the formula log (a ✕ b) = log a + log b log 4 + log 2 + log 10  (1) log10 10 = 1 log 4 = 0.6020 log 2 = ½ log 4 = ½ ✕ 0.6020 = 0.3010 substituting the values in equation (1) we get the answer 1.9030 slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved