Question

please answer the question..​

Answer 1

Question:- If y = x⁴e²ˣ and if y₁₀(0)= N × 2¹⁰ , where N is an odd integer. then the sum of digits of N is ?

Answer:

9

Step-by-step explanation:

Given,

$y=x^4e^{2x}$

Now,

Applying the Leibnitz's Theorem for the nth derivative of y.

We have,

$y_{10}=(D^{10}e^{2x})x^4+^{10}C_1(D^9e^{2x})Dx^4+^{10}C_2(D^8e^{2x})D^2x^4+^{10}C_3(D^7e^{2x})D^3x^4+C_4(D^6e^{2x})D^4x^4+C_5(D^5e^{2x})D^5x^4+.....................$

$y_{10}=(2^{10}e^{2x})x^4+^{10}C_1(2^{9}e^{2x})4x^3+^{10}C_2(2^8e^{2x})12x^2+^{10}C_3(2^7e^{2x})24x+^{10}C_4(2^6e^{2x})24+0+.....................$

Now putting x = 0 We get,

$y_{10}(0)=0+0+0+0+^{10}C_4(2^6e^0)24$

$y_{10}(0)=\frac{10!}{4!\times6!}\times2^6\times24\\\;\\y_{10}(0)=\frac{10\times9\times8\times7}{4\times3\times2\times1}\times2^6\times24\\\;\\y_{10}(0)=315\times2^{10}$

So thus comparing this expression by the expression given in question,

We get,

N = 315

Thus sum of its Digit = 3 + 1 + 5 = 9

Note:-

1) If there are two functions u and v be two functions of x then, by Leibnitz's Theorem,

$D^n(uv)=(D^nu)v+^nC_1(D^{n-1}u)Dv+^nC_2(D^{n-2}u)D^2v+............+^nC_r(D^{n-r}u)D^rv+........+uD^nv$

2) nth derivative of $e^{ax}$ is given by,

$y_n=a^ne^{ax}$