UPSEE Solved Model Paper 6

We can solve this problem using the definition of Binomial theorem for a rational index. Binomial Theorem for a rational Index If n is a nonzero rational number and –1 < x < 1 then $(1+x)^n$ = $1+\frac{n}{1!}x + \frac{n(n-1)}{2!}x^2$ + $\frac{n(n-1)(n-2)}{3!} x^3$ + ... + $\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} x^r$ + ... to ∞ In the above expansion, the first term must be unity. In the expansion of $(a+x)^n$ , where n is either a negative integer or a fraction, we proceed as follows $(a+x)^n$ = $[a(1+x/a)]^n$ = $a^n(1+x/a)^n$ = $a^n\left[1+\frac{n x}{a} + \frac{n(n-1)}{2!} \left(\frac{x}{a}\right)^2+\cdots\right]$ By using this definition we can find the power of x is valid We have $\frac{1}{\sqrt{6-3x}}$ = $\frac{1}{\sqrt{6}\sqrt{1-\frac{3x}{6}}}$ = $\frac{1}{\sqrt{6}\sqrt{1-\frac{x}{2}}}$ The expansion in powers of x is valid if $\left|\frac{x}{2}\right|$ < 1 i.e., - 2 < x < 2 (value inside the square root should be positive) Therefore The expansion of $\frac{1}{\sqrt{6-3x}}$ in powers of x is valid if - 2 < x < 2