If L and M are respectively the coefficient of x-7 in ax+bx211 and the coefficient of x7 in bx2+ax11 then L+M=

(B) 1acoefficient of x6 in ax2+bx12

(A) 1bcoefficient of x-6 in ax+bx212

(B) 1acoefficient of x6 in ax2+bx12

(C) acoefficient of x-10 in ax+bx211

(D) bcoefficient of x4 in ax2+bx11

TS EAMCET · Maths · Binomial Theorem

If $L$ and $M$ are respectively the coefficient of $x^{- 7}$ in ${(a x + \frac{b}{x^{2}})}^{11}$ and the coefficient of $x^{7}$ in ${(b x^{2} + \frac{a}{x})}^{11}$ then $L + M =$

A $\frac{1}{b} [coefficient of x^{- 6} in {(a x + \frac{b}{x^{2}})}^{12}]$
B $\frac{1}{a} [coefficient of x^{6} in {(a x^{2} + \frac{b}{x})}^{12}]$
C $a [coefficient of x^{- 10} in {(a x + \frac{b}{x^{2}})}^{11}]$
D $b [coefficient of x^{4} in {(a x^{2} + \frac{b}{x})}^{11}]$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{1}{a} [coefficient of x^{6} in {(a x^{2} + \frac{b}{x})}^{12}]$

Step-by-step Solution

Detailed explanation

General term of a+bx2 is Tr+1=Cr11ax11-r×bx2r ⇒Tr+1=Cr11 a11-r·brx11-3r For coefficient of x-7, 11-3r=-7 ⇒ 3r=18 ⇒ r=6 So, T6+1=C611 a11-6 b6 x-7 Tr+1=C611 a5b6x-7 So, L=C611a5b6 Similarly, general term…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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