If the coefficient of x^7 in ( ax - 1 \over bx^2 )^13 and the coefficient of x^ - 5 in ( ax + 1 \over bx^2 )^13 are equal, then a^4b^4 is equal to :

Binomial Theorem PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q29 with Solution

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Correct answer: A

For $\left(ax-\frac{1}{bx^2}\right)^{13},$ the general term is $T_{r+1}=\binom{13}{r}(ax)^{13-r}\left(-\frac{1}{bx^2}\right)^r.$

Simplifying,

=\binom{13}{r}a^{13-r}(-1)^r b^{-r} x^{13-3r}.$$ So the power of $x$ is $13-3r$. --- 2. **Find the coefficient of $x^7$ in** $$\left(ax-\frac{1}{bx^2}\right)^{13}.$$ We need $$13-3r=7.$$ So, $$3r=6\Rightarrow r=2.$$ Thus the required coefficient is $$\binom{13}{2}a^{11}(-1)^2 b^{-2}=\binom{13}{2}\frac{a^{11}}{b^2}.$$ Since $$\binom{13}{2}=78,$$ the coefficient is $$78\frac{a^{11}}{b^2}.$$ --- 3. **Write the general term for the second expansion** For $$\left(ax+\frac{1}{bx^2}\right)^{13},$$ the general term is $$T_{r+1}=\binom{13}{r}(ax)^{13-r}\left(\frac{1}{bx^2}\right)^r.$$ Simplifying, $$T_{r+1}=\binom{13}{r}a^{13-r}b^{-r}x^{13-r-2r} =\binom{13}{r}a^{13-r}b^{-r}x^{13-3r}.$$ --- 4. **Find the coefficient of $x^{-5}$ in** $$\left(ax+\frac{1}{bx^2}\right)^{13}.$$ We need $$13-3r=-5.$$ So, $$3r=18\Rightarrow r=6.$$ Thus the required coefficient is $$\binom{13}{6}a^{7}b^{-6}=\binom{13}{6}\frac{a^7}{b^6}.$$ Now, $$\binom{13}{6}=1716.$$ So the coefficient is $$1716\frac{a^7}{b^6}.$$ --- 5. **Use the condition that these coefficients are equal** Given, $$78\frac{a^{11}}{b^2}=1716\frac{a^7}{b^6}.$$ Multiply both sides by $b^6$: $$78a^{11}b^4=1716a^7.$$ Divide by $78a^7$: $$a^4b^4=\frac{1716}{78}=22.$$ --- 6. **Match with the options** $$a^4b^4=22.$$ So the correct option is **A**.

View written solutionFree

Correct answer: A

For $\left(ax-\frac{1}{bx^2}\right)^{13},$ the general term is $T_{r+1}=\binom{13}{r}(ax)^{13-r}\left(-\frac{1}{bx^2}\right)^r.$

Simplifying,

=\binom{13}{r}a^{13-r}(-1)^r b^{-r} x^{13-3r}.$$ So the power of $x$ is $13-3r$. --- 2. **Find the coefficient of $x^7$ in** $$\left(ax-\frac{1}{bx^2}\right)^{13}.$$ We need $$13-3r=7.$$ So, $$3r=6\Rightarrow r=2.$$ Thus the required coefficient is $$\binom{13}{2}a^{11}(-1)^2 b^{-2}=\binom{13}{2}\frac{a^{11}}{b^2}.$$ Since $$\binom{13}{2}=78,$$ the coefficient is $$78\frac{a^{11}}{b^2}.$$ --- 3. **Write the general term for the second expansion** For $$\left(ax+\frac{1}{bx^2}\right)^{13},$$ the general term is $$T_{r+1}=\binom{13}{r}(ax)^{13-r}\left(\frac{1}{bx^2}\right)^r.$$ Simplifying, $$T_{r+1}=\binom{13}{r}a^{13-r}b^{-r}x^{13-r-2r} =\binom{13}{r}a^{13-r}b^{-r}x^{13-3r}.$$ --- 4. **Find the coefficient of $x^{-5}$ in** $$\left(ax+\frac{1}{bx^2}\right)^{13}.$$ We need $$13-3r=-5.$$ So, $$3r=18\Rightarrow r=6.$$ Thus the required coefficient is $$\binom{13}{6}a^{7}b^{-6}=\binom{13}{6}\frac{a^7}{b^6}.$$ Now, $$\binom{13}{6}=1716.$$ So the coefficient is $$1716\frac{a^7}{b^6}.$$ --- 5. **Use the condition that these coefficients are equal** Given, $$78\frac{a^{11}}{b^2}=1716\frac{a^7}{b^6}.$$ Multiply both sides by $b^6$: $$78a^{11}b^4=1716a^7.$$ Divide by $78a^7$: $$a^4b^4=\frac{1716}{78}=22.$$ --- 6. **Match with the options** $$a^4b^4=22.$$ So the correct option is **A**.