Let the sixth term in the binomial expansion of ( \sqrt 2^\log _2( 10 - 3^x ) + \sqrt[5]2^(x - 2)\log _23 )^m in the increasing powers of 2^(x-2) \log _2 3 , be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q42 with Solution

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Correct answer: 4

Simplify the two terms inside the binomial

Given $\left( \sqrt {2^{{\log _2}(10-3^x)}}+\sqrt[5]{2^{(x-2)\log_2 3}}\right)^m.$

Using $2^{\log_2 a}=a,$ we get $\sqrt{2^{\log_2(10-3^x)}}=\sqrt{10-3^x}.$

Also, $\sqrt[5]{2^{(x-2)\log_2 3}}=2^{\frac{(x-2)\log_2 3}{5}}=(2^{\log_2 3})^{\frac{x-2}{5}}=3^{\frac{x-2}{5}}.$

So the expansion is $\left(\sqrt{10-3^x}+3^{\frac{x-2}{5}}\right)^m.$

But the question says the terms are arranged in increasing powers of $2^{(x-2)\log_2 3}=3^{x-2}.$

Let $t=3^{\frac{x-2}{5}}.$ Then $t^5=3^{x-2}=2^{(x-2)\log_2 3}.$

Also, $3^x=9\cdot 3^{x-2}=9t^5.$ Hence $\sqrt{10-3^x}=\sqrt{10-9t^5}.$

Use the condition on binomial coefficients to find $m$

In the expansion of $(a+b)^m$ , the binomial coefficients of the 2nd, 3rd, and 4th terms are: $\binom m1,\binom m2,\binom m3.$

These are the first, third, and fifth terms of an A.P. Therefore, $\binom m2-\binom m1=\binom m3-\binom m2,$ so $2\binom m2=\binom m1+\binom m3.$

Now,

\quad \binom m2=\frac{m(m-1)}2, \quad \binom m3=\frac{m(m-1)(m-2)}6.$$ Thus, $$2\cdot \frac{m(m-1)}2=m+\frac{m(m-1)(m-2)}6.$$ So, $$m(m-1)=m+\frac{m(m-1)(m-2)}6.$$ Multiplying by 6, $$6m(m-1)=6m+m(m-1)(m-2).$$ For $m\neq 0$, divide by $m$: $$6(m-1)=6+(m-1)(m-2).$$ Expand: $$6m-6=6+m^2-3m+2.$$ $$m^2-9m+14=0.$$ $$(m-7)(m-2)=0.$$ So, $$m=2 \text{ or } 7.$$ But the **sixth term** exists only if $m\ge 5$, so $$m=7.$$ 3. **Write the sixth term** For $(a+b)^m$, the $(r+1)$-th term is $$T_{r+1}=\binom mr a^{m-r}b^r.$$ Here, $$a=\sqrt{10-3^x},\quad b=3^{\frac{x-2}{5}},\quad m=7.$$ The sixth term corresponds to $r=5$: $$T_6=\binom75\left(\sqrt{10-3^x}\right)^{2}\left(3^{\frac{x-2}{5}}\right)^5.$$ Now, $$\binom75=21, \quad \left(\sqrt{10-3^x}\right)^2=10-3^x, \quad \left(3^{\frac{x-2}{5}}\right)^5=3^{x-2}.$$ Therefore, $$T_6=21(10-3^x)3^{x-2}.$$ Given that the sixth term is $21$, $$21(10-3^x)3^{x-2}=21.$$ Hence, $$(10-3^x)3^{x-2}=1.$$ 4. **Solve for $x$** Let $$y=3^{x-2}.$$ Then $$3^x=9y.$$ So the equation becomes $$(10-9y)y=1.$$ $$10y-9y^2=1.$$ $$9y^2-10y+1=0.$$ Factorizing, $$(9y-1)(y-1)=0.$$ Thus, $$y=1 \quad \text{or} \quad y=\frac19.$$ That is, $$3^{x-2}=1 \Rightarrow x-2=0 \Rightarrow x=2,$$ or $$3^{x-2}=3^{-2} \Rightarrow x-2=-2 \Rightarrow x=0.$$ Both satisfy the domain condition $10-3^x>0$. 5. **Sum of squares of all possible values of $x$** Possible values are $$x=0,2.$$ So the required sum is $$0^2+2^2=4.$$ 6. **Comparison with stored answer** Derived answer = $4$. Stored correct answer = $4$. So they agree.

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View written solutionFree

Correct answer: 4

Simplify the two terms inside the binomial

Given $\left( \sqrt {2^{{\log _2}(10-3^x)}}+\sqrt[5]{2^{(x-2)\log_2 3}}\right)^m.$

Using $2^{\log_2 a}=a,$ we get $\sqrt{2^{\log_2(10-3^x)}}=\sqrt{10-3^x}.$

Also, $\sqrt[5]{2^{(x-2)\log_2 3}}=2^{\frac{(x-2)\log_2 3}{5}}=(2^{\log_2 3})^{\frac{x-2}{5}}=3^{\frac{x-2}{5}}.$

So the expansion is $\left(\sqrt{10-3^x}+3^{\frac{x-2}{5}}\right)^m.$

But the question says the terms are arranged in increasing powers of $2^{(x-2)\log_2 3}=3^{x-2}.$

Let $t=3^{\frac{x-2}{5}}.$ Then $t^5=3^{x-2}=2^{(x-2)\log_2 3}.$

Also, $3^x=9\cdot 3^{x-2}=9t^5.$ Hence $\sqrt{10-3^x}=\sqrt{10-9t^5}.$

Use the condition on binomial coefficients to find $m$

In the expansion of $(a+b)^m$ , the binomial coefficients of the 2nd, 3rd, and 4th terms are: $\binom m1,\binom m2,\binom m3.$

These are the first, third, and fifth terms of an A.P. Therefore, $\binom m2-\binom m1=\binom m3-\binom m2,$ so $2\binom m2=\binom m1+\binom m3.$

Now,

\quad \binom m2=\frac{m(m-1)}2, \quad \binom m3=\frac{m(m-1)(m-2)}6.$$ Thus, $$2\cdot \frac{m(m-1)}2=m+\frac{m(m-1)(m-2)}6.$$ So, $$m(m-1)=m+\frac{m(m-1)(m-2)}6.$$ Multiplying by 6, $$6m(m-1)=6m+m(m-1)(m-2).$$ For $m\neq 0$, divide by $m$: $$6(m-1)=6+(m-1)(m-2).$$ Expand: $$6m-6=6+m^2-3m+2.$$ $$m^2-9m+14=0.$$ $$(m-7)(m-2)=0.$$ So, $$m=2 \text{ or } 7.$$ But the **sixth term** exists only if $m\ge 5$, so $$m=7.$$ 3. **Write the sixth term** For $(a+b)^m$, the $(r+1)$-th term is $$T_{r+1}=\binom mr a^{m-r}b^r.$$ Here, $$a=\sqrt{10-3^x},\quad b=3^{\frac{x-2}{5}},\quad m=7.$$ The sixth term corresponds to $r=5$: $$T_6=\binom75\left(\sqrt{10-3^x}\right)^{2}\left(3^{\frac{x-2}{5}}\right)^5.$$ Now, $$\binom75=21, \quad \left(\sqrt{10-3^x}\right)^2=10-3^x, \quad \left(3^{\frac{x-2}{5}}\right)^5=3^{x-2}.$$ Therefore, $$T_6=21(10-3^x)3^{x-2}.$$ Given that the sixth term is $21$, $$21(10-3^x)3^{x-2}=21.$$ Hence, $$(10-3^x)3^{x-2}=1.$$ 4. **Solve for $x$** Let $$y=3^{x-2}.$$ Then $$3^x=9y.$$ So the equation becomes $$(10-9y)y=1.$$ $$10y-9y^2=1.$$ $$9y^2-10y+1=0.$$ Factorizing, $$(9y-1)(y-1)=0.$$ Thus, $$y=1 \quad \text{or} \quad y=\frac19.$$ That is, $$3^{x-2}=1 \Rightarrow x-2=0 \Rightarrow x=2,$$ or $$3^{x-2}=3^{-2} \Rightarrow x-2=-2 \Rightarrow x=0.$$ Both satisfy the domain condition $10-3^x>0$. 5. **Sum of squares of all possible values of $x$** Possible values are $$x=0,2.$$ So the required sum is $$0^2+2^2=4.$$ 6. **Comparison with stored answer** Derived answer = $4$. Stored correct answer = $4$. So they agree.

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