If the term independent of \(x\) in the expansion of \[ \left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10} \] is 105, then \(a^2\) is equal to:
- 4
- 9
- 6
- 2
The Correct Option is A
Approach Solution - 1
Consider the given expression:
\(\left(\sqrt{a}x^{2} + \frac{1}{2x^{3}}\right)^{10}\)
Step 1: General term
The general term of the expansion is given by:
\[
T_{r+1} = {}^{10}C_{r} (\sqrt{a}x^{2})^{10 - r} \left(\frac{1}{2x^{3}}\right)^{r}
\]
Step 2: Determine the power of \(x\)
Power of \(x\) in the general term is:
\[
2(10 - r) - 3r = 20 - 2r - 3r = 20 - 5r
\]
For the term independent of \(x\), we set the power of \(x\) to 0:
\[
20 - 5r = 0 \Rightarrow r = 4
\]
Step 3: Substitute \(r = 4\) in the general term
\[
T_{5} = {}^{10}C_{4} (\sqrt{a})^{6} \left(\frac{1}{2}\right)^{4}
\]
Simplifying further:
\[
{}^{10}C_{4} \cdot a^{3} \cdot \frac{1}{16} = 105
\]
Step 4: Solving for \(a\)
\[
105 = 210 \cdot \frac{a^{3}}{16} \Rightarrow a^{3} = 8
\]
Taking cube root on both sides:
\[
a^{2} = 4
\]
Approach Solution -2
Consider the given expression:
\[\left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10}\]
The general term in the binomial expansion of \((x + y)^n\) is given by:
\[T_{r+1} = \binom{n}{r} x^{n-r} y^r.\]
For the given expression, the general term becomes:
\[T_{r+1} = \binom{10}{r} \left( \sqrt{ax^2} \right)^{10-r} \left( \frac{1}{2x^3} \right)^r.\]
Simplify the powers of \(x\):
\[T_{r+1} = \binom{10}{r} \left( \sqrt{a} \right)^{10-r} x^{(20-2r)} \times \frac{1}{(2x^3)^r}\]
Combine the powers of \(x\):
\[T_{r+1} = \binom{10}{r} \left( \sqrt{a} \right)^{10-r} \times \frac{1}{2^r} \times x^{20-5r}\]
To find the term independent of \(x\), set the power of \(x\) to zero:
\[20 - 5r = 0\]
Solve for \(r\):
\[r = 4\]
Substitute \(r = 4\) into the general term:
\[T_5 = \binom{10}{4} \left( \sqrt{a} \right)^{10-4} \times \frac{1}{2^4}\]
Simplify:
\[T_5 = \binom{10}{4} \left( \sqrt{a} \right)^6 \times \frac{1}{16}\]
Substitute \(\binom{10}{4} = 210\):
\[T_5 = 210 \times \left( \sqrt{a} \right)^6 \times \frac{1}{16}\]
\[T_5 = 210 \times \frac{a^3}{16}\]
The value of \(T_5\) is given as 105:
\[210 \times \frac{a^3}{16} = 105\]
Solve for \(a^3\):
\[a^3 = \frac{105 \times 16}{210}\]
\[a^3 = 8\]
Take the cube root of both sides:
\[a = \sqrt[3]{8}\]
\[a = 2\]
Finally, \(a^2 = 4\).
Answer: \((1) \ 4\)
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Binomial theorem Questions
- The number of integral terms in the expansion of (31/2 + 51/4)680 is equal to:
- Suppose \( 2 - p \), \( p \), \( 2 - \alpha \), \( \alpha \) are the coefficients of four consecutive terms in the expansion of \( (1 + x)^n \). Then the value of \( p^2 - \alpha^2 + 6\alpha + 2p \) equals
- Let \( \alpha = \sum_{k=0}^{n} \left( \frac{\binom{n}{k}}{k+1} \right)^2 \) and \( \beta = \sum_{k=0}^{n-1} \left( \frac{\binom{n}{k} \binom{n}{k+1}}{k+2} \right) \).
If \( 5\alpha = 6\beta \), then \( n \) equals __________. - \(^{n-1}C_r = (k^2 - 8) ^{n}C_{r+1}\) if and only if:
- If \( A \) denotes the sum of all the coefficients in the expansion of \( (1 - 3x + 10x^2)^n \) and \( B \) denotes the sum of all the coefficients in the expansion of \( (1 + x^2)^n \), then:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.