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NIMCET 2007 Mathematics PYQ — Find the term in the expansion of :… | Mathem Solvex

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NIMCET 2007 Mathematics PYQ — Find the term in the expansion of :… | Mathem Solvex | Mathem Solvex

Explanation

Formula and Concepts

According to the binomial theorem for any rational index $n$ , the expansion of $(1 - y)^{n}$ is given by:

$(1 - y)^{n} = 1 - n y + \frac{n ( n - 1 )}{2 !} y^{2} - \frac{n ( n - 1 ) ( n - 2 )}{3 !} y^{3} + \dots$

The $(r + 1)^{th}$ general term in the binomial expansion of $(1 - y)^{n}$ is:

$T_{r + 1} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} (- y)^{r}$

Alternatively, comparing directly with $(1 + X)^{n}$ where $X = - 2 x^{2}$ :

$T_{r + 1} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} X^{r}$

Step 1: Identify Key Variables

From the given expression $(1 - 2 x^{2})^{1/2}$ :

$X = - 2 x^{2}$
Index ( $n$ ) = $\frac{1}{2}$

We need to find the $6^{th}$ term ( $T_{6}$ ). This means:

$r + 1 = 6 ⟹ r = 5$

Step 2: Substitute Values into the Formula

Substitute $n = \frac{1}{2}$ , $r = 5$ , and $X = - 2 x^{2}$ into the general term equation:

$T_{6} = \frac{\frac{1}{2} ( \frac{1}{2} - 1 ) ( \frac{1}{2} - 2 ) ( \frac{1}{2} - 3 ) ( \frac{1}{2} - 4 )}{5 !} (- 2 x^{2})^{5}$

Step 3: Simplify the Numerator Terms

Let's simplify the bracket components individually:

$(\frac{1}{2} - 1) = - \frac{1}{2}$
$(\frac{1}{2} - 2) = - \frac{3}{2}$
$(\frac{1}{2} - 3) = - \frac{5}{2}$
$(\frac{1}{2} - 4) = - \frac{7}{2}$

Now combine them inside the numerator expression:

$Numerator = (\frac{1}{2}) (- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2}) (- \frac{7}{2}) = \frac{1 \times ( - 1 ) \times ( - 3 ) \times ( - 5 ) \times ( - 7 )}{2 ^{5}}$

Since there are 4 negative signs, the product turns positive:

$Numerator = \frac{105}{32}$

Step 4: Expand the Remaining Terms

Now compute the factorial and power elements:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120</li><li>$ (-2x^2)^5 = (-2)^5 \cdot (x^2)^5 = -32x^{10} $</li></ul><h3>Step 5: Put It All Together</h3>Substitute the evaluated parts back into the expression for $ T_6 $:T_6 = \frac{\frac{105}{32}}{120} \times (-32x^{10})Cancel out the common $ 32 $ from the numerator and denominator:T_6 = \frac{105}{120} \times (-1) \times x^{10}</p><p>T_6 = -\frac{105}{120} x^{10}Reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor ($ 15 $):T_6 = -\frac{105 \div 15}{120 \div 15} x^{10} = -\frac{7}{8} x^{10}$ $

Explanation

Formula and Concepts

According to the binomial theorem for any rational index $n$ , the expansion of $(1 - y)^{n}$ is given by:

$(1 - y)^{n} = 1 - n y + \frac{n ( n - 1 )}{2 !} y^{2} - \frac{n ( n - 1 ) ( n - 2 )}{3 !} y^{3} + \dots$

The $(r + 1)^{th}$ general term in the binomial expansion of $(1 - y)^{n}$ is:

$T_{r + 1} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} (- y)^{r}$

Alternatively, comparing directly with $(1 + X)^{n}$ where $X = - 2 x^{2}$ :

$T_{r + 1} = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !} X^{r}$

Step 1: Identify Key Variables

From the given expression $(1 - 2 x^{2})^{1/2}$ :

$X = - 2 x^{2}$
Index ( $n$ ) = $\frac{1}{2}$

We need to find the $6^{th}$ term ( $T_{6}$ ). This means:

$r + 1 = 6 ⟹ r = 5$

Step 2: Substitute Values into the Formula

Substitute $n = \frac{1}{2}$ , $r = 5$ , and $X = - 2 x^{2}$ into the general term equation:

$T_{6} = \frac{\frac{1}{2} ( \frac{1}{2} - 1 ) ( \frac{1}{2} - 2 ) ( \frac{1}{2} - 3 ) ( \frac{1}{2} - 4 )}{5 !} (- 2 x^{2})^{5}$

Step 3: Simplify the Numerator Terms

Let's simplify the bracket components individually:

$(\frac{1}{2} - 1) = - \frac{1}{2}$
$(\frac{1}{2} - 2) = - \frac{3}{2}$
$(\frac{1}{2} - 3) = - \frac{5}{2}$
$(\frac{1}{2} - 4) = - \frac{7}{2}$

Now combine them inside the numerator expression:

$Numerator = (\frac{1}{2}) (- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2}) (- \frac{7}{2}) = \frac{1 \times ( - 1 ) \times ( - 3 ) \times ( - 5 ) \times ( - 7 )}{2 ^{5}}$

Since there are 4 negative signs, the product turns positive:

$Numerator = \frac{105}{32}$

Step 4: Expand the Remaining Terms

Now compute the factorial and power elements:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120</li><li>$ (-2x^2)^5 = (-2)^5 \cdot (x^2)^5 = -32x^{10} $</li></ul><h3>Step 5: Put It All Together</h3>Substitute the evaluated parts back into the expression for $ T_6 $:T_6 = \frac{\frac{105}{32}}{120} \times (-32x^{10})Cancel out the common $ 32 $ from the numerator and denominator:T_6 = \frac{105}{120} \times (-1) \times x^{10}</p><p>T_6 = -\frac{105}{120} x^{10}Reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor ($ 15 $):T_6 = -\frac{105 \div 15}{120 \div 15} x^{10} = -\frac{7}{8} x^{10}$ $

NIMCET 2007 — Mathematics PYQ

Choose the correct answer:

Explanation

Formula and Concepts

Step 1: Identify Key Variables

Step 2: Substitute Values into the Formula

Step 3: Simplify the Numerator Terms

Step 4: Expand the Remaining Terms

Explanation

Formula and Concepts

Step 1: Identify Key Variables

Step 2: Substitute Values into the Formula

Step 3: Simplify the Numerator Terms

Step 4: Expand the Remaining Terms

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