AIEEE 2006 | Binomial Theorem Question 196 | Mathematics

For natural numbers $$m$$ , $$n$$, if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,\,n} \right)$$ is

$$\left( {20,\,45} \right)$$

$$\left( {35,\,20} \right)$$

$$\left( {45,\,35} \right)$$

$$\left( {35,\,45} \right)$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

Out of Syllabus

The value of $$\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$$ is

$${}^{55}{C_4}$$

$${}^{55}{C_3}$$

$${}^{56}{C_3}$$

$${}^{56}{C_4}$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation

$$a - b = 1$$

$$a + b = 1$$

$${a \over b} = 1$$

$$ab = 1$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as

$$1 - {3 \over 8}{x^2}$$

$$3x + {3 \over 8}{x^2}$$

$$ - {3 \over 8}{x^2}$$

$${x \over 2} - {3 \over 8}{x^2}$$

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