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Functions: Range

JEE Maths question with a full step-by-step solution.

Question

Range of

f(x)={}^{16-x}C_{2x-1}+{}^{20-3x}C_{4x-5}

is:

\{728,1474\}

\{728,1617\}

correct

\{0,728\}

DNone of these

Solution

Step 1: For

{}^{n}C_{r}

to be defined:

n\ge 0

r\ge 0

and

n\ge r

, with

n,r\in\mathbb{Z}

. For

{}^{16-x}C_{2x-1}

: (i)

16-x\ge 0\;\Rightarrow\;x\le 16

(ii)

2x-1\ge 0\;\Rightarrow\;x\ge\dfrac{1}{2}

(iii)

16-x\ge 2x-1\;\Rightarrow\;17\ge 3x\;\Rightarrow\;x\le\dfrac{17}{3}

(iv)

16-x

is a non-negative integer, so

x

must be an integer. Step 2: For

{}^{20-3x}C_{4x-5}

: (v)

20-3x\ge 0\;\Rightarrow\;x\le\dfrac{20}{3}

(vi)

4x-5\ge 0\;\Rightarrow\;x\ge\dfrac{5}{4}

(vii)

20-3x\ge 4x-5\;\Rightarrow\;25\ge 7x\;\Rightarrow\;x\le\dfrac{25}{7}

Step 3:

x

must be an integer. Combining all conditions:

\dfrac{5}{4}\le x\le\dfrac{25}{7}

and

x

is an integer.

\dfrac{25}{7}\approx 3.57

, so

x\in\{2,3\}

. Step 4: Compute:

x=2

{}^{14}C_{3}+{}^{14}C_{3}=364+364=728

x=3

{}^{13}C_{5}+{}^{11}C_{7}=1287+330=1617

. Range

=\{728,1617\}

. Answer: (2)

\{728,1617\}

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