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STD 11 - 4.1 complex nubers — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 4.1 complex nubers4 Marks
MCQ
If $|{a_k}| < 1,{\lambda _k} \ge 0$ for $k = 1,\,2,....n$ and ${\lambda _1} + {\lambda _2} + ... + {\lambda _n} = 1,$ then the value of $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ....{\lambda _n}{a_n}|$ is
  • A
    Equal to one
  • B
    Greater than one
  • C
    Zero
  • ✓
    Less than one
✓

Answer

Correct option: D.
Less than one
d
(d) We have $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ..... + {\lambda _n}{a_n}|$

$ \le |{\lambda _1}{a_1}| + |{\lambda _2}{a_2}| + ..... + |{\lambda _n}{a_n}|$

$ = |{\lambda _1}||{a_1}| + ..... + |{\lambda _n}||{a_n}|$

$ = {\lambda _1}|{a_1}| + ..... + {\lambda _n}|{a_n}|\,$[ each ${\lambda _k}_. \ge 0$]

$ < {\lambda _1} + ..... + {\lambda _n}$

[ $|{a_k}| < $ $1$ and so ${\lambda _k}|{a_k}| < {\lambda _k}$for all $k = 1,2,....n$]

Hence $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ..... + {\lambda _n}{a_n}| < 1$.

Thus $|{\lambda _1}{a_1} + ..... + {\lambda _n}{a_n}| < 1$.

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