Ai - x - ai-x - ai- i - 1- 2- - - n- and a1 - a2 - a3 - - - an-If 1 - m - n- m - N- then the value of R- limx - am- A1 A2 - An is

The correct option is D ( 1)^n m lim_x→ a_mA_i =lim_x→ a_mx a_i|x a_i| =1 a_m≥ a_i = 1 a_m lt; a_i ........... (1) lim_x→ a_m^+ ...

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Byju's Answer

Standard XII

Physics

Finding Significant Figures

Ai = x - ai/|...

Question

$A_{i} = \frac{x - a_{i}}{| x - a_{i} |}, i = 1, 2, . . . . ., n,$ and $a_{1} < a_{2} < a_{3} < . . . . . < a_{n} .$

If $1 \leq m \leq n, m \in N,$ then the value of $R = lim x \to a_{m}^{+} (A_{1} A_{2} . . . . A_{n})$ is

always 1

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always -1

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(- 1)^{m + 1}

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(- 1)^{n - m}

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Solution

The correct option is D $(- 1)^{n - m}$
${lim}_{x \to a_{m}} A_{i}$
$= {lim}_{x \to a_{m}} \frac{x - a_{i}}{| x - a_{i} |} = 1 a_{m} \geq a_{i}$
$= - 1 a_{m} < a_{i}$ ........... $(1)$

${lim}_{x \to a_{m^{+}}} (A_{1} A_{2} . . . A_{n})$

$= {lim}_{x \to a_{m^{+}}} (A_{1} A_{2} . . . A_{m}) \times {lim}_{x \to a_{m^{+}}} (A_{m + 1} A_{m + 2} . . . A_{n})$

$= (1^{m}) \times ((- 1)^{n - m})$ ......by $e q^{n} 1$

$= (- 1)^{n - m}$

So, the answer is option $(D)$ .

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Ai=x−ai|x−ai|,i=1,2,.....,n, and a1<a2<a3<.....<an.If 1≤m≤n,m∈N, then the value of R=limx→a+m(A1A2....An) is

The correct option is D (−1)n−mlimx→amAi=limx→amx−ai|x−ai|=1am≥ai=−1am<ai...........(1)limx→am+(A1A2...An)=limx→am+(A1A2...Am)×limx→am+(Am+1Am+2...An)=(1m)×((−1)n−m)......by eqn1=(−1)n−mSo, the answer is option (D).

$A_{i} = \frac{x - a_{i}}{| x - a_{i} |}, i = 1, 2, . . . . ., n,$ and $a_{1} < a_{2} < a_{3} < . . . . . < a_{n} .$

If $1 \leq m \leq n, m \in N,$ then the value of $R = lim x \to a_{m}^{+} (A_{1} A_{2} . . . . A_{n})$ is