खालीलपैकी कोणते विधान योग्य आहे?

I. जर \(\mathrm{K}+\frac{1}{\mathrm{~K}}=12\), तर \(\mathrm{K}^2+\frac{1}{\mathrm{~K}^2}=142\)

II. \(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right) \) चे मूल्य 

\(\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}\) आहे.

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SSC CGL 2022 Tier-I Official Paper (Held On : 07 Dec 2022 Shift 4)
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  1. फक्त I 
  2. I किंवा II कोणतेही नाही.
  3. I आणि II दोन्हीही
  4. फक्त II 

Answer (Detailed Solution Below)

Option 1 : फक्त I 
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Detailed Solution

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दिलेले:

k + 1/k = 12

उपयोजित सूत्र:

(x + 1/x)2 = (x2 + 1/x2) + 2

(a - b)(a + b) = a2 - b2

गणना:

पहिले विधान विचारात घेऊ,

k + 1/k = 12

चा वापर करून,

(x + 1/x)2 = (x2 + 1/x2) + 2

(k2 + 1/k2) = 144 - 2

(k2 + 1/k2) = =142

म्हणून, पहिले विधान योग्य आहे.

आता दुसरे विधान विचारात घेऊ,​​​​

\(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right) \)

(a - b)(a + b) = a2 - bचा वापर करु,

\(\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k^2}-\frac{1}{\mathrm{k^2}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)\)

\(\left(\mathrm{k^4}-\frac{1}{\mathrm{k^4}}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right) \)

\(\mathrm{k}^{8}-\frac{1}{\mathrm{k}^{8}}\)

म्हणून, दुसरे विधान अयोग्य आहे.

म्हणून, फक्त विधान I योग्य आहे.

∴ पर्याय 1 हे योग्य उत्तर आहे.

Mistake Points

कृपया नोंद घ्या,

(K4)2  = K4 * 2 = K8

म्हणून, विधान 2 योग्य नाहे.

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