⇒ 1 - 1/( 1 + √2 ) + 1/( 1 - √2 ) = [ 1 - ( √2 )² - ( 1 - √2 ) + ( 1 + √2 ) ]/[ 1 - ( √2 )² ]
= [ 1 - 2 - 1 + √2 + 1 + 1 + √2 ]/( 1-2 ) = [ -1 + 2√2 ]/-1 = [ 1 - 2√2 ]

सही विकल्प: B

⇒ 1 - 1/( 1 + √2 ) + 1/( 1 - √2 ) = [ 1 - ( √2 )² - ( 1 - √2 ) + ( 1 + √2 ) ]/[ 1 - ( √2 )² ]
= [ 1 - 2 - 1 + √2 + 1 + 1 + √2 ]/( 1-2 ) = [ -1 + 2√2 ]/-1 = [ 1 - 2√2 ]
∴ 1 - 1/( 1 + √2 ) + 1/( 1 - √2 ) = [ 1 - 2√2 ]

( 2 , 3 , 6 ) का ल.स. = 6
√2 = ( 2 )^1/2 = ( 2 )^{1/2 × 1/6} = ( 2⁶)^1/12 = ( 64 )^1/12
∛3 = ( 3 )^1/3 = ( 1/3 )^{1/3 × 1/4} = ( 3⁴ )^1/12 = ( 81 )^1/12
तथा ( 6 )^1/6 = ( 6² )^1/12 = ( 36 )^1/12

सही विकल्प: D

( 2 , 3 , 6 ) का ल.स. = 6
√2 = ( 2 )^1/2 = ( 2 )^{1/2 × 1/6} = ( 2⁶)^1/12 = ( 64 )^1/12
∛3 = ( 3 )^1/3 = ( 1/3 )^{1/3 × 1/4} = ( 3⁴ )^1/12 = ( 81 )^1/12
तथा ( 6 )^1/6 = ( 6² )^1/12 = ( 36 )^1/12
अब , ( 81 )^1/12 > ( 64)^1/12 (36 )^1/12
⇒ ∛3 > √2 > ( 6 )^1/6
अतः इससे स्पष्ट है कि दोनों कथन गलत है |

विकल्प ( a )लेने पर ,
√5 + √3 > √6 + √2

सही विकल्प: A

विकल्प ( a )लेने पर ,
√5 + √3 > √6 + √2
दोनों ओर वर्ग करने पर
⇒ 5 + 3 + 2 × √5 × √3 > 6 + 2 + 2 × √6 × √2
⇒ 8 + 2√15 > 8 + 2√12
⇒ √15 > √12
अतः स्पष्ट है कि √15 , √12 से बड़ा है |अतः विकल्प ( a ) सत्य होगा ।

[ 2^{n + 4} - 2 × 2ⁿ ]/2 × 2^{n + 3} + 2^-3 = x
⇒ x = [ 2ⁿ ×2⁴ - 2 × 2ⁿ ]/2 × 2ⁿ × 2³ + 1/2³ [ ∴ a^{b + c} = a^b × a^c तथा a^-x = 1/a^x ]

सही विकल्प: B

[ 2^{n + 4} - 2 × 2ⁿ ]/2 × 2^{n + 3} + 2^-3 = x
⇒ x = [ 2ⁿ ×2⁴ - 2 × 2ⁿ ]/2 × 2ⁿ × 2³ + 1/2³ [ ∴ a^{b + c} = a^b × a^c तथा a^-x = 1/a^x ]
⇒ x = [ 2ⁿ ( 2⁴ - 2 ) ]/2 × 2ⁿ × 2³ + 1/2³
⇒ x = ( 16 - 2 )/2 × 8 + 1/8 = 7/8 + 1/8 = 1
∴ x = 1
अतः x का मान 1होगा ।

[ ( 243 )^n/5 × 3^{2n + 1} ]/[ 9ⁿ × 3^{n - 1} ]
= [ ( 3⁵ )^n/5 × 3^{2n + 1} ]/[ 3²ⁿ × 3^{n - 1} ]

सही विकल्प: B

[ ( 243 )^n/5 × 3^{2n + 1} ]/[ 9ⁿ × 3^{n - 1} ]
= [ ( 3⁵ )^n/5 × 3^{2n + 1} ]/[ 3²ⁿ × 3^{n - 1} ]
= [ ( 3 )ⁿ × 3^{2n + 1} ]/[ 3²ⁿ × 3^{n - 1} ]
= ( 3 )^{n + 2n + 1}/( 3 )^{2n + n -1}
= 3^{3n + 1}/( 3 )^{3n -1} = 3^{( 3n + 1 ) - ( 3n - 1 )}
= 3^{1 +1} = 3²
⇒ 3² = 9
∴ [ ( 243 )^n/5 × 3^{2n + 1} ]/[ 9ⁿ × 3^{n - 1} ] = 9