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CAT Exam Model Paper 2 with solutions for free online practice

Directions (Q. No. 74-75) A function

f ( z _ { 1 } , z _ { 2 } , z _ { 3 } , ... , z _ { n } )

f ( z _ { 1 } , z _ { n } ) + f ( z _ { 2 } , z _ { 3 } , ... , z _ { n - 1 } ) + ( z _ { 1 } + z _ { 2 } +... + z _ { n } );

forn > 0

f(y, z) = f(z, 0) +f( 0, y)

f(y, 0) = y + f(y - 1,0)

f(0,y)= y - f(0,y-1)

f(0, 0) = 1

Question : 75 of 100

Marks: +1, -0

f(9,2,k,0,9, 4) = 124, then the value of k is (where, k is a positive integer)

Solution:

f(9, 2, k, 0, 9, 4)

= f(9, 4) + f(2, k, 0, 9)+ (9 + 2 + k + 0 + 9 + 4)

= f(4, 0) + f(0, 9)+ [f(2, 9) + f(k, 0)+ (2 + k + 0 + 9)]+ 24 + k

= 11 + 4 + [{f(9, 0) + f(0,2)} + f(k, 0) + (11 + k)]+ 24+ k

= 15 + [48 + f(k, 0) + (11 + k)] + 24 + k

= 98 + 2k + f(k, 0)

= 98 + 2 k +

\left[ \frac{k(k+1)}{2} + 1 \right] = 124

⇒ k

^{2}

+5k -50 = 0

⇒ k = − 10 or k = 5

Since k is a positive integer, hence k = 5.

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