Question

Find m and n if following function is continuous

f(x) = \(\begin{cases} x^2 + mx + n; &0\leq x < 2\\ 4x - 1;& 2 \leq x \leq 4\\ mx^2 + 17n;&4 < x \leq 6 \end{cases} \)

Answer 1

For continuity at x = 2,

Left hand limit

f (2 – 0) = lim_h→0 f(2 – A)

= lim_h→0 [(2 – h)² + m(2 – h) + n]

= (2 – 0)² + m(2 – 0) + n

= 4 + 2m + n

Right hand limit

f(2 + 0) = lim_h→0 f(2 + h)

= lim_h→0 [4(2 + h) – 1 ]

= 4(2 + 0) – 1

= 8 – 1

= 7

At x = 2, value of f (x),

f(2) = 4 × 2 – 1 = 8 – 1 = 7

Given function is continuous at x = 2, then

f(2 – 0) = f(2 + 0) = f(2)

4 + 2m + n = 7 = 7

⇒ 4 + 2m + n = 7

⇒ 2m + n = 7 – 4

or

2m + n = 3

For continuity at x = 4,

Left hand limit

f (4 – 0) = lim_h→0 f (4-h)

= lim_h→0 [4(4 – h) – 1]

= 4 (4 – 0) – 1

= 16 – 1

= 15

Right hand limit

f (4 + 0) = lim_h→0 f (4 – h)

= lim_h→0 [m(4 + h)² + 17n]

= m (4 + 0)² + 17n

= 16m + 17n

At x = 4, value of function

f(4) = 4 × 4 – 1 = 16 – 1 = 15

Given function is continuous at x = 4,

f (4 – 0) = f (4 + 0) = f (4)

15 = 16m + 17n = 15

⇒ 16m + 17n = 15 ……(ii)

Putting n = 3 – 2m from (i) in

16m + 17(3 – 2m) = 15

16m + 51 – 34m = 15

– 18m = 15 – 51

– 18m = – 36

m = 2

From equation (i),

2 × 2 + n = 3

n = 3 – 4

Hence, m = 2,n = – 1.