Question

Question is in attachment​

Answer 1

Answer:

option A is correct answer

$\text{(a) }a+b=0$

Step-by-step explanation:

Given function is :

f(x) = a [x + 1] + b [x - 1], where, [.] is the GIF

f(x) is continuous at x = 1

So, its Left hand limit (LHL) and right hand limit (RHL) should be equal.

• Considering LHL, that is x<1

f(x) = a [x + 1] + b [x - 1]

If x < 1, (assume 0.9)

=> [x+1] tends to 1

because, [0.9+1] = [1.9] = 1

=> [x-1] tends to -1

because, [0.9 - 1] = [-0.1] = -1

So, the function turns into,

LHL : f(x) = a(1) + b(-1)

=> f(x) = a - b at LHL {x < 1}

• Considering RHL, that is x>1

f(x) = a [x + 1] + b [x - 1]

If x > 1, (assume 1.1)

=> [x+1] tends to 2

because, [1.1+1] = [2.1] = 2

=> [x-1] tends to 0

because, [1.1 - 1] = [0.1] = 0

So, the function turns into,

RHL : f(x) = a(2) + b(0)

=> f(x) = 2a  at RHL {x > 1}

As, LHL = RHL

a - b = 2a

=> a + b = 0

Hope it helps!!

Answer 2

Answer:

Given function is :

f(x) = a [x + 1] + b [x - 1], where, [.] is the GIF

f(x) is continuous at x = 1

So, its Left hand limit (LHL) and right hand limit (RHL) should be equal.

Considering LHL, that is x<1

f(x) = a [x + 1] + b [x - 1]

If x < 1, (assume 0.9)

=> [x+1] tends to 1

because, [0.9+1] = [1.9] = 1

=> [x-1] tends to -1

because, [0.9 - 1] = [-0.1] = -1

So, the function turns into,

LHL : f(x) = a(1) + b(-1)

=> f(x) = a - b at LHL {x < 1}

Considering RHL, that is x>1

f(x) = a [x + 1] + b [x - 1]

If x > 1, (assume 1.1)

=> [x+1] tends to 2

because, [1.1+1] = [2.1] = 2

=> [x-1] tends to 0

because, [1.1 - 1] = [0.1] = 0

So, the function turns into,

RHL : f(x) = a(2) + b(0)

=> f(x) = 2a at RHL {x > 1}

As, LHL = RHL

a - b = 2a

a-2a=b

-a=b

a+b=0

♠ Option A ♠