Question

For what value of k, the function defined by

f(x) = `{{:((log(1 + 2x)sin^0)/x^2",", "for"  x ≠ 0),(k",", "for"  x = 0):}`

is continuous at x = 0 ?

पर्याय

• 2

• `1/2`

• `π/90`

• `90/π`

Answer 1

`bb(π/90)`

Explanation:

Since,

p(x) = `{{:((log(1 + 2x)sin^0)/x^2",", "for"  x ≠ 0),(k",", "for"  x = 0):}`

= `{{:((log(1 + 2x)sin  (πx)/180)/x^2",", "for"  x ≠ 0),(k",", "for"  x = 0):}`

As, f(x) is continuous at x = 0

So, LHL = f(0)

∵ LHL = `lim_(x rightarrow 0^-) f(x) = lim_(h rightarrow 0) f(0 - h)`

= `lim_(h rightarrow 0) (log(1 + 2(0 - h)sin  π/180^circ (0 - h)))/(0 - h)^2`

= `lim_(h rightarrow 0) (log(1 - 2h){-sin  (πh)/180})/h^2`

= `lim_(h rightarrow 0) (-2) (log(1 - 2h))/(-2h) xx lim_(h rightarrow 0) ((-)sin  (πh)/180)/((πh)/180) xx π/180`

= `(-2) xx (-1) xx 1 xx π/180`  ...`[∵ lim_(x rightarrow 0) log  (1 + x)/x = 1 and lim_(x rightarrow 0) sinx/x = 1]`

= `π/90`