Question

If f(x) = (fig.)
Is continuous at x = 0, then the ordered pair (p, d) is equal to
(A) (-3/2, -1/2) (B) (5/2, 1/2)
(C) (-1/2, 3/2)
(D) (-3/2, 1/2)

Answer 1

The ordered pair ( p , d ) is equal to ( - 3 / 2 , 1 / 2 )

f(x) = (  sin ( p + 1 ) x + sin x ) / x ,  x < 0

        d , x = 0

       (  $\sqrt{x + x^{2} } - \sqrt{x}$ ) / $x^{3/2}$  , x > 0

Its given that f is continuous at 0.

• Therefore Left hand limit  at 0 = Right hand limit at  0 = value of f at 0
• lim x - > 0 - ( f ( x ) ) = f ( 0 ) = lim x - > 0 + ( f ( x ) )

If we directly give x = 0 then Lhl and Rhl becomes 0 / 0 form.

Hence by applying L'Hospital Rule,

We can derivative of numerator and denominator and then give x = 0.

• Left Hand limit = = >
• (( p + 1 ) cos ( p + 1 ) x + cos x ) / 1 at x = 0 becomes:
• p + 1 + 1 = p + 2 - - - ( 1 )

We know f ( 0 ) = q

• Right Hand limit = = >
• We can multiply and divide by conjugate of   $\sqrt{x + x^{2} } - \sqrt{x}$  
• Rhl becomes  ( x + $x^{2}$ - x )/ $x^{3/2}$ (  $\sqrt{x + x^{2} } + \sqrt{x}$ )

• = = >
• $\sqrt{x}$ /  $\sqrt{x + x^{2} } + \sqrt{x}$ = 1 / $\sqrt{x + 1}$ + 1  = = >
• At x = 0 ,
• Rhl = 1/2

Therefore Rhl = f ( 0 ) = Lhl

• q = 1/2

• p + 2 = q =  1 / 2 = = >

• p = - 3 / 2