Show that the function f x = x m sin 1 x , x ≠ 0 0 , x = 0 neither continuous nor differentiable, if m ≤ 0 .

Given: f x = x m sin 1 x , x ≠ 0 0 , x = 0 Let m = 0 , m ≤ 0 . Then the function becomes f x = x 0 sin 1 x , x ≠ 0 0 , x = 0 ⇒ f x = sin 1 x , x ≠ 0 0 , x = 0 Since we know during finding continuity and differentiability we will come across the term sin 1 0 and it is not defined. So given function is neither continuous nor differentiable at x = 0 if m ≤ 0 .

Show that the function f x = x m sin 1 x , x ≠ 0 0 , x = 0 neither continuous nor differentiable, if m ≤ 0 .

Given: f x = x m sin 1 x , x ≠ 0 0 , x = 0 Let m = 0 , m ≤ 0 . Then the function becomes f x = x 0 sin 1 x , x ≠ 0 0 , x = 0 ⇒ f x = sin 1 x , x ≠ 0 0 , x = 0 Since we know during finding continuity and differentiability we will come across the term sin 1 0 and it is not defined. So given function is neither continuous nor differentiable at x = 0 if m ≤ 0 .

Find the values of a and b so that the function f x = x 2 + 3 x + a , if x ≤ 1 b x + 2 , if x > 1 is differentiable at each x ∈ R .

Given: f x = x 2 + 3 x + a , if x ≤ 1 b x + 2 , if x > 1 It is given that the function is differentiable at each x ∈ R and every differentiable function is continuous. So f ( x ) is continuous at x = 1 Therefore, lim x → 1 - f x = lim x → 1 + f x = f 1 ⇒ lim x → 1 x 2 + 3 x + a = lim x → 1 b x + 2 = a + 4 ⇒ a + 4 = b + 2 = a + 4 → 1 Since, f ( x ) is differentiable at x = 1 . So, ( LHD at x = 1 ) = ( RHD at x = 1 ) Thus, lim x → 1 - f ( x ) - f ( 1 ) x - 1 = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ⇒ lim x → 1 x 2 + 3 x + a - a - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 x 2 + 3 x - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 x 2 + 4 x - x - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 ( x + 4 ) ( x - 1 ) x - 1 = lim x → 1 b x - 2 - a x - 1 ⇒ lim x → 1 ( x + 4 ) ( x - 1 ) x - 1 = lim x → 1 b x - b x - 1 [Using 1 ] ⇒ lim x → 1 x + 4 = lim x → 1 b ⇒ 5 = b From 1 , we have: a + 4 = b + 2 ⇒ a + 4 = 5 + 2 ⇒ a = 7 - 4 ⇒ a = 3 Hence, a = 3 , b = 5 .

Show that the function f x = | 2 x - 3 | [ x ] , x ≥ 1 sin π x 2 , x < 1 is continuous but not differentiable at x = 1 .

Given: f x = | 2 x - 3 | [ x ] , x ≥ 1 sin π x 2 , x < 1 Now, check continuity at x = 1 ( LHL at x = 1 ) = lim x → 1 - f x = lim h → 0 f 1 - h = lim h → 0 sin π ( 1 - h ) 2 = sin π 2 = 1 And ( RHL at x = 1 ) = lim x → 1 + f x = lim h → 0 f 1 + h = lim h → 0 | 2 ( 1 + h ) - 3 | [ 1 + h ] = lim h → 0 | 2 ( 1 + h ) - 3 | = 1 And also, f 1 = 2 1 - 3 1 = 1 Hence, L H L = R H L = f 1 Similarly, check differentiability at x = 1 Now, ( LHD at x = 1 ) = lim x → 1 - f ( x ) - f ( 1 ) x - 1 ( LHD at x = 1 ) = lim h → 0 f ( 1 - h ) - f ( 1 ) 1 - h - 1 ( LHD at x = 1 ) = lim h → 0 f ( 1 - h ) - f ( 1 ) - h ( LHD at x = 1 ) = lim h → 0 sin π ( 1 - h ) 2 - 1 - h ( LHD at x = 1 ) = lim h → 0 cos π h 2 - 1 - h ( LHD at x = 1 ) = - π 2 lim h → 0 cos π h 2 - 1 π h 2 = 0 Similarly, ( RHD at x = 1 ) = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ( RHD at x = 1 ) = lim h → 0 f ( 1 + h ) - f ( 1 ) 1 + h - 1 ( RHD at x = 1 ) = lim h → 0 f ( 1 + h ) - f ( 1 ) h ( RHD at x = 1 ) = lim h → 0 - ( 2 ( 1 + h ) - 3 ) - 1 h ( RHD at x = 1 ) = lim h → 0 - 2 h h = - 2 Thus, LHD ≠ RHD Hence, the function is continuous but not differentiable at x = 1 .

If f x = a x 2 - b , if | x | < 1 1 | x | , if | x | ≥ 1 is differentiable at x = 1 , find a , b .

Given, f x = a x 2 - b , if | x | < 1 1 | x | , if | x | ≥ 1 ⇒ f x = - 1 x , if x < - 1 a x 2 + b , if - 1 < x < 1 1 x , if x ≥ 1 It is given that the given function is differentiable at x = 1 . We know every differentiable function is continuous. Therefore, it is continuous at x = 1 . Then, lim x → 1 - f x = lim x → 1 + f x ⇒ lim x → 1 a x 2 - b = lim x → 1 1 x ⇒ a - b = 1 … ( i ) It is also differentiable at x = 1 . Therefore, L H D = R H D ⇒ lim x → 1 - f ( x ) - f ( 1 ) x - 1 = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ⇒ lim x → 1 a x 2 - b - 1 x - 1 = lim x → 1 1 x - 1 x - 1 ⇒ lim x → 1 a x 2 + 1 - a - 1 x - 1 = lim x → 1 - ( x - 1 ) x x - 1 [Using ( i ) ] ⇒ lim x → 1 a x + 1 = lim x → 1 - 1 ⇒ 2 a = - 1 ⇒ a = - 1 2 From ( i ) , we have: a - b = 1 ⇒ - 1 2 - b = 1 ⇒ b = - 3 2 Hence, when a = - 1 2 and b = - 3 2 , the function is differentiable at x = 1 .

Find the values of a and b , if the function f defined by f x = x 2 + 3 x + a , x ⩽ 1 b x + 2 , x > 1 .

Given that, we have: f x = x 2 + 3 x + a , x ⩽ 1 b x + 2 , x > 1 Given that f ( x ) is differentiable at x = 1 . We know every differentiable function is continuous. Therefore, f ( x ) is continuous at x = 1 as well Thus, lim x → 1 - f x = lim x → 1 + f x = f 1 ⇒ lim x → 1 x 2 + 3 x + a = lim x → 1 ( b x + 2 ) ⇒ 1 + 3 + a = b + 2 ⇒ a - b + 2 = 0 → 1 Again, f ( x ) is differentiable at x = 1 . So, ( LHD at x = 1 ) = ( RHD at x = 1 ) ⇒ lim x → 1 - f ( x ) - f ( 1 ) x - 1 = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ⇒ lim x → 1 x 2 + 3 x + a - ( 4 + a ) x - 1 = lim x → 1 ( b x + 2 ) - ( 4 + a ) x - 1 ⇒ lim x → 1 x 2 + 3 x - 4 x - 1 = lim x → 1 ( b x - 2 - a ) x - 1 ⇒ lim x → 1 x 2 + 4 x - x - 4 x - 1 = lim x → 1 ( b x - 2 - a ) x - 1 ⇒ lim x → 1 ( x + 4 ) ( x - 1 ) x - 1 = lim x → 1 b x - b x - 1 ⇒ lim x → 1 ( x + 4 ) = lim x → 1 b ⇒ 5 = b Putting b = 5 in 1 , we get: a = 3 Hence, a = 3 and b = 5 .

If f is defined by f ( x ) = x 2 - 4 x + 7 , show that f ' 5 = 2 f ' 7 2 .

Given: f ( x ) = x 2 - 4 x + 7 Clearly, f ( x ) being a polynomial function, is everywhere differentiable. The derivative of f at x is given by f ' x = lim h → 0 f ( x + h ) - f ( x ) h ⇒ f ' x = lim h → 0 ( x + h ) 2 - 4 ( x + h ) + 7 - x 2 - 4 x + 7 h ⇒ f ' x = lim h → 0 x 2 + h 2 + 2 x h - 4 x - 4 h + 7 - x 2 + 4 x - 7 h ⇒ f ' x = lim h → 0 h 2 + 2 x h - 4 h h ⇒ f ' x = lim h → 0 h ( h + 2 x - 4 ) h ⇒ f ' x = 2 x - 4 Now, f ' 5 = 2 × 5 - 4 = 6 Similarly, f ' 7 2 = 2 × 7 2 - 4 = 3 Therefore, f ' 5 = 2 × 3 = 2 f ' 7 2 Hence, proved.