Find the values of a and b so that the function f given by f x = 1 , if x ≤ 3 a x + b , if 3 < x < 5 7 , if x ≥ 5 is continuous at x = 3 and x = 5 .

Given that, the function is f x = 1 , if x ≤ 3 a x + b , if 3 < x < 5 7 , if x ≥ 5 A function f x is continuous at x = a if L H L = R H L = f a i.e., lim x → a - f x = lim x → a + f x = f a Now, L H L at x = 3 is lim x → 3 - f x = lim h → 0 f 3 - h = lim h → 0 1 = 1 And R H L at x = 3 is lim x → 3 + f x = lim h → 0 f 3 + h = lim h → 0 a 3 + h + b = 3 a + b Similarly, L H L at x = 5 is lim x → 5 - f x = lim h → 0 f 5 - h = lim h → 0 a 5 - h + b = 5 a + b And R H L at x = 5 is lim x → 5 + f x = lim h → 0 f 5 + h = lim h → 0 7 = 7 Since, f ( x ) is continuous at x = 3 and x = 5 , then lim x → 3 + f x = lim x → 3 + f x and lim x → 5 - f x = lim x → 5 + f x ⇒ 1 = 3 a + b → ( 1 ) and 5 a + b = 7 → ( 2 ) On solving equations ( 1 ) and ( 2 ) , we get: a = 3 and b = - 8 . Hence, the required values are: a = 3 and b = - 8 .

If f x = x 2 2 , if 0 ≤ x ≤ 1 2 x 2 - 3 x + 3 2 , if 1 < x ≤ 2 . Show that f is continuous at x = 1 .

Given that, the function is f x = x 2 2 , if 0 ≤ x ≤ 1 2 x 2 - 3 x + 3 2 , if 1 < x ≤ 2 A function f x is continuous at x = a if L H L = R H L = f a i.e., lim x → a - f x = lim x → a + f x = f a Now, L H L = lim x → 1 - f x = lim h → 0 f 1 - h = lim h → 0 ( 1 - h ) 2 2 = 1 2 Similarly, R H L = lim x → 1 + f x = lim h → 0 f 1 + h = lim h → 0 2 ( 1 + h ) 2 - 3 ( 1 + h ) + 3 2 = 2 - 3 + 3 2 = 1 2 Also, f 1 = ( 1 ) 2 2 = 1 2 ∴ lim x → 1 - f x = lim x → 1 + f x = f 1 Hence, the given function is continuous at x = 1 .

Prove that f x = x - | x | x , x ≠ 0 2 , x = 0 is discontinuous at x = 0 .

Given that, the function is f x = x - | x | x , x ≠ 0 2 , x = 0 The given function can be rewritten as f x = x - x x , when x > 0 x + x x , when x < 0 2 , when x = 0 ⇒ f x = 0 , when x > 0 2 , when x < 0 2 , when x = 0 A function f x is continuous at x = a if L H L = R H L = f a i.e., lim x → a - f x = lim x → a + f x = f a Now, L H L = lim x → 0 - f x = lim h → 0 f 0 - h = lim h → 0 f - h = lim h → 0 2 = 2 And R H L = lim x → 0 + f x = lim h → 0 f 0 + h = lim h → 0 f h = lim h → 0 0 = 0 ∴ lim x → 0 - f x ≠ lim x → 0 + f x Thus, f ( x ) is discontinuous at x = 0 .

If f x = 2 x 2 + k , if x ≥ 0 - 2 x 2 + k , if x < 0 , then what should be the value of k so that f ( x ) is continuous at x = 0 .

The given function is f x = 2 x 2 + k , if x ≥ 0 - 2 x 2 + k , if x < 0 A function f x is continuous at x = a if L H L = R H L = f a i.e., lim x → a - f x = lim x → a + f x = f a Now, L H L = lim x → 0 − f x = lim h → 0 f 0 − h = lim h → 0 f − h = lim h → 0 − 2 − h 2 + k = k Similarly, R H L = lim x → 0 + f x = lim h → 0 f 0 + h = lim h → 0 f h = lim h → 0 2 h 2 + k = k If f x is continuous at x = 0 , then lim x → 0 − f x = lim x → 0 + f x = f 0 ⇒ lim x → 0 − f x = lim x → 0 + f x = k The function f x is continuous at x = 0 , for any real number of k .

For what value of λ is the function f x = λ x 2 - 2 x if x ≤ 0 4 x + 1 if x > 0 , continuous at x = 0 ? What about continuity at x = 1 ?

The given function f is f x = λ x 2 - 2 x if x ≤ 0 4 x + 1 if x > 0 A function f x is continuous at x = a if L H L = R H L = f a i.e., lim x → a - f x = lim x → a + f x = f a Here, function is said to be continuous at x = 0 if lim x → 0 - f x = lim x → 0 + f x = f 0 Now, L H L = lim x → 0 - f x = lim x → 0 λ x 2 - 2 x = λ 0 2 - 2 × 0 = 0 R H L = lim x → 0 + f x = lim x → 0 4 x + 1 = 4 × 0 + 1 = 1 Also f 0 = 0 Thus, Clearly, lim x → 0 - f x ≠ lim x → 0 + f x ≠ f 0 Hence, function f is discontinuous at x = 0 for any value of λ . Now, continuity at x = 1 L H L = lim x → 1 - 4 x + 1 = 4 × 1 + 1 = 5 R H L = lim x → 1 + 4 x + 1 = 4 × 1 + 1 = 5 Also, f 1 = 4 × 1 + 1 = 5 ∴ lim x → 1 - f x = lim x → 1 + f x = f 1 Hence, function f is continuous at x = 1 .

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 9 - Continuity>EXERCISE>Q 36

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Find the value of the constant $k$ so that the given function is continuous at the indicated point; $f (x) = \{\begin{cases} k (x^{2} + 2), & if x \leq 0 \\ 3 x + 1 & , if x > 0 \end{cases}$ at $x = 0$ .

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Important Questions on Continuity

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12th CBSE

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MATHEMATICS CLASS XII VOLUME-1>Chapter 9 - Continuity>EXERCISE>Q 36

Find the value of the constant $k$ so that the given function is continuous at the indicated point; $f (x) = \{\begin{cases} \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}}, & x \neq 2 \\ k, x = 2 \end{cases}$ at $x = 2$ .

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MATHEMATICS CLASS XII VOLUME-1>Chapter 9 - Continuity>EXERCISE>Q 37

Find the values of $a$ and $b$ so that the function $f$ given by $f (x) = \{\begin{matrix} 1, if x \leq 3 \\ a x + b, if 3 < x < 5 \\ 7, if x \geq 5 \end{matrix}$ is continuous at $x = 3$ and $x = 5$ .

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12th CBSE

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MATHEMATICS CLASS XII VOLUME-1>Chapter 9 - Continuity>EXERCISE>Q 38

If $f (x) = \{\begin{cases} \frac{x^{2}}{2}, & if 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, & if 1 < x \leq 2 \end{cases}$ . Show that $f$ is continuous at $x = 1$ .