Maharashtra Board Class 11 Maths Part 1 Chapter 8 Continuity 8.1 Solutions

Std 11 Maths 1 Exercise 8.1 Solutions Commerce Maths

Question 1. Examine the continuity of
(i) f(x) = x³ + 2x² - x - 2 at x = -2
Answer: Solution:
f(x) = x³ + 2x² - x - 2
Here f(x) is a polynomial function and hence it is continuous for all x ∈ R.
∴ f(x) is continuous at x = -2
In simple words: A polynomial function is continuous for all real numbers. Since the given function is a polynomial, it is continuous at the specified point x = -2.

🎯 Exam Tip: Remember that all polynomial functions are continuous over their entire domain (all real numbers). This property often simplifies continuity checks.

(ii) f(x) = \( \frac{x^2-9}{x-3} \) on R
Answer: Solution:
f(x) = \( \frac{x^2-9}{x-3} \); x ∈ R
f(x) is a rational function and is continuous for all x ∈ R, except at the points where denominator becomes zero.
Here, denominator x - 3 = 0 when x = 3.
∴ Function f is continuous for all x ∈ R, except at x = 3, where it is not defined.
In simple words: A rational function is continuous everywhere except where its denominator is zero. For this function, the denominator is zero at x=3, so it is discontinuous at x=3.

🎯 Exam Tip: For rational functions, always check for points where the denominator is zero. These points are potential discontinuities, unless the discontinuity is removable.

Question 2. Examine whether the function is continuous at the points indicated against them.
(i) f(x) = x³ - 2x + 1, for x ≤ 2
= 3x - 2, for x > 2, at x = 2
Answer: Solution:
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3-2x+1) \)
= \( (2)^3-2(2)+1 = 5 \)
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x-2) \)
= \( 3(2)-2 = 4 \)
\( \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x) \)
∴ Function f is discontinuous at x = 2
In simple words: For a function to be continuous at a point, its left-hand limit, right-hand limit, and the function's value at that point must all be equal. Here, the left-hand limit (5) is not equal to the right-hand limit (4) at x = 2, so the function is discontinuous.

🎯 Exam Tip: When dealing with piecewise functions, always evaluate the left-hand and right-hand limits at the breakpoint. If they are not equal, the function is discontinuous.

(ii) f(x) = \( \frac{x^2+18x-19}{x-1} \) for x ≠ 1
= 20, for x = 1, at x = 1
Answer: Solution:
\( \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x^2+18x-19}{x-1} \)
= \( \lim_{x \to 1} \frac{x^2+19x-x-19}{x-1} \)
= \( \lim_{x \to 1} \frac{x(x+19)-1(x+19)}{(x-1)} \)
= \( \lim_{x \to 1} \frac{(x-1)(x+19)}{(x-1)} \)
= \( \lim_{x \to 1} (x+19) \)
...[x \( \to \) 1, x \( \neq \) 1, x - 1 \( \neq \) 0]
= 1+19=20
Also, f(1) = 20
\( \lim_{x \to 1} f(x) = f(1) \)
∴ f(x) is continuous at x = 1
In simple words: The function's limit as x approaches 1 is 20, and the function's value at x=1 is also 20. Since the limit equals the function's value, the function is continuous at x = 1.

🎯 Exam Tip: For removable discontinuities, algebraic simplification (like factoring) can help evaluate the limit. Continuity is established if this limit matches the defined function value at that point.

Question 3. Test the continuity of the following functions at the points indicated against them.
(i) f(x) = \( \frac{\sqrt{x-1}-(x-1)^{\frac{1}{3}}}{x-2} \) for x \( \neq \) 2
= \( \frac{1}{5} \) for x = 2, at x = 2
Answer: Solution:
f(2)= \( \frac{1}{5} \) ...(given)
\( \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{\sqrt{x-1}-(x-1)^{\frac{1}{3}}}{x-2} \)
Put x - 1 = y
x = 1+y
As x \( \to \) 2, y \( \to \) 1
∴ \( \lim_{x \to 2} f(x) = \lim_{y \to 1} \frac{y^{\frac{1}{2}}-y^{\frac{1}{3}}}{1+y-2} \)
= \( \lim_{y \to 1} \frac{y^{\frac{1}{2}}-1-y^{\frac{1}{3}}+1}{y-1} \)
= \( \lim_{y \to 1} [\frac{y^{\frac{1}{2}}-1}{y-1} - \frac{y^{\frac{1}{3}}-1}{y-1}] \)
= \( \lim_{y \to 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1} - \lim_{y \to 1} \frac{y^{\frac{1}{3}}-1^{\frac{1}{3}}}{y-1} \)
= \( \frac{1}{2} (1)^{\frac{1}{2}-1} - \frac{1}{3} (1)^{\frac{1}{3}-1} \)
...[ \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = n.a^{n-1} \)]
= \( \frac{1}{2} - \frac{1}{3} \)
= \( \frac{1}{6} \)
∴ \( \lim_{x \to 2} f(x) \neq f(2) \)
∴ f(x) is discontinuous at x = 2
In simple words: After simplifying the limit using a substitution and standard limit formulas, the limit of the function as x approaches 2 is found to be 1/6. However, the function's value at x=2 is given as 1/5. Since the limit does not equal the function's value, the function is discontinuous at x=2.

🎯 Exam Tip: When evaluating limits of complex expressions, substitution can simplify the form to match standard limit theorems (like \( \lim_{x \to a} \frac{x^n-a^n}{x-a} \)). Always compare the calculated limit with the function's explicitly defined value at the point.

(ii) f(x) = \( \frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}} \) for x \( \neq \) 2
= -24 for x = 2, at x = 2
Answer: Solution:
f(2) = - 24 ...(given)
\( \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}} \)
= \( \lim_{x \to 2} \frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}} \times \frac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{x+2}+\sqrt{3x-2}} \)
= \( \lim_{x \to 2} \frac{(x^3-8)(\sqrt{x+2}+\sqrt{3x-2})}{(x+2)-(3x-2)} \)
= \( \lim_{x \to 2} \frac{(x-2)(x^2+2x+4)(\sqrt{x+2}+\sqrt{3x-2})}{-2x+4} \)
= \( \lim_{x \to 2} \frac{(x-2)(x^2+2x+4)(\sqrt{x+2}+\sqrt{3x-2})}{-2(x-2)} \)
...[x \( \to \) 2, x \( \neq \) 2]
...[x - 2 \( \neq \) 0]
= \( -\frac{1}{2} \lim_{x \to 2} (x^2+2x+4) \lim_{x \to 2} (\sqrt{x+2}+\sqrt{3x-2}) \)
= \( -\frac{1}{2} \times [2^2+2(2)+4] \times (\sqrt{2+2}+\sqrt{3(2)-2}) \)
= \( -\frac{1}{2} \times 12 \times (2+2) \)
= -24
∴ \( \lim_{x \to 2} f(x) = f(2) \)
∴ f(x) is continuous at x = 2
In simple words: To evaluate the limit, we rationalized the denominator and factored the numerator. After simplification, the limit as x approaches 2 is -24, which matches the defined function value f(2)=-24. Thus, the function is continuous at x = 2.

🎯 Exam Tip: When evaluating limits involving square roots leading to an indeterminate form, rationalizing the expression by multiplying by the conjugate is a common and effective technique.

(iii) f(x) = 4x + 1 for x \( \le \frac{8}{3} \)
= \( \frac{59-9x}{3} \), for x > \( \frac{8}{3} \), at x = \( \frac{8}{3} \)
Answer: Solution:
\( \lim_{x \to (\frac{8}{3})^-} f(x) = \lim_{x \to (\frac{8}{3})^-} (4x+1) \)
= \( 4(\frac{8}{3})+1 \)
= \( \frac{32}{3}+1 \)
= \( \frac{35}{3} \)
\( \lim_{x \to (\frac{8}{3})^+} f(x) = \lim_{x \to (\frac{8}{3})^+} \frac{59-9x}{3} \)
= \( \frac{59-9(\frac{8}{3})}{3} \)
= \( \frac{59-24}{3} \)
= \( \frac{35}{3} \)
f(x) = 4x + 1, x \( \le \frac{8}{3} \)
∴ f(\( \frac{8}{3} \)) = \( 4(\frac{8}{3})+1 \)
= \( \frac{32}{3}+1 \)
= \( \frac{35}{3} \)
∴ \( \lim_{x \to (\frac{8}{3})^-} f(x) = \lim_{x \to (\frac{8}{3})^+} f(x) = f(\frac{8}{3}) \)
∴ f(x) is continuous at x = \( \frac{8}{3} \)
In simple words: The left-hand limit, right-hand limit, and the function's value at x = 8/3 all evaluate to 35/3. Since these three values are equal, the function is continuous at x = 8/3.

🎯 Exam Tip: For piecewise functions defined with inequality conditions, check continuity at the boundary point by calculating the left-hand limit, right-hand limit, and the function value directly.

(iv) f(x) = \( \frac{x^3-27}{x^2-9} \) for 0 \( \le \) x < 3
= \( \frac{9}{2} \), for 3 \( \le \) x \( \le \) 6, at x = 3
Answer: Solution:
f(3) = \( \frac{9}{2} \) ...(given)
\( \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^3-27}{x^2-9} \)
= \( \lim_{x \to 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} \)
...[As x \( \to \) 3, x \( \neq \) 3]
...[x - 3 \( \neq \) 0]
= \( \lim_{x \to 3} \frac{x^2+3x+9}{x+3} \)
= \( \frac{(3)^2+3(3)+9}{3+3} \)
= \( \frac{9+9+9}{6} \)
= \( \frac{27}{6} \)
= \( \frac{9}{2} \)
∴ \( \lim_{x \to 3} f(x) = f(3) \)
∴ Function f is continuous at x = 3
In simple words: We simplify the function by factoring the numerator and denominator, cancelling the common term (x-3). The limit as x approaches 3 is 9/2, which matches the given function value at x=3. Therefore, the function is continuous at x = 3.

🎯 Exam Tip: Always look for opportunities to factor and simplify rational functions before evaluating limits, especially when direct substitution results in an indeterminate form (0/0).

Question 4.
(i) If f(x) = \( \frac{2^{4x}-8^x-3^x+1}{12^x-4^x-3^x+1} \), for x \( \neq \) 0
= k, for x = 0
is continuous at x = 0, find k.
Answer: Solution:
Function f is continuous at x = 0
f(0) = \( \lim_{x \to 0} f(x) \)
∴ k = \( \lim_{x \to 0} \frac{2^{4x}-8^x-3^x+1}{12^x-4^x-3^x+1} \)
= \( \lim_{x \to 0} \frac{(8^x)(3^x)-8^x-3^x+1}{(4^x)(3^x)-4^x-3^x+1} \)
= \( \lim_{x \to 0} \frac{8^x(3^x-1)-1(3^x-1)}{4^x(3^x-1)-1(3^x-1)} \)
= \( \lim_{x \to 0} \frac{(3^x-1)(8^x-1)}{(3^x-1)(4^x-1)} \)
...[x \( \to \) 0, 3\(^x\) \( \to \) 3\(^0\)]
...[3\(^x\) \( \to \) 1, 3\(^x\) \( \neq \) 1]
...[3\(^x\)-1 \( \neq \) 0]
= \( \lim_{x \to 0} \frac{\frac{8^x-1}{x}}{\frac{4^x-1}{x}} \)
...[x \( \to \) 0, x \( \neq \) 0]
= \( \frac{\lim_{x \to 0} \frac{8^x-1}{x}}{\lim_{x \to 0} \frac{4^x-1}{x}} \)
...[ \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \)]
= \( \frac{\log 8}{\log 4} \)
= \( \frac{\log (2)^3}{\log (2)^2} \)
= \( \frac{3 \log 2}{2 \log 2} \)
∴ f(0) = \( \frac{3}{2} \)
In simple words: Since the function is continuous at x=0, k must be equal to the limit of f(x) as x approaches 0. By factoring the expression and using the standard limit formula for \( \frac{a^x-1}{x} \), we find that k = log 8 / log 4, which simplifies to 3/2.

🎯 Exam Tip: For continuity problems involving exponential terms, factor common factors and apply the fundamental limit \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \). Ensure to simplify logarithmic expressions fully.

(ii) If f(x) = \( \frac{5^{x}+5^{-x}-2}{x^2} \), for x \( \neq \) 0
= k for x = 0
is continuous at x = 0, find k.
Answer: Solution:
Function f is continuous at x = 0
∴ f(0) = \( \lim_{x \to 0} f(x) \)
∴ k = \( \lim_{x \to 0} \frac{5^{x}+5^{-x}-2}{x^2} \)
= \( \lim_{x \to 0} \frac{5^{x}+\frac{1}{5^x}-2}{x^2} \)
= \( \lim_{x \to 0} \frac{(5^x)^2+1-2(5^x)}{5^x.x^2} \)
= \( \lim_{x \to 0} \frac{(5^x-1)^2}{5^x.x^2} \)
...[∴ a²-2ab + b² = (a - b)²]
= \( \lim_{x \to 0} \frac{(\frac{5^x-1}{x})^2}{5^x} \)
= \( \lim_{x \to 0} (\frac{5^x-1}{x})^2 \times \lim_{x \to 0} \frac{1}{5^x} \)
= \( (\log 5)^2 \times \frac{1}{5^0} \)
...[ \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \)]
∴ k = \( (\log 5)^2 \)
In simple words: Since the function is continuous at x=0, k equals the limit of f(x) as x approaches 0. By rewriting and simplifying the expression using the algebraic identity (a-b)^2 and the standard limit for \( \frac{a^x-1}{x} \), we find that k is equal to \( (\log 5)^2 \).

🎯 Exam Tip: Recognize algebraic identities like \( (a-b)^2 \) within limit problems. This often helps transform the expression into a form where standard limit formulas (like \( \lim_{x \to 0} \frac{a^x-1}{x} \)) can be applied.

(iii) For what values of a and b is the function
f(x) = ax + 2b + 18 for x \( \le \) 0
= x² + 3a - b for 0 < x \( \le \) 2
= 8x - 2 for x > 2,
continuous for every x?
Answer: Solution:
Function f is continuous for every x.
∴ Function f is continuous at x = 0 and x = 2
As f is continuous at x = 0.
∴ \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \)
∴ \( \lim_{x \to 0^-} (ax + 2b +18) = \lim_{x \to 0^+} (x^2 + 3a - b) \)
∴ a(0) + 2b + 18 = (0)² + 3a - b
∴ 3a - 3b = 18
∴ a - b = 6 .....(i)
Also, Function f is continuous at x = 2
∴ \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \)
∴ \( \lim_{x \to 2^-} (x^2 + 3a - b) = \lim_{x \to 2^+} (8x - 2) \)
∴ (2)² + 3a - b = 8(2) - 2
∴ 4 + 3a - b = 14
∴ 3a - b = 10 .....(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2-b = 6
∴ b = -4
∴ a = 2 and b = -4
In simple words: For the piecewise function to be continuous everywhere, it must be continuous at the breakpoints x=0 and x=2. By setting the left-hand and right-hand limits equal at both points, we establish a system of two linear equations for 'a' and 'b'. Solving these equations yields a=2 and b=-4.

🎯 Exam Tip: For piecewise functions to be continuous over their entire domain, they must be continuous at all points, especially at the transition points (breakpoints). Set up equations by equating the one-sided limits and the function's value at these points.

(iv) For what values of a and b is the function
f(x) = \( \frac{x^2-4}{x-2} \) for x < 2
= ax² - bx + 3 for 2 \( \le \) x < 3
= 2x - a + b for x \( \ge \) 3
continuous in its domain.
Answer: Solution:
Function f is continuous for every x on R.
∴ Function f is continuous at x = 2 and x = 3.
As f is continuous at x = 2.
∴ \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \)
\( \lim_{x \to 2^-} \frac{x^2-4}{x-2} = \lim_{x \to 2^+} (ax^2-bx+3) \)
∴ \( \lim_{x \to 2^-} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2^+} (ax^2-bx+3) \)
∴ \( \lim_{x \to 2^-} (x+2) = \lim_{x \to 2^+} (ax^2-bx+3) \)
...[x \( \to \) 2, x \( \neq \) 2]
...[x-2 \( \neq \) 0]
2 + 2 = a(2)² - b(2) + 3
∴ 4 = 4a - 2b + 3
∴ 4a - 2b = 1 .....(i)
Also function f is continuous at x = 3
∴ \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \)
∴ \( \lim_{x \to 3^-} (ax^2 - bx + 3) = \lim_{x \to 3^+} (2x - a + b) \)
∴ a(3)² - b(3) + 3 = 2(3) - a + b
∴ 9a - 3b + 3 = 6 - a + b
∴ 10a - 4b = 3 .....(ii)
Multiplying (i) by 2, we get
8a - 4b = 2 .....(iii)
Subtracting (iii) from (ii), we get
2a = 1
∴ a = \( \frac{1}{2} \)
Substituting a = \( \frac{1}{2} \) in (i), we get
\( 4(\frac{1}{2}) - 2b = 1 \)
∴ 2 - 2b = 1
∴ 1 = 2b
∴ b = \( \frac{1}{2} \)
∴ a = \( \frac{1}{2} \) and b = \( \frac{1}{2} \)
In simple words: For the function to be continuous across its domain, it must be continuous at the breakpoints x=2 and x=3. By equating the left and right limits at these points, we get a system of linear equations. Solving these equations yields a = 1/2 and b = 1/2.

🎯 Exam Tip: Continuity for multi-piece functions requires careful evaluation of limits at each boundary point. Simplify expressions before taking limits, and use the resulting equations to solve for unknown constants.