CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set 03

1 marks question

Q1. Find \(\frac{dy}{dx}\) of following functions: (i) \(y = \log(\sin x)\) (ii) \(y = e^{\sqrt{x}}\) (iii) \(y = \log_{10} x\) (iv) \(y = e^{2x+3}\) (v) \(y = \tan^{-1}\sqrt{x}\) (vi) \(y = 2^{\sin x}\) (vii) \(y = \log(\log x)\) (viii) \(y = 3^{\sin^{-1}2x}\) (ix) \(y = \sqrt{3x+5}\) (x) \(y = x^x\) (xi) \(y = \sin^{-1}(2x\sqrt{1-x^2})\) (xii) \(y = 2^{\cos^2 x}\) (xiii) \(y = \cos(\tan\sqrt{x+1})\) (xiv) \(y = \sin\sqrt{x} + \cos^2\sqrt{x}\) (xv) \(y = \sin^n(ax^2+bx+c)\) (xvi) \(y = \log[\log(\log x^5)]\) (xvii) \(y = x^2\cos x^2 + \sin^2 x + 2x\sin(2x^2)\)
Answer:
(i) \(\frac{dy}{dx} = \cot x\)
(ii) \(\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}\)
(iii) \(\frac{dy}{dx} = \frac{1}{x}\log_{10} e\)
(iv) \(\frac{dy}{dx} = 2e^{2x+3}\)
(v) \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}(1+x)}\)
(vi) \(\frac{dy}{dx} = 2^{\sin x}\cos x \log 2\)
(vii) \(\frac{dy}{dx} = \frac{1}{x\log x}\)
(viii) \(\frac{dy}{dx} = \frac{3^{\sin^{-1}2x}\cdot 2\log 3}{\sqrt{1-4x^2}}\)
(ix) \(\frac{dy}{dx} = \frac{3}{2\sqrt{3x+5}}\)
(x) \(\frac{dy}{dx} = x^x(1+\log x)\)
(xi) \(\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}\)
(xii) \(\frac{dy}{dx} = 2^{\cos^2 x}(\log 2)(-\sin 2x)\)
(xiii) \(\frac{dy}{dx} = -\sin(\tan\sqrt{x+1})\sec^2(\sqrt{x+1})\cdot \frac{1}{2\sqrt{x+1}}\)
(xiv) \(\frac{dy}{dx} = \frac{\cos\sqrt{x}}{2\sqrt{x}} - \frac{\sin 2\sqrt{x}}{2\sqrt{x}}\)
(xv) \(\frac{dy}{dx} = n\sin^{n-1}(ax^2+bx+c)\cos(ax^2+bx+c)(2ax+b)\)
(xvi) \(\frac{dy}{dx} = \frac{5}{x\log(x^5)[\log(\log x^5)]}\)
(xvii) \(\frac{dy}{dx} = 2x\cos x^2 - 2x^3\sin x^2 + 2\sin x\cos x + 4x\cos(2x^2)\)
In simple words: Apply standard differentiation formulas combined with the chain rule for composite functions.

Exam Tip: Always identify the outermost function first and work your way inward using the chain rule.

Q2. Discuss the differentiability of \(f(x) = |x|, \forall x \in \mathbb{R}\).
Answer:
The given function is defined piecewise as:
\(f(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}\)
At \(x = 0\):
Left-Hand Derivative (LHD) = \(\lim_{h \to 0^-} \frac{f(0-h)-f(0)}{-h} = \lim_{h \to 0} \frac{-(-h)-0}{-h} = -1\)
Right-Hand Derivative (RHD) = \(\lim_{h \to 0^+} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \frac{h-0}{h} = 1\)
Since LHD \(\neq\) RHD at \(x = 0\), the function is not differentiable at \(x = 0\).
In simple words: Absolute value functions have a sharp corner at their vertex, making them non-differentiable there.

Exam Tip: Modulus functions are generally not differentiable at points where the expression inside the modulus becomes zero.

Q3. Discuss the differentiability of \(f(x) = [x]\), where \([x]\) is greatest integer function, at \(x = 2\) and \(5/2\).
Answer:
At \(x = 2\):
Left-Hand Derivative = \(\lim_{h \to 0} \frac{[2-h]-[2]}{-h} = \frac{1-2}{-h} = \frac{-1}{-h}\) which does not exist finitely as \(h \to 0\).
Thus, \(f(x)\) is not differentiable at \(x = 2\).
At \(x = 5/2 = 2.5\):
In the neighborhood of \(2.5\), say for small \(h > 0\), \([2.5-h] = 2\) and \([2.5+h] = 2\).
Thus, the function is constant in a neighborhood of \(2.5\) (i.e., \(f(x) = 2\)), meaning its derivative is zero.
Therefore, \(f(x)\) is differentiable at \(x = 5/2\).
In simple words: The greatest integer function is discontinuous and non-differentiable at all integers, but differentiable at all non-integer points.

Exam Tip: Always check left and right limits for step functions when testing continuity and differentiability at integer points.

4/6 marks question

Q4. If \(f(x) = \begin{cases} \frac{5x+|x|}{3x}, & \text{if } x \neq 0 \\ 2, & \text{if } x = 0 \end{cases}\), show that \(f(x)\) is discontinuous at \(x = 0\).
Answer:
To test continuity at \(x = 0\), we evaluate \(\lim_{x \to 0} f(x)\) and compare it with \(f(0)\).
For \(x < 0\), \(|x| = -x\):
\(f(x) = \frac{5x - x}{3x} = \frac{4x}{3x} = \frac{4}{3}\)
Therefore, Left-Hand Limit (LHL) = \(\lim_{x \to 0^-} f(x) = \frac{4}{3}\)
For \(x > 0\), \(|x| = x\):
\(f(x) = \frac{5x + x}{3x} = \frac{6x}{3x} = 2\)
Therefore, Right-Hand Limit (RHL) = \(\lim_{x \to 0^+} f(x) = 2\)
Since LHL \(\neq\) RHL (\(\frac{4}{3} \neq 2\)), \(\lim_{x \to 0} f(x)\) does not exist.
Hence, the function is discontinuous at \(x = 0\).
In simple words: A function is discontinuous if the approach from the left does not match the approach from the right.

Exam Tip: Split modulus functions into separate cases for positive and negative values before evaluating limits.

Q5. Show that function \(f\), defined by \(f(x) = \begin{cases} x^2, & \text{if } x < 2 \\ 2, & \text{if } x = 2 \\ 2^x, & \text{if } x > 2 \end{cases}\), is discontinuous at \(x = 2\).
Answer:
We evaluate the limits at \(x = 2\):
Left-Hand Limit (LHL) = \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4\)
Right-Hand Limit (RHL) = \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 2^x = 2^2 = 4\)
Value of the function at \(x = 2\) is \(f(2) = 2\).
Since \(\lim_{x \to 2} f(x) = 4\) but \(f(2) = 2\), we have \(\lim_{x \to 2} f(x) \neq f(2)\).
Therefore, the function is discontinuous at \(x = 2\).
In simple words: Even though the left and right limits are equal, the function value at that point is different, causing a discontinuity (specifically, a removable discontinuity).

Exam Tip: Always verify all three conditions of continuity: LHL exists, RHL exists, and both equal the functional value \(f(a)\).

Q6. Find the value of \(k\) if \(f(x) = \begin{cases} \frac{\sin 5x}{3x}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}\), is continuous at \(x = 0\).
Answer:
Since \(f(x)\) is continuous at \(x = 0\), the limit of \(f(x)\) as \(x \to 0\) must equal \(f(0)\).
\(f(0) = k\)
\(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 5x}{3x}\)
Using the standard limit formula \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(\lim_{x \to 0} \frac{\sin 5x}{5x} \times \frac{5}{3} = 1 \times \frac{5}{3} = \frac{5}{3}\)
Equating limit to \(f(0)\):
\(k = \frac{5}{3}\)
In simple words: For a function to be continuous at a point, the defined value at that point must match the limit value.

Exam Tip: Remember and apply standard trigonometric limits like \(\lim_{x \to 0} \frac{\sin ax}{ax} = 1\) to quickly solve such problems.

Q7. Check the continuity of \(f(x) = \begin{cases} \sin x, & \text{if } x < 0 \\ x+1, & \text{if } x \ge 0 \end{cases}\).
Answer:
We check continuity at the boundary point \(x = 0\):
Left-Hand Limit (LHL) = \(\lim_{x \to 0^-} \sin x = \sin(0) = 0\)
Right-Hand Limit (RHL) = \(\lim_{x \to 0^+} (x+1) = 0 + 1 = 1\)
Since LHL \(\neq\) RHL (\(0 \neq 1\)), the limit does not exist at \(x = 0\).
Thus, the function is discontinuous at \(x = 0\). (Note: The printed answer key entry "Continuous" appears to contain a typographical error regarding this boundary case).
In simple words: Check the limit at the point where the function definition changes to determine continuity.

Exam Tip: Always test continuity at points where the definition of piecewise functions changes.

Q8. Find \(\frac{dy}{dx}\) of following functions: (i) \(y = \tan^{-1}\left(\frac{1-\cos x}{\sin x}\right)\) (ii) \(y = \sqrt{\frac{\sec x-1}{\sec x+1}}\) (iii) \(y = \log\sqrt{\frac{1+\sin x}{1-\sin x}}\) (iv) \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) (v) \(y = \log\sqrt{\frac{1+\cos^2 x}{1-e^{2x}}}\) (vi) \(y = 5^{\log(\sin x)} + \sin^x x\) (vii) \(y = (\log x)^x + x^{\log x}\) (viii) \(y = \log\left(\frac{x+\sqrt{x^2+a^2}}{x-\sqrt{x^2-a^2}}\right)\) (ix) \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)\) (x) \(y = (x^x)^x\)
Answer:
(i) \(\frac{dy}{dx} = \frac{1}{2}\)
(ii) \(\frac{dy}{dx} = \frac{1}{2}\sec^2\frac{x}{2}\)
(iii) \(\frac{dy}{dx} = \sec x\)
(iv) \(\frac{dy}{dx} = \frac{1}{2(1+x^2)}\)
(v) \(\frac{dy}{dx} = \frac{-\sin 2x}{1+\cos^2 2x} + \frac{2e^{2x}}{1-e^{2x}}\)
(vi) \(\frac{dy}{dx} = 5^{\log(\sin x)}\log 5\cdot \frac{\cos x}{\sin x} + \sin^x x(x\cot x + \log(\sin x))\)
(vii) \(\frac{dy}{dx} = (\log x)^x\left(\frac{1}{\log x}+\log(\log x)\right) + x^{\log x}\left(\frac{2\log x}{x}\right)\)
(viii) \(\frac{dy}{dx} = \frac{1}{\sqrt{x^2+a^2}} + \frac{1}{\sqrt{x^2-a^2}}\)
(ix) \(\frac{dy}{dx} = \frac{1}{2\sqrt{1-x^4}}\)
(x) \(\frac{dy}{dx} = (x^x)^x x^x(2\log x + 1)\)
In simple words: Simplify trigonometric expressions using identities or apply logarithmic differentiation where functions are raised to variable powers.

Exam Tip: Simplify inverse trigonometric functions using appropriate substitutions (like \(x = \tan\theta\) or \(x = \cos 2\theta\)) before differentiating.

Q9. (i) If \(y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\), then prove that \((1-x^2)\frac{dy}{dx} - xy = 1\).
(ii) If \(y\sqrt{1-x^2} + x\sqrt{1-y^2} = 1\), then prove that \(\frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}\).
(iii) If \(y = \sqrt{\sin x + \sqrt{\sin x + \dots}},\) then prove that \(\frac{dy}{dx} = \frac{\cos x}{2y-1}\).
(iv) If \(\sqrt{1-x^6} + \sqrt{1-y^6} = a^3(x^3-y^3)\), then prove that \(\frac{dy}{dx} = \frac{x^2}{y^2}\sqrt{\frac{1-y^6}{1-x^6}}\).
(v) If \(x = a(\cos t + \log(\tan \frac{t}{2})), y = a\sin t\), then show that \(\frac{dy}{dx} = 1\) at \(t = \frac{\pi}{4}\).
(vi) If \(y = \frac{x}{\log(a+bx)},\) then show that \(x^3\frac{d^2y}{dx^2} = \left(x\frac{dy}{dx}-y\right)^2\).
Answer:
(i) Substitute \(x = \tan\theta\), leading to \(y = \sin^{-1}(\cos 2\theta) = \frac{\pi}{2} - 2\theta\). Differentiating gives \(\frac{dy}{dx} = \frac{-2}{1+x^2}\), which satisfies the differential equation.
(ii) Put \(x = \sin A\) and \(y = \sin B\), reducing the equation to \(\sin(A+B) = 1\), hence \(A+B = \frac{\pi}{2}\). Differentiating implicitly yields the desired result.
(iii) Write \(y = \sqrt{\sin x + y}\), square both sides to get \(y^2 = \sin x + y\), and differentiate implicitly with respect to \(x\).
(iv) Substitute \(x^3 = \sin A\) and \(y^3 = \sin B\), simplify using inverse trigonometric identities, differentiate, and substitute back.
(v) Find \(\frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan(t/2)}\sec^2(t/2)\cdot \frac{1}{2}\right) = a\frac{\cos^2 t}{\sin t}\) and \(\frac{dy}{dt} = a\cos t\). Then \(\frac{dy}{dx} = \tan t\). At \(t = \frac{\pi}{4}\), \(\frac{dy}{dx} = \tan\left(\frac{\pi}{4}\right) = 1\).
(vi) Differentiate \(y\) with respect to \(x\) using quotient rule, compute \(\frac{d^2y}{dx^2}\), and substitute into both sides to verify equality.
In simple words: Use substitution methods for inverse trigonometric functions and implicit differentiation for equations where y is not isolated.

Exam Tip: For infinite series equations like part (iii), substitute the repeating part back as \(y\) to simplify before differentiating.

Q10. Find \(\frac{d^2y}{dx^2}\) if:
(i) \(x = \frac{3at}{1+t^3}, y = \frac{3at^2}{1+t^3}\)
(ii) \(x = a(\theta-\sin\theta), y = a(1-\cos\theta)\) at \(\theta = \frac{\pi}{2}\)
(iii) \(x = t^3, y = 2at^2\)
Answer:
(i) \(\frac{d^2y}{dx^2} = \frac{-\left(3t(1+t^2)\right)}{2(1+t^2)^3}\)
(ii) \(\frac{d^2y}{dx^2} = -\frac{1}{a}\)
(iii) \(\frac{d^2y}{dx^2} = \frac{-4a}{9t^4}\)
In simple words: For parametric equations, compute \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) and then differentiate with respect to t, dividing by \(\frac{dx}{dt}\).

Exam Tip: Be careful with the chain rule when finding the second derivative of parametric functions: \(\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}\).

Q11. Verify Rolle's Theorem if (i) \(f(x) = \cos 2x\) in \(0 \le x \le 2\pi\), (ii) \(f(x) = (x-a)^3(x-b)^4\) in \((a,b)\).
Answer:
(i) For \(f(x) = \cos 2x\) on \([0, 2\pi]\):
- \(f(x)\) is continuous on \([0, 2\pi]\) and differentiable on \((0, 2\pi)\).
- \(f(0) = \cos(0) = 1\) and \(f(2\pi) = \cos(4\pi) = 1\), so \(f(0) = f(2\pi)\).
- Thus, Rolle's theorem applies. There exists some \(c \in (0, 2\pi)\) such that \(f'(c) = 0\).
\(f'(x) = -2\sin 2x = 0 \implies 2x = \pi, 2\pi, 3\pi \implies c = \frac{\pi}{2}\).
(ii) For \(f(x) = (x-a)^3(x-b)^4\) on \([a,b]\):
- Function is continuous on \([a,b]\) and differentiable on \((a,b)\).
- \(f(a) = 0\) and \(f(b) = 0\), so \(f(a) = f(b)\).
- Differentiating using product rule and setting \(f'(c) = 0\) yields \(c = \frac{4a+3b}{7}\).
In simple words: Rolle's theorem states that if a smooth curve starts and ends at the same height, it must have at least one horizontal tangent in between.

Exam Tip: Verify all three conditions (continuity, differentiability, and equal endpoints) before finding \(c\).

Q12. Verify Lagrange's Mean Value Theorem if \(f(x) = \sqrt{25-x^2}\) in \([-3, 4]\).
Answer:
- \(f(x) = \sqrt{25-x^2}\) is continuous on \([-3, 4]\) and differentiable on \((-3, 4)\).
- According to LMVT, there exists at least one \(c \in (-3, 4)\) such that:
\(f'(c) = \frac{f(b)-f(a)}{b-a}\)
Here, \(a = -3\) and \(b = 4\):
\(f(-3) = \sqrt{25 - (-3)^2} = 4\)
\(f(4) = \sqrt{25 - 4^2} = 3\)
\(\frac{f(4)-f(-3)}{4 - (-3)} = \frac{3-4}{4 - (-3)} = \frac{-1}{7}\)
Derivative \(f'(x) = \frac{-x}{\sqrt{25-x^2}}\). Setting \(f'(c) = -\frac{1}{7}\):
\(\frac{-c}{\sqrt{25-c^2}} = \frac{-1}{7} \implies 7c = \sqrt{25-c^2}\)
Squaring both sides: \(49c^2 = 25 - c^2 \implies 50c^2 = 25 \implies c^2 = \frac{1}{2} \implies c = \pm\frac{1}{\sqrt{2}}\).
Since both values lie within \((-3, 4)\), LMVT is verified.
In simple words: LMVT states that at some point on the curve, the instantaneous slope equals the average slope between the endpoints.

Exam Tip: Always check whether the calculated value of \(c\) strictly falls inside the given open interval \((a, b)\).