Miscellaneous Exercise, Question 1
\((3x^2 - 9x + 5)^9\)
Show solution
- \(9(3x^2 - 9x + 5)^8(6x - 9)\).
Answer: \(27(3x^2 - 9x + 5)^8(2x - 3)\)
NCERT Solutions for Class 12 Maths Chapter 5 Miscellaneous Exercise
The Miscellaneous Exercise on Continuity and Differentiability. Mixed practice: chain rule with powers, logarithmic differentiation for variable powers, simplifying inverse trigonometric expressions before differentiating, implicit and parametric second derivatives, and the link between continuity and differentiability.
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Miscellaneous Exercise questions and solutions
Miscellaneous Exercise, Question 2
\(\sin^3 x + \cos^6 x\)
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- \(3\sin^2 x\cos x + 6\cos^5 x(-\sin x)\).
Answer: \(3\sin x\cos x(\sin x - 2\cos^4 x)\)
Miscellaneous Exercise, Question 3
\((5x)^{3\cos 2x}\)
Show solution
- \(\log y = 3\cos 2x\log 5x\): \[\dfrac{y'}{y} = -6\sin 2x\log 5x + \dfrac{3\cos 2x}{x}\]
Answer: \[(5x)^{3\cos 2x}\left[\dfrac{3\cos 2x}{x} - 6\sin 2x\log 5x\right]\]
Miscellaneous Exercise, Question 4
\(\sin^{-1}(x\sqrt x)\), \(0 \le x \le 1\)
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- \[\dfrac{1}{\sqrt{1 - x^3}} \cdot \tfrac32x^{1/2}\]
Answer: \(\dfrac{3\sqrt x}{2\sqrt{1 - x^3}}\)
Miscellaneous Exercise, Question 5
\(\dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x + 7}}\), \(-2 < x < 2\)
Show solution
- \[\left(\cos^{-1}\tfrac{x}{2}\right)' = -\dfrac{1}{\sqrt{4 - x^2}}\], \[\left(\sqrt{2x + 7}\right)' = \dfrac{1}{\sqrt{2x + 7}}\]
- Quotient rule, then split the fraction.
Answer: \[-\left[\dfrac{1}{\sqrt{4 - x^2}\sqrt{2x + 7}} + \dfrac{\cos^{-1}\frac{x}{2}}{(2x + 7)^{3/2}}\right]\]
Miscellaneous Exercise, Question 6
\(\cot^{-1}\left[\dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right]\), \(0 < x < \tfrac{\pi}{2}\)
Show solution
- \[\sqrt{1 \pm \sin x} = \cos\tfrac{x}{2} \pm \sin\tfrac{x}{2}\] (both positive for \(0 < x < \tfrac{\pi}{2}\)).
- The fraction is \[\dfrac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\tfrac{x}{2}\], so \(y = \tfrac{x}{2}\).
Answer: \(\tfrac12\)
Miscellaneous Exercise, Question 7
\((\log x)^{\log x}\), \(x > 1\)
Show solution
- \(\log y = \log x \cdot \log(\log x)\): \[\dfrac{y'}{y} = \dfrac{\log(\log x)}{x} + \log x \cdot \dfrac{1}{x\log x}\]
Answer: \[(\log x)^{\log x}\cdot\dfrac{1 + \log(\log x)}{x}\]
Miscellaneous Exercise, Question 8
\(\cos(a\cos x + b\sin x)\)
Show solution
- \[-\sin(a\cos x + b\sin x) \cdot (-a\sin x + b\cos x)\]
Answer: \[(a\sin x - b\cos x)\sin(a\cos x + b\sin x)\]
Miscellaneous Exercise, Question 9
\((\sin x - \cos x)^{\sin x - \cos x}\), \(\tfrac{\pi}{4} < x < \tfrac{3\pi}{4}\)
Show solution
- \(u = \sin x - \cos x > 0\) here; \(\log y = u\log u\), \(\dfrac{y'}{y} = u'(\log u + 1)\) with \(u' = \cos x + \sin x\).
Answer: \[(\sin x - \cos x)^{\sin x - \cos x}(\sin x + \cos x)(1 + \log(\sin x - \cos x))\]
Miscellaneous Exercise, Question 10
\(x^x + x^a + a^x + a^a\), fixed \(a > 0\), \(x > 0\)
Show solution
- \((x^x)' = x^x(1 + \log x)\); \((x^a)' = ax^{a-1}\); \((a^x)' = a^x\log a\); \(a^a\) is a constant.
Answer: \(x^x(1 + \log x) + ax^{a-1} + a^x\log a\)
Miscellaneous Exercise, Question 11
\(x^{x^2 - 3} + (x - 3)^{x^2}\), \(x > 3\)
Show solution
- \(u = x^{x^2 - 3}\): \[\dfrac{u'}{u} = 2x\log x + \dfrac{x^2 - 3}{x}\]
- \(v = (x - 3)^{x^2}\): \[\dfrac{v'}{v} = 2x\log(x - 3) + \dfrac{x^2}{x - 3}\]
Answer: \[x^{x^2 - 3}\left[\dfrac{x^2 - 3}{x} + 2x\log x\right] + (x - 3)^{x^2}\left[\dfrac{x^2}{x - 3} + 2x\log(x - 3)\right]\]
Miscellaneous Exercise, Question 12
\(y = 12(1 - \cos t),\ x = 10(t - \sin t)\), \(-\tfrac{\pi}{2} < t < \tfrac{\pi}{2}\)
Show solution
- \(\dfrac{dy}{dt} = 12\sin t\), \(\dfrac{dx}{dt} = 10(1 - \cos t)\).
- \[\dfrac{12 \cdot 2\sin\frac{t}{2}\cos\frac{t}{2}}{10 \cdot 2\sin^2\frac{t}{2}}\]
Answer: \[\dfrac{dy}{dx} = \dfrac65\cot\dfrac{t}{2}\] (\(t \ne 0\))
Miscellaneous Exercise, Question 13
\(y = \sin^{-1}x + \sin^{-1}\sqrt{1 - x^2}\), \(0 < x < 1\)
Show solution
- For \(0 < x < 1\), \(\sin^{-1}\sqrt{1 - x^2} = \cos^{-1}x\) (put \(x = \cos\theta\), \[\theta \in \left(0, \tfrac{\pi}{2}\right)\]).
- So \[\begin{aligned}y &= \sin^{-1}x + \cos^{-1}x \\ &= \tfrac{\pi}{2}\end{aligned}\], a constant.
Answer: \(\dfrac{dy}{dx} = 0\)
Miscellaneous Exercise, Question 14
\(x\sqrt{1 + y} + y\sqrt{1 + x} = 0\), \(-1 < x < 1\). Prove \(\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}\).
Show solution
- \(x\sqrt{1 + y} = -y\sqrt{1 + x}\); squaring, \(x^2(1 + y) = y^2(1 + x)\), i.e. \(x^2 - y^2 = xy(y - x)\), so \((x - y)(x + y + xy) = 0\).
- \(x = y\) would give \(2x\sqrt{1 + x} = 0\), only \(x = 0\); so on the curve \(x + y + xy = 0\), \(y = -\dfrac{x}{1 + x}\).
- \[\dfrac{dy}{dx} = -\dfrac{(1 + x) - x}{(1 + x)^2}\]
Answer: Proved
Miscellaneous Exercise, Question 15
\((x - a)^2 + (y - b)^2 = c^2\), \(c > 0\). Prove \(\dfrac{\left[1 + (y')^2\right]^{3/2}}{y''}\) is a constant independent of a and b.
Show solution
- \[\begin{aligned}&2(x - a) + 2(y - b)y' = 0 \\ \Rightarrow\ &y' = -\dfrac{x - a}{y - b}\end{aligned}\]
- \[\begin{aligned}y'' &= -\dfrac{(y - b) - (x - a)y'}{(y - b)^2} \\ &= -\dfrac{(y - b)^2 + (x - a)^2}{(y - b)^3} \\ &= -\dfrac{c^2}{(y - b)^3}\end{aligned}\]
- \(1 + (y')^2 = \dfrac{c^2}{(y - b)^2}\), so \[\left[1 + (y')^2\right]^{3/2} = \dfrac{c^3}{|y - b|^3}\] and the ratio is \(-c\) (for \(y > b\); \(+c\) on the lower half).
Answer: The ratio is \(-c\) (magnitude c), independent of a and b
Miscellaneous Exercise, Question 16
\(\cos y = x\cos(a + y)\), \(\cos a \ne \pm1\). Prove \(\dfrac{dy}{dx} = \dfrac{\cos^2(a + y)}{\sin a}\).
Show solution
- \(x = \dfrac{\cos y}{\cos(a + y)}\).
- \[\begin{aligned}\dfrac{dx}{dy} &= \dfrac{-\sin y\cos(a + y) + \cos y\sin(a + y)}{\cos^2(a + y)} \\ &= \dfrac{\sin(a + y - y)}{\cos^2(a + y)} \\ &= \dfrac{\sin a}{\cos^2(a + y)}\end{aligned}\]
- Invert.
Answer: Proved
Miscellaneous Exercise, Question 17
\(x = a(\cos t + t\sin t)\), \(y = a(\sin t - t\cos t)\). Find \(\dfrac{d^2y}{dx^2}\).
Show solution
- \(\dfrac{dx}{dt} = at\cos t\), \(\dfrac{dy}{dt} = at\sin t\), so \(\dfrac{dy}{dx} = \tan t\).
- \[\begin{aligned}\dfrac{d^2y}{dx^2} &= \dfrac{d}{dt}(\tan t) \cdot \dfrac{dt}{dx} \\ &= \sec^2 t \cdot \dfrac{1}{at\cos t}\end{aligned}\]
Answer: \[\dfrac{d^2y}{dx^2} = \dfrac{\sec^3 t}{at}\], \(0 < t < \tfrac{\pi}{2}\)
Miscellaneous Exercise, Question 18
\(f(x) = |x|^3\). Show \(f''(x)\) exists for all x and find it.
Show solution
- \(f(x) = x^3\) for \(x \ge 0\) and \(-x^3\) for \(x < 0\), so \(f'(x) = 3x^2\) for \(x > 0\), \(-3x^2\) for \(x < 0\); at 0 both one-sided derivatives are \(\lim \dfrac{|h|^3}{h} = 0\). So \(f'(x) = 3x|x|\).
- \(f''(x) = 6x\) for \(x > 0\), \(-6x\) for \(x < 0\); at 0, \(\lim\dfrac{3h|h|}{h} = 0\) from both sides.
Answer: \(f''(x) = 6|x|\) (i.e. \(6x\) for \(x \ge 0\), \(-6x\) for \(x < 0\))
Miscellaneous Exercise, Question 19
Using \(\sin(A + B) = \sin A\cos B + \cos A\sin B\) and differentiation, obtain the sum formula for cosines.
Show solution
- Treat B as a constant and differentiate both sides with respect to A.
- Left: \(\cos(A + B)\). Right: \(\cos A\cos B - \sin A\sin B\).
Answer: \[\cos(A + B) = \cos A\cos B - \sin A\sin B\]
Miscellaneous Exercise, Question 20
Is there a function continuous everywhere but not differentiable at exactly two points? Justify.
Show solution
- Yes. Take \(f(x) = |x| + |x - 1|\): a sum of continuous functions, so continuous everywhere.
- At 0 the left and right derivatives are \(-2\) and \(0\); at 1 they are \(0\) and \(2\). Everywhere else f is linear near the point, so differentiable.
Answer: Yes, e.g. \(f(x) = |x| + |x - 1|\) (not differentiable only at 0 and 1)
Miscellaneous Exercise, Question 21
\(y = \begin{vmatrix}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{vmatrix}\). Prove \(\dfrac{dy}{dx} = \begin{vmatrix}f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c\end{vmatrix}\).
Show solution
- Expand along row 1: \[y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)\]; the brackets are constants.
- \[y' = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)\], which is the expansion of the stated determinant along row 1.
Answer: Proved
Miscellaneous Exercise, Question 22
\(y = e^{a\cos^{-1}x}\), \(-1 \le x \le 1\). Show \((1 - x^2)y'' - xy' - a^2y = 0\).
Show solution
- \(y' = -\dfrac{ay}{\sqrt{1 - x^2}}\), so \(\sqrt{1 - x^2}\,y' = -ay\) and \((1 - x^2)(y')^2 = a^2y^2\).
- Differentiate: \((1 - x^2)2y'y'' - 2x(y')^2 = 2a^2yy'\); divide by \(2y'\) (non-zero).
Answer: Shown
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