CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 09

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 9. The Continuity And Differentiability questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Continuity And Differentiability concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Continuity And Differentiability worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Continuity And Differentiability chapter and other subjects too. Use them for better understanding of the subjects.

Rolle’s And Mean Value Theorem

Question. If \( f : [-5, 5] \rightarrow \mathbb{R} \) is differentiable and if \( f'(x) \) does not vanish anywhere, then show that \( f(-5) \neq f(5) \).
Answer: Here \( a = -5 \) and \( b = 5 \)
let us assume that \( f(-5) = f(5) \dots \{i.e.\ f(a) = f(b)\} \)
we are given ; \( f(x) \) is differentiable function
\( \therefore f(x) \) is also continuous
Hence the three conditions of Rolle's theorem are satisfied.
\( \therefore \) there must exist a value \( c \in (-5, 5) \) such that \( f'(c) = 0 \)
But we are given that \( f'(x) \) does not vanish anywhere (i.e. \( f'(x) \neq 0 \))
\( \dots \) our assumption is wrong
\( \Rightarrow f(-5) \neq f(5) \) (Proved)

Miscellaneous – Type - Questions

Question. If \( y = x \log \left(\frac{x}{a+bx}\right) \), show that \( x^3 \cdot \frac{d^2y}{dx^2} = \left(x\frac{dy}{dx} - y\right)^2 \).
Answer: \( y = x \log \left(\frac{x}{a+bx}\right) \)
\( \Rightarrow y = x[\log x - \log(a + bx)] \)
Diff w.r.t. x (product rule)
\( \frac{dy}{dx} = x \left[\frac{1}{x} - \frac{b}{a+bx}\right] + (\log x - \log(a + bx))\cdot 1 \)
\( \frac{dy}{dx} = x \left[\frac{a+bx-bx}{x(a+bx)}\right] + \log x - \log(a + bx) \)
\( \frac{dy}{dx} = \frac{a}{a+bx} + \log x - \log(a + bx) \)
Diff again
\( \frac{d^2y}{dx^2} = \frac{(a+bx)(0) - a(b)}{(a+bx)^2} + \frac{1}{x} - \frac{b}{a+bx} \)
\( \frac{d^2y}{dx^2} = \frac{-ab}{(a+bx)^2} + \frac{1}{x} - \frac{b}{a+bx} \)
\( \frac{d^2y}{dx^2} = \frac{-abx + (a+bx)^2 - b(a+bx)x}{x(a+bx)^2} \)
\( \frac{d^2y}{dx^2} = \frac{-abx + a^2 + b^2x^2 + 2abx - abx - b^2x^2}{x(a+bx)^2} \)
\( \frac{d^2y}{dx^2} = \frac{a^2}{x(a+bx)^2} \)
LHS \( x^3 \cdot \frac{d^2y}{dx^2} = x^3 \cdot \frac{a^2}{x(a+bx)^2} = \frac{a^2x^2}{(a+bx)^2} \)
Now RHS \( \left(x\cdot\frac{dy}{dx} - y\right)^2 \)
\( = \left[x \left(\frac{a}{a+bx} + \log x - \log(a + bx)\right) - x\log\left(\frac{x}{a+bx}\right)\right]^2 \)
\( = \left[\frac{ax}{a+bx} + x\log\left(\frac{x}{a+bx}\right) - x\log\left(\frac{x}{a+bx}\right)\right]^2 \)
\( = \left(\frac{ax}{a+bx}\right)^2 = \frac{a^2x^2}{(a+bx)^2} = \text{LHS Proved} \)

Question. If \( y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots \infty \), then write \( \frac{d^2y}{dx^2} \) in terms of y.
Answer: Diff w.r.t. x
\( \frac{dy}{dx} = -1 + \frac{2x}{2!} - \frac{3x^2}{3!} + \frac{4x^3}{4!} - \dots \infty \)
\( \frac{dy}{dx} = -1 + x - \frac{x^2}{2} + \frac{x^3}{6} - \dots \infty \)
\( \frac{d^2y}{dx^2} = 1 - \frac{2x}{2} + \frac{3x^2}{6} - \dots \infty \)
\( \frac{d^2y}{dx^2} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots \infty \)
\( \frac{d^2y}{dx^2} = y \) Ans.

Question. If \( y = | \log x | \). Find \( \frac{d^2y}{dx^2} \).
Answer: Now point \( \log x < 0 \) when \( 0 < x < 1 \)
\( \log x > 0 \) when \( x > 1 \)
\( \therefore y = \begin{cases} -\log x & ; \ 0 < x < 1 \\ \log x & ; \ x > 1 \end{cases} \)
Diff w.r.t x
\( \frac{dy}{dx} = \begin{cases} -\frac{1}{x} & ; \ 0 < x < 1 \\ \frac{1}{x} & ; \ x > 1 \end{cases} \)
Diff again
\( \frac{d^2y}{dx^2} = \begin{cases} \frac{1}{x^2} & ; \ 0 < x < 1 \\ -\frac{1}{x^2} & ; \ x > 1 \end{cases} \) Ans.

Question. If \( y = f\left(\frac{2x-1}{x^2+1}\right) \) and \( f'(x) = \sin(x^2) \). Find \( \frac{dy}{dx} \).
Answer: \( y = f\left(\frac{2x-1}{x^2+1}\right) \)
Diff w.r.t. x \( \quad \frac{dy}{dx} = f'\left(\frac{2x-1}{x^2+1}\right) \cdot \frac{d}{dx}\left(\frac{2x-1}{x^2+1}\right) \)
\( \frac{dy}{dx} = \sin \left(\frac{2x-1}{x^2+1}\right)^2 \cdot \left[\frac{(x^2+1)(2)-(2x-1)(2x)}{(x^2+1)^2}\right] \)
\( \dots\dots \left\{\text{since } f'(x) = \sin(x^2), f'\left(\frac{2x-1}{x^2+1}\right) = \sin\left(\frac{2x-1}{x^2+1}\right)^2\right\} \)
\( \frac{dy}{dx} = \sin \left(\frac{2x-1}{x^2+1}\right)^2 \cdot \left[\frac{2x^2+2-4x^2+2x}{(x^2+1)^2}\right] \)
\( \frac{dy}{dx} = \sin \left(\frac{2x-1}{x^2+1}\right)^2 \cdot \left(\frac{2+2x-2x^2}{(x^2+1)^2}\right) \) Ans.

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