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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 101–106 (part 11 of 11)

  1. Fill in the blanks in each of the Exercises $\displaystyle 97$ to $\displaystyle 101$:

    Exercise 101

    For the curve x+y=1,dydx\displaystyle \sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} at (14,14)\displaystyle \left(\frac{1}{4}, \frac{1}{4}\right) is ____.

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    NCERT’s answer
    \(\displaystyle -1\)
    \(\displaystyle -1\)\[\sqrt x+\sqrt y=1 \]\[\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=0 \]\[\frac{dy}{dx}=-\frac{\sqrt y}{\sqrt x} \]\[\left.\frac{dy}{dx}\right|_{\left(\frac14,\frac14\right)}=-\frac{1/2}{1/2}=-1 \]
  2. State True or False for the statements in each of the Exercises $\displaystyle 102$ to 106.

    Exercise 102

    Rolle's theorem is applicable for the function f(x)=∣x−1∣\displaystyle f(x)=|x-1| in [0\displaystyle 0, 2\displaystyle 2].

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    NCERT’s answer
    False
    False\(\displaystyle f\) is continuous on \(\displaystyle [0,2]\) and \(\displaystyle f(0)=f(2)=1\), but at \(\displaystyle x=1\):NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q102\[f'(1^-)=-1 \ne 1=f'(1^+) \]\(\displaystyle f\) is not differentiable at \(\displaystyle 1\in(0,2)\), so Rolle's theorem does not apply.

  3. Exercise 103

    If f\displaystyle f is continuous on its domain D, then ∣f∣\displaystyle |f| is also continuous on D.

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    NCERT’s answer
    True
    TrueFor any \(\displaystyle c\in D\):\[\big|\,|f(x)|-|f(c)|\,\big|\le |f(x)-f(c)| \]\[f \text{ continuous at } c \;\Rightarrow\; |f(x)-f(c)|\to 0 \text{ as } x\to c \]\[\lim_{x\to c}|f(x)|=|f(c)| \]

  4. Exercise 104

    The composition of two continuous function is a continuous function.

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    NCERT’s answer
    True
    TrueLet \(\displaystyle f\) be continuous at \(\displaystyle c\) and \(\displaystyle g\) continuous at \(\displaystyle f(c)\):\[\lim_{x\to c}g\big(f(x)\big)=g\Big(\lim_{x\to c}f(x)\Big)=g\big(f(c)\big) \]So \(\displaystyle g\circ f\) is continuous at \(\displaystyle c\).

  5. Exercise 105

    Trigonometric and inverse - trigonometric functions are differentiable in their respective domain.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    False — \(\displaystyle \sin^{-1}x\) is a counterexample: it is defined on \(\displaystyle [-1,1]\) but is not differentiable at both end points.\[f(x)=\sin^{-1}x,\quad \text{domain } [-1,1] \]\[f'(x)=\frac{1}{\sqrt{1-x^{2}}},\quad |x|<1 \]\[\lim_{x\to 1^{-}}\frac{1}{\sqrt{1-x^{2}}}=\infty \]\[\Rightarrow\; f'(1)\ \text{does not exist, likewise } f'(-1) \]

  6. Exercise 106

    If f.g\displaystyle f . g is continuous at x=a\displaystyle x=a, then f\displaystyle f and g\displaystyle g are separately continuous at x=a\displaystyle x=a.

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    NCERT’s answer
    False
    False — a counterexample at \(\displaystyle x=0\) is enough.\[f(x)=g(x)=\begin{cases}1, & x\ge 0\\ -1, & x<0\end{cases} \]\[(f\cdot g)(x)=\big(f(x)\big)^{2}=1\ \text{ for all } x \;\Rightarrow\; f\cdot g \text{ is continuous at } 0 \]\[\lim_{x\to 0^{-}}f(x)=-1\neq 1=f(0) \]\[\Rightarrow\; f \text{ and } g \text{ are not continuous at } x=0 \]