Question 23:

If $f (x) = \sqrt{x^{2} - 10 x + 25}$ , then the derivative of f (x) in the interval [0, 7] is
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(d) none of these

$We have, f (x) = \sqrt{x^{2} - 10 x + 25} = \sqrt{{(x - 5)}^{2}} = |x - 5| = \{\begin{cases} x - 5 for x > 5 \\ - (x - 5) for x < 5 \end{cases} LHD = \lim_{x \to 5^{-}} \frac{f (x) - f (a)}{x - a} = \lim_{x \to 5^{-}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5} = \lim_{x \to 5^{-}} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^{-}} \frac{- (x - 5)}{x - 5} = - 1 RHD = \lim_{x \to 5^{+}} \frac{f (x) - f (a)}{x - a} = \lim_{x \to 5^{+}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5} = \lim_{x \to 5^{+}} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^{+}} \frac{x - 5}{x - 5} = 1 Here, LHD \neq RHD Thus, the functionis not differentiable at x = 5$

Page No 10.119:

Question 24:

If $f (x) = |x - 3| and g (x) = f o f (x)$ , then for x > 10, g ' (x) is equal to
(a) 1
(b) −1
(c) 0
(d) none of these

$Let, y = \sin^{- 1} \{\frac{\sin x + \cos x}{\sqrt{2}}\} \Rightarrow y = \sin^{- 1} \{\sin x (\frac{1}{\sqrt{2}}) + \cos x (\frac{1}{\sqrt{2}})\} \Rightarrow y = \sin^{- 1} \{\sin x \cos \frac{π}{4} + \cos x \sin \frac{π}{4}\} \Rightarrow y = \sin^{- 1} \{\sin (x + \frac{π}{4})\} . . . (i) Here, \frac{- 3 π}{4} < x < \frac{π}{4} \Rightarrow \frac{- 3 π}{4} + \frac{π}{4} < x + \frac{π}{4} < \frac{π}{4} + \frac{π}{4} \Rightarrow \frac{- π}{2} < x + \frac{π}{4} < \frac{π}{2} From (i) we get, \Rightarrow y = x + \frac{π}{4} [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [\frac{- π}{2}, \frac{π}{2}]] Differentiating it with respect to x, \frac{d y}{d x} = 1 + 0 ∴ \frac{d y}{d x} = 1$

Page No 10.63:

Question 11:

Differentiate

$\cos^{- 1} \{\frac{\cos x + \sin x}{\sqrt{2}}\}, - \frac{π}{4} < x < \frac{π}{4}$

Answer:

$Let, y = \cos^{- 1} \{\frac{\cos x + \sin x}{\sqrt{2}}\} y = \cos^{- 1} \{(\frac{1}{\sqrt{2}}) \cos x + (\frac{1}{\sqrt{2}}) \sin x\} y = \cos^{- 1} \{\cos \frac{π}{4} \cos x + \sin \frac{π}{4} \sin x\} y = \cos^{- 1} \{\cos (\frac{π}{4} - x)\} . . . (i) Here, - \frac{π}{4} < x < \frac{π}{4} \Rightarrow \frac{π}{4} > - x > - \frac{π}{4} \Rightarrow - \frac{π}{4} < - x < \frac{π}{4} \Rightarrow (- \frac{π}{4} + \frac{π}{4}) < (- x + \frac{π}{4}) < (\frac{π}{4} + \frac{π}{4}) \Rightarrow 0 < (\frac{π}{4} - x) < \frac{π}{2} So, from equation (i), y = \frac{π}{4} - x [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] Differentiating it with respect to x, \frac{d y}{d x} = 0 - 1 \frac{d y}{d x} = - 1$

Page No 10.63:

Question 12:

Differentiate

(i) $\tan^{- 1} \{\frac{x}{1 + \sqrt{1 - x^{2}}}\}, - 1 < x < 1$

(ii)

If $y = \sin^{- 1} (\frac{x}{1 + x^{2}}) + \cos^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}), 0 < x < \infty$ , prove that $\frac{d y}{d x} = \frac{2}{1 + x^{2}} .$

Answer:

$Let, y = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}) + \cos^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}) Put x = \tan θ ∴ y = \sin^{- 1} (\frac{\tan θ}{\sqrt{1 + \tan^{2} θ}}) + \cos^{- 1} (\frac{1}{\sqrt{1 + \tan^{2} θ}}) \Rightarrow y = \sin^{- 1} (\frac{\frac{\sin θ}{\cos θ}}{s e c θ}) + \cos^{- 1} (\frac{1}{s e c θ}) \Rightarrow y = \sin^{- 1} (\frac{\frac{\sin θ}{\cos θ}}{\frac{1}{\cos θ}}) + \cos^{- 1} (\cos θ) \Rightarrow y = \sin^{- 1} (\sin θ) + \cos^{- 1} (\cos θ) . . . (i) Here, 0 < x < \infty \Rightarrow 0 < \tan θ < \infty \Rightarrow 0 < θ < \frac{π}{2} So, from equation (i), y = θ + θ [\begin{matrix} Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}] \\ \cos^{- 1} (\cos θ) = θ, if θ \in [0, π] \end{matrix}] \Rightarrow y = 2 θ \Rightarrow y = 2 \tan^{- 1} x [Since, x = \tan θ]$

Differentiate it with respect to x,

$∴ \frac{d y}{d x} = \frac{2}{1 + x^{2}}$

Page No 10.64:

Question 37:

Differentiate the following with respect to x:

$(i) \cos^{- 1} (\sin x)$

(ii) $\cot^{- 1} (\frac{1 - x}{1 + x})$

Answer:

(i) Let, f (x) = \cos^{- 1} (\sin x) \Rightarrow f (x) = \cos^{- 1} [\cos (\frac{π}{2} - x)] \Rightarrow f (x) = \frac{π}{2} - x Thus, f' (x) = \frac{d}{d x} (\frac{π}{2} - x) = - 1

(ii)

\cot^{- 1} (\frac{1 - x}{1 + x}) = \frac{d}{d x} [\tan^{- 1} (\frac{1 + x}{1 - x})] = \frac{1}{1 + {(\frac{1 + x}{1 - x})}^{2}} \times \frac{1 - x + 1 + x}{{(1 - x)}^{2}} = \frac{{(1 - x)}^{2}}{{(1 - x)}^{2} + {(1 + x)}^{2}} \times \frac{2}{{(1 - x)}^{2}} = \frac{{(1 - x)}^{2}}{1 - 2 x + x^{2} + 1 + 2 x + x^{2}} = \frac{{(1 - x)}^{2}}{2 (1 + x^{2})} \times \frac{2}{{(1 - x)}^{2}} = \frac{1}{1 + x^{2}}

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