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MATHS M.C.Q (1 Marks)

Differentiation · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 33 questions.

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M.C.Q (1 Marks)

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33 questions · timed · auto-graded

Question 11 Mark
$\frac{\text{d}}{\text{dx}}\Big\{\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)\Big\}$ equals:
  1. $\frac{1}{2}$
  2. $-\frac{1}{2}$
  3. $1$
  4. $-1$
Answer
  1. $-\frac{1}{2}$
Solution:
Let $\text{u}=\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$
$\Rightarrow\text{u}=\tan^{-1}\bigg(\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}+\sin^2\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\bigg)$
$\Rightarrow\text{u}=\tan^{-1}\frac{\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\big)\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^2}$
$\Rightarrow\text{u}=\tan^{-1}\bigg(\frac{\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}}\bigg)$
dividing by $\cos\frac{\text{x}}{2}$
$\Rightarrow\text{u}=\tan^{-1}\bigg[\frac{1-\tan\frac{\text{x}}{2}}{1+\tan\frac{\text{x}}{2}}\bigg]$
$\Rightarrow\text{u}=\tan^{-1}\bigg[\frac{\tan\frac{\pi}{2}-\tan\frac{\text{x}}{2}}{1+\tan\frac{\text{x}}{2}\times\tan\frac{\text{x}}{2}}\bigg]$
$\Rightarrow\text{u}=\tan^{-1}\Big[\tan\Big(\frac{\pi}{2}-\frac{\text{x}}{2}\Big)\Big]$
$\Rightarrow\text{u}=\frac{\pi}{4}-\frac{\text{x}}{2}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=0-\Big(\frac{1}{2}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=-\frac{1}{2}$
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Question 21 Mark
Differential coefficient of $\sec(\tan^{-1}\text{x})$ is:
  1. $\frac{\text{x}}{1+\text{x}^2}$
  2. $\text{x}\sqrt{1+\text{x}^2}$
  3. $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
  4. $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
Answer
  1. $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
Solution:
We have, $\text{y}=\sec(\tan^{-1}\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sec\big(\tan^{-1}\text{x}\big)\tan\big(\tan^{-1}\text{x}\big)\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sec\big(\tan^{-1}\text{x}\big)\tan\big(\tan^{-1}\text{x}\big)\times\frac{1}{\sqrt{1+\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)\text{y}$
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Question 31 Mark
Let $\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{V}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ then $\frac{\text{dU}}{\text{dV}}=$
  1. $\frac{1}{2}$
  2. $\text{x}$
  3. $\frac{1-\text{x}^2}{\text{x}^2-4}$
  4. $1$
Answer
  1. 1
Solution:
$\text{U}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{ and V}=\tan^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Put, $\text{x}=\tan\theta$
$\text{U}=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)\text{ and V}=\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)$
$\text{U}=\sin^{-1}(\sin2\theta)\text{ and V}=\tan^{-1}(\tan2\theta)$
$\text{U}=2\theta\text{ and V}=2\theta$
$\text{U}=2\tan^{-1}\text{ x and V}=2\tan^{-1}\text{x}$
$\frac{\text{dU}}{\text{dx}}=\frac{\text{dV}}{\text{dx}}=\frac{2}{1+\text{x}^2}$
$\frac{\text{dU}}{\text{dV}}=\frac{\frac{\text{dU}}{\text{dx}}}{\frac{\text{dU}}{\text{dx}}}=1$
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Question 41 Mark
If $\text{f}(\text{x})=\tan^{-1}\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}},0\leq\text{x}\leq\frac{\pi}{2},$ then $\text{f}'\Big(\frac{\pi}{6}\Big)$ is:
  1. $-\frac{1}{4}$
  2. $-\frac{1}{2}$
  3. $\frac{1}{4}$
  4. $\frac{1}{2}$
Answer
  1. $\frac{1}{2}$
Solution:
$\text{f}(\text{x})=\tan^{-1}\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
$\text{f}(\text{x})=\tan^{-1}\sqrt{\frac{\Big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)^2}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)^2}}$
$\text{f}(\text{x})=\tan^{-1}\frac{\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}}$
$\text{f}(\text{x})=\tan^{-1}\Bigg(\tan\bigg(\frac{1+\tan\frac{\text{x}}{2}}{1+\tan\frac{\text{x}}{2}}\bigg)\Bigg)$
$\text{f}(\text{x})=\tan^{-1}\Big(\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big)$
$\text{f}(\text{x})=\frac{\pi}{4}+\frac{\text{x}}{2}$
$\text{f}'(\text{x})=\frac{1}{2}$
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Question 51 Mark
The derivative of $\cos^{-1}(2\text{x}^2-1)$ with respect to $\cos^{-1}\text{x}$ is:
  1. $2$
  2. $\frac{1}{2\sqrt{1+\text{x}^2}}$
  3. $\frac{2}{\text{x}}$
  4. $1-\text{x}^2$
Answer
  1. 2
Solution:
Let, $\text{u}=\cos^{-1}(2\text{x}^2-1)\text{ and v}=\cos^{-1}\text{x}$
Put $\text{x}=\cos\theta\Rightarrow\theta=\cos^{-1}\text{x}$
$\text{u}=\cos^{-1}(2\cos^2\theta-1)\text{ and v}=\cos^-1(\cos\theta)$
$\text{u}=\cos^{-1}(\cos2\theta)\text{ and v}=\theta$
$\text{u}=2\theta$
$\text{u}=2\cos^{-1}\text{x}\text{ and v}=\cos^{-1}\text{x}$
$\frac{\text{du}}{\text{dx}}=\frac{-2}{\sqrt{1-\text{x}^2}}\text{ and }\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{du}}{\text{dv}}=\frac{\frac{-2}{\sqrt{1-\text{x}^2}}}{\frac{-1}{\sqrt{1-\text{x}^2}}}=2$
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MCQ 61 Mark
If $x^y = e^{x-y}$ then $\frac{\text{dy}}{\text{dx}}$ is:
  • A
    $\frac{1+\text{x}}{1+\log\text{x}}$
  • B
    $\frac{1-\log\text{x}}{1+\log\text{x}}$
  • C
    $\text{Not defined.}$
  • ✓
    $\frac{\log\text{x}}{(1+\log\text{x})^2}$
Answer
Correct option: D.
$\frac{\log\text{x}}{(1+\log\text{x})^2}$
We have, $x^y = e^{x-y}$
Taking $\log$ on both sides we get,
$\Rightarrow\text{y}\log\text{x}=(\text{x}-\text{y})\log)_\text{e}\text{e}$
$\Rightarrow\text{y}\log\text{x}=\text{x}-\text{y}$
$\Rightarrow\text{y}\log\text{x}+\text{y}=\text{x}$
$\Rightarrow\text{y}(1+\log\text{x})=\text{x}$
$\Rightarrow\text{y}=\frac{\text{x}}{(1+\log\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{x})\times1-\text{x}\times\Big(1+\frac{1}{\text{x}}\Big)}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\log\text{x}-1}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{(1+\log\text{x})^2}$
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Question 71 Mark
The derivative of the function $\cot^{-1}\Big|(\cos2\text{x})^{\frac{1}{2}}\Big|\text{ at }\text{x}=\frac{\pi}{6}$ is:
  1. $\Big(\frac{2}{3}\Big)^\frac{1}{2}$
  2. $\Big(\frac{1}{3}\Big)^\frac{1}{2}$
  3. $3^\frac{1}{2}$
  4. $6^\frac{1}{2}$
Answer
  1. $\Big(\frac{2}{3}\Big)^\frac{1}{2}$
Solution:
We have, $\text{y}=\cot^{-1}\Big(\sqrt{\cos2\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{1+\cot2\text{x}}\frac{\text{d}}{\text{dx}}\sqrt{\cos2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\frac{\text{d}}{\text{dx}}(\cos2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\cos^2\text{x}}\times\frac{1}{2\sqrt{\cos2\text{x}}}\times-2\sin2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin2\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\sin\text{x}\cos\text{x}}{\cos^2\text{x}\times2\sqrt{\cos2\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\tan\text{x}}{\sqrt{\cos2\text{x}}}$
So, at $\text{x}=\frac{\pi}{6},$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{6}}=\frac{\tan\big(\frac{\pi}{6}\big)}{\sqrt{\cos2\big(\frac{\pi}{6}\big)}}=\frac{\big(\frac{1}{\sqrt{3}}\big)}{\sqrt{\frac{1}{2}}}=\Big(\frac{2}{3}\Big)^\frac{1}{2}$
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Question 81 Mark
If $\text{x}=\text{a}\cos^3\theta,\text{y}=\text{a}\sin^3,$ then $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=$
  1. $\tan^2\theta$
  2. $\sec^2\theta$
  3. $\sec^2\theta$
  4. $|\sec\theta|$
Answer
  1. $|\sec\theta|$
Solution:
We have, $\text{x}=\text{a}\cos^2\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\frac{\text{d}}{\text{d}\theta}(\cos^2\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\cos^2\theta\frac{\text{d}}{\text{d}\theta}(\cos\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta\ .....(\text{i})$
And,
$\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}\frac{\text{d}}{\text{d}\theta}(\sin^3\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\sin^2\theta\frac{\text{d}}{\text{d}\theta}(\sin\theta)$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=3\text{a}\sin^2\theta\cos\theta\ .....(\text{ii})$
Dividing (ii) by (i), we get,
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\theta}{-\cos\theta}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\tan\theta$
Now, $\sqrt{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}=\sqrt{1+\tan^2\theta}$
$=\sqrt{\sec^2\theta}=|\sec\theta|$
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Question 91 Mark
If $\text{y}=\sqrt{\sin\text{x}+\text{y}},$ then $\frac{\text{dy}}{\text{dx}}$ equals:
  1. $\frac{\cos\text{x}}{2\text{y}-1}$
  2. $\frac{\cos\text{x}}{1-2\text{y}}$
  3. $\frac{\sin\text{x}}{1-2\text{y}}$
  4. $\frac{\sin\text{x}}{2\text{y}-1}$
Answer
  1. $\frac{\cos\text{x}}{2\text{y}-1}$
Solution:
$\text{y}=\sqrt{\sin\text{x}+\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\times\Big(\cos\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}+\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\Big(\frac{2\text{y}-1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}-1}$
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Question 101 Mark
If $\text{f}\text{(x)}=\sqrt{\text{x}^2-10\text{x}+25},$ then the derivative of f(x) in the intereval [0, 7] is:
  1. 1
  2. -1
  3. 0
  4. None of these.
Answer
  1. None of these.
Solution:
$\text{f}(\text{x})=\sqrt{\text{x}^2-10\text{x}+25}$
$\text{f}(\text{x})=\sqrt{(\text{x}-5)^2}=|\text{x}-5|$
$\text{f}(\text{x})=\text{x}-5\ \text{x}\geq5$
$=-(\text{x}-5)\ \text{x}<5$
$\text{f}'\text{(x)}=1\ \text{x}\geq5$
$=-1\text{ x}<5$
Hence, we can not define derivative of the function on [0, 7].
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Question 111 Mark
If $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}=$
  1. 2
  2. -2
  3. 1
  4. -1
Answer
  1. -1
Solution:
We have, $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\frac{1}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\cos(\text{x}+\text{y})+\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}-\cos(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$
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Question 121 Mark
Let $\text{y}=\sqrt{\sin\text{x}+\text{y}},$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{\sin\text{x}}{2\text{y}-1}$
  2. $\frac{\sin\text{x}}{1-2\text{y}}$
  3. $\frac{\cos\text{x}}{1-2\text{y}}$
  4. $\frac{\cos\text{x}}{2\text{y}-1}$
Answer
  1. $\frac{\cos\text{x}}{2\text{y}-1}$
Solution:
$\text{y}=\sqrt{\sin\text{x}+\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\times\Big(\cos\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}+\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\Big(\frac{2\text{y}-1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}-1}$
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MCQ 131 Mark
If $f(x) = |x^2 - 9x + 20|,$ then $f(x)$ is equal to:
  • A
    $-2x + 9$ for all $\text{x}\in\text{R}$
  • B
    $2x - 9$ if $4 < x < 5$
  • ✓
    $-2x + 9,$ if $4 < x < 5$
  • D
    None of these.
Answer
Correct option: C.
$-2x + 9,$ if $4 < x < 5$
We have, $f(x) = |x^2 - 9x + 20|$
$\text{f}(\text{x})=\begin{Bmatrix} \text{x}^2-9\text{x}+20, & -\infty<\text{x}\leq4 \\ -\big(\text{x}^2-9\text{x}+20\big), & 4<\text{x}<5 \\ \text{x}^2-9\text{x}+20, & 5\leq\text{x}<\infty \end{Bmatrix}$
$\Rightarrow\text{f}'(\text{x})=\begin{Bmatrix} 2\text{x}-9\text{x}, & -\infty<\text{x}\leq4 \\ 2\text{x}-9, & 4<\text{x}<5 \\ 2\text{x}-9, & 5\leq\text{x}<\infty \end{Bmatrix}$
$\therefore f'(x) = -2x + 9$ for $4 < x < 5$
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Question 141 Mark
If f(x) = |x - 3| and g(x) = fof(x), then for x > 10, g'(x) is equal to:
  1. 1
  2. -1
  3. 0
  4. None of these.
Answer
  1. 0
Solution:
For, x > 10
f(x) = |x - 3| = x - 3
g(x) = fof (x) = |x - 3| - 3
= x - 3 - 3
= x - 6
$\therefore$ g'(x) = 1
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Question 151 Mark
If $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x},$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
  2. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
  3. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log(\text{x}+1)-\frac{\text{x}}{\text{x}+1}\Big\}$
  4. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}\Big\}$
Answer
  1. $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
Solution:
Let $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}$
Taking log on both sides,
$\log\text{y}=\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\bigg(\frac{1}{1+\frac{1}{\text{x}}}\bigg)\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\times\frac{\text{x}}{\text{x}+1}\Big(-\frac{1}{\text{x}^2}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{x}+1}\times-\frac{1}{\text{x}^2}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)-\frac{-1}{\text{x}+1}\Big]$
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Question 161 Mark
$\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$ equals:
  1. $\frac{\text{x}^2-1}{\text{x}^2-4}$
  2. $1$
  3. $\frac{\text{x}^2+1}{\text{x}^2-4}$
  4. $\text{e}^\text{x}\frac{\text{x}^2-1}{\text{x}^2-4}$
Answer
  1. $\frac{\text{x}^2-1}{\text{x}^2-4}$
Solution:
Let, $\text{y}=\frac{\text{d}}{\text{dx}}\bigg[\log\bigg\{\text{e}^\text{x}\Big(\frac{\text{x}-2}{\text{x}+2}\Big)^\frac{3}{4}\bigg\}\bigg]$
$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\log\text{e}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$
$\Rightarrow\text{y}=\frac{\text{d}}{\text{dx}}\Big[\text{x}+\frac{3}{4}\log\Big(\frac{\text{x}-2}{\text{x}+2}\Big)\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3}{4\Big(\frac{\text{x}-2}{\text{x}+2}\Big)}\times\frac{(\text{x}+2)\times1-(\text{x}-2)\times1}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{\text{x}+2-\text{x}+2}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{d}}{\text{dx}}=1+\frac{3(\text{x}+2)}{4(\text{x}-2)}\times\frac{4}{(\text{x}+2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\frac{3}{(\text{x}^2-4)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-4+3}{\text{x}^2-4}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-1}{\text{x}^2-4}$
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Question 171 Mark
Let $\sin\text{y}=\text{x}\sin(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is:
  1. $\frac{\sin\text{a}}{\sin\text{a}\sin^2(\text{a}+\text{y})}$
  2. $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
  3. $\sin\text{a}\sin^2(\text{a}+\text{y})$
  4. $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
Answer
  1. $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
Solution:
We have, $\sin\text{y}=\text{x}\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big\{\sin(\text{a}+\text{y})\big\}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\times1+\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})+\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\big\{\cos\text{y}-\text{x}\cos(\text{a}+\text{y})\big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\Big\{\cos\text{y}-\frac{\sin\text{y}}{\sin(\text{a}+\text{y})}\times\cos(\text{a}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\begin{bmatrix} \because \sin\text{y}=2\sin\text{c}\cos\text{x} \\ \therefore\text{x}=\frac{\sin\text{y}}{\sin(\text{a}+\text{y})} \end{bmatrix}$
$\Rightarrow\Big\{\frac{\sin(\text{a}+\text{y})\cos\text{y}-\sin\text{y}\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\sin(\text{a}+\text{y}-\text{y})}{\sin(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
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Question 181 Mark
If $\sin\text{y}=\text{x}\cos(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  1. $\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
  2. $\frac{\cos\text{a}}{\cos^2(\text{a}+\text{y})}$
  3. $\frac{\sin^2\text{y}}{\cos\text{a}}$
  4. $\text{None of these.}$
Answer
  1. $\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
Solution:
We have, $\sin\text{y}=\text{x}\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})\big]$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=1\times\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\big[\cos\text{y}+\text{x}\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\cos\text{y}+\frac{\sin\text{y}}{\cos(\text{a}+\text{y})}\times\sin(\text{a}+\text{y})\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\begin{bmatrix}\because\sin\text{y}=\text{x}\cos(\text{a}+\text{y}) \\ \because\text{x}=\frac{\sin\text{y}}{\cos(\text{a}+\text{y})} \end{bmatrix}$
$\Rightarrow\Big[\frac{\cos(\text{a}+\text{y})\cos\text{y}+\sin\text{y}\sin(\text{a}+\text{y})}{\cos(\text{a}+\text{y})}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\cos(\text{a}+\text{y}-\text{y})}{\cos(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
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Question 191 Mark
If $\text{y}=\log\sqrt{\tan\text{x}},$ then the value of $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$ is givne by:
  1. $\infty$
  2. $1$
  3. $0$
  4. $\frac{1}{2}$
Answer
  1. 1
Solution:
We have, $\text{y}=\log\sqrt{\tan\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{\tan\text{x}}}\times\frac{1}{2\sqrt{\tan\text{x}}}\times\frac{\text{d}}{\text{dx}}\big(\sqrt{\tan\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sec^2\text{x}}{2\tan\text{x}}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\Big[\sec\big(\frac{\pi}{4}\big)\Big]^2}{2\tan\big(\frac{\pi}{4}\big)}=\frac{2}{2\times1}=1$
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Question 201 Mark
If $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}^3(\text{x}^3-\text{y})^3,$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  1. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
  2. $\frac{\text{y}^2}{\text{x}^2}\sqrt{\frac{1-\text{y}^6}{1+\text{x}^6}}$
  3. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{x}^6}{1-\text{y}^6}}$
  4. $\text{None of these.}$
Answer
  1. $\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
Solution:
We have, $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}(\text{x}^3-\text{y}^3)$
Putting $\text{x}^3=\sin\text{A}\text{ and y}^3=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin2\text{B}}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{b}}{2}\Big) \\ =2\text{a}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)$
$\Rightarrow\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{a}$
$\Rightarrow\frac{\text{A}+\text{B}}{2}=\cot^{-1}(\text{a})$
$\Rightarrow\text{A}+\text{B}=2\cot^{-1}(\text{a})$
$\Rightarrow\sin^{-1}\text{x}^{3}-\sin^{-1}\text{y}^3=2\cot^{-1}(\text{a})$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}\times\frac{\text{d}}{\text{dx}}(\text{x}^3)-\frac{1}{\sqrt{1-\text{y}^6}}\times\frac{\text{d}}{\text{dx}}(\text{y}^3)=0$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}3\text{x}^2-\frac{1}{\sqrt{1-\text{y}^6}}\times3\text{y}^2\times\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
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MCQ 211 Mark
If $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1},$ the $f\ '(x)$ is equal to:
  • A
    $1$
  • ✓
    $0$
  • C
    $x^{l+m+n}$
  • D
    None of these.
Answer
Correct option: B.
$0$
We have $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}-\text{m})(\text{l}+\text{m})}\times\text{x}^{(\text{m}-\text{n})(\text{m}+\text{n})}\times\text{x}^{(\text{n}-\text{l})(\text{n}-\text{l})}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{\text{l}^2-\text{m}^2}\times\text{x}^{\text{m}^2-\text{n}^2}\times\text{x}^{\text{n}^2-\text{l}^2}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}^2-\text{m}^2+\text{m}^2-\text{n}^2+\text{n}^2-\text{l}^2)}$
$\Rightarrow\text{f}(\text{x})=\text{x}^0$
$\Rightarrow\text{f}(\text{x})=1$
$\Rightarrow\text{f}\ '(\text{x})=0$
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Question 221 Mark
The differential coefficient of $\text{f}(\log\text{x})$ w.r.t. x, where $\text{f(x)}=\log\text{x}$ is:
  1. $\frac{\text{x}}{\log\text{x}}$
  2. $\frac{\log\text{x}}{\text{x}}$
  3. $(\text{x}\log\text{x})^{-1}$
  4. $\text{None of these.}$
Answer
  1. $(\text{x}\log\text{x})^{-1}$
Solution:
$\text{f(x)}=\log\text{x}$
$\Rightarrow\text{f}(\log\text{x})=\log(\log\text{x})$
$\Rightarrow\text{f}'(\log\text{x})=\frac{1}{\text{x}\log\text{x}}=(\text{x}\log\text{x})^{-1}$
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Question 231 Mark
The derivative of $\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$ w.r.t. $\sqrt{1+3\text{x}}$ at $\text{x}=\frac{-1}{3}$
  1. $\text{Does not exist.}$
  2. $0$
  3. $\frac{1}{2}$
  4. $\frac{1}{3}$
Answer
  1. $\text{Does not exist.}$
Solution:
Put, $\text{u}=\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$ and $\text{v}=\sqrt{1+3\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+3\text{x}}}\times3$
But at $\text{x}=\frac{-1}{3}\frac{\text{dv}}{\text{dx}}$ does not exist
Hence, derivative of $\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$
With respect to $\sqrt{1+3\text{x}}$ does not exist.
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Question 241 Mark
For the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=1,\frac{\text{dy}}{\text{dx}}$ at $\Big(\frac{1}{4},\frac{1}{4}\Big)$ is:
  1. $\frac{1}{2}$
  2. 1
  3. -1
  4. 0
Answer
  1. -1
Solution:
We have, $\sqrt{\text{x}}+\sqrt{\text{y}}=1$
Differentiating with respect to x, we get,
$\frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}\times\frac{2\sqrt{\text{y}}}{1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\sqrt{\text{y}}}{\sqrt{\text{x}}}$
Now, $\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\Big(\frac{1}{4},\frac{1}{4}\Big)}=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$
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Question 251 Mark
If $\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  1. $\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}$
  2. $\frac{\text{y}}{\text{x}}$
  3. $\frac{\text{x}}{\text{y}}$
  4. $\text{None of these.}$
Answer
  1. $\frac{\text{y}}{\text{x}}$
Solution:
$\sin^{-1}\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}\Big)=\log\text{a}$
$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}=\sin(\log\text{a})=\text{k}$
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
$\text{x}^2-\text{y}^2=\text{kx}^2+\text{ky}^2$
$\text{x}^2-\text{kx}^2=\text{ky}^2+\text{y}^2$
$(1-\text{k})\text{x}^2=(\text{x}+1)\text{y}^2$
$\frac{1-\text{x}}{\text{k}+1}=\frac{\text{y}^2}{\text{x}^2}\ .....(\text{i})$
Consider,
$\text{x}^2-\text{y}^2=\text{k}(\text{x}^2+\text{y}^2)$
Differentaiting with resepect to x,
$2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=2\text{kx}+2\text{ky}\frac{\text{dy}}{\text{dx}}$
$2\text{x}-2\text{kx}=2\text{ky}\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}$
$2\text{x}(1-\text{k})=2\text{y}(\text{k}+1)\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}(1-\text{k})}{\text{y}(\text{k}+1)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{y}^2}{\text{x}^2}\ .....(\because\text{from(i)})$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
 
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Question 261 Mark
If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{4\text{x}^3}{1-\text{x}^4}$
  2. $-\frac{4\text{x}}{1-\text{x}^4}$
  3. $\frac{1}{4-\text{x}^4}$
  4. $-\frac{4\text{x}^3}{1-\text{x}^4}$
Answer
  1. $-\frac{4\text{x}}{1-\text{x}^4}$
Solution:
We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{x}^2}{1-\text{x}^2}\Big[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1-\text{x}^2}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{1-\text{x}^4}$
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Question 271 Mark
If $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9},$ then f'(x) is equal to:
  1. 1 for x < -3
  2. -1 for x < -3
  3. 1 for all $\text{x}\in\text{R}$
  4. None of these.
Answer
  1. -1 for x < -3
Solution:
We have, $\text{f}(\text{x})=\sqrt{\text{x}^2+6\text{x}+9}$
$=\sqrt{(\text{x}+3)^2}$
$=|\text{x}+3|$
$\text{f}(\text{x})=\begin{cases}\text{x}+ 3,\text{x}\geq-3\\ -\text{x}-3,\text{x}<-3\end{cases}$
$\Rightarrow\text{f}'(\text{x})=\begin{cases} 1,\text{x}\geq-3\\ -1,\text{x}<-3\end{cases}$
$\therefore$ f'(x) = -1 for x < -3
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Question 281 Mark
Let $3\sin(\text{xy})+4\cos(\text{xy})=5,$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $-\frac{\text{y}}{\text{x}}$
  2. $\frac{3\sin(\text{xy})+4\cos(\text{xy})}{3\cos(\text{xy})-4\sin(\text{xy})}$
  3. $\frac{3\cos(\text{xy})+4\sin(\text{xy})}{4\cos(\text{xy})-3\sin(\text{xy})}$
  4. $\text{None of these.}$
Answer
  1. $-\frac{\text{y}}{\text{x}}$
Solution:
We have, $3\sin(\text{xy})+4\cos(\text{xy})=5$
$\Rightarrow3\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]-4\sin(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]\big[3\cos(\text{xy})-4\sin(\text{xy})\big]=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
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Question 291 Mark
If $\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}},$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  1. 1
  2. $(\text{a}+\text{b}+\text{c})^{\text{x}^{\text{a}+\text{b}+\text{c}-1}}$
  3. 0
  4. None of these.
Answer
  1. 0
Solution:
$\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}$
$\text{y}=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}+\frac{\text{x}^\text{c}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{c}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}}}$
$\text{y}=\frac{\text{x}^\text{b}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=\frac{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=1$
$\frac{\text{dy}}{\text{dx}}=0$
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Question 301 Mark
If $\text{y}=\tan^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big),$ then $\frac{\text{dy}}{\text{dx}}$ is equals to:
  1. $\frac{1}{2}$
  2. 0
  3. 1
  4. None of these.
Answer
  1. 1
Solution:
We have, $\text{y}=\tan^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}\times \\ \bigg[\frac{(\cos\text{x}-\sin\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})-(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x}-\sin\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times\\ \bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(-\sin\text{x}-\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times\\ \bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})(-\sin\text{x}+\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}\times\frac{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}{(\cos\text{x}-\sin\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
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MCQ 311 Mark
If $f(x) = 4x^8,$ then:
  • A
    $\text{f}'\Big(\frac{1}{2}\Big)=\text{f}'\Big(-\frac{1}{2}\Big)$
  • B
    $\text{f}\Big(\frac{1}{2}\Big)=-\text{f}'\Big(-\frac{1}{2}\Big)$
  • ✓
    $\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
  • D
    $\text{f}\Big(\frac{1}{2}\Big)=\text{f}'\Big(-\frac{1}{2}\Big)$
Answer
Correct option: C.
$\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
$f(x) = 4x^8$
$\Rightarrow\text{f}\Big(\frac{1}{2}\Big)=4\Big(\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$
$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=4\Big(-\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$
$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
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Question 321 Mark
If $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $-\frac{2}{1+\text{x}^2}$
  2. $\frac{2}{1+\text{x}^2}$
  3. $\frac{1}{2-\text{x}^2}$
  4. $\frac{2}{2-\text{x}^2}$
Answer
  1. $-\frac{2}{1+\text{x}^2}$
Solution:
Let $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Differentiating with respect to x using chain rule, we get,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{(1+\text{x}^2)^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{2\text{x}}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{2\text{x}(1+\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{1+\text{x}^2}$
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Question 331 Mark
If $\text{f(x)}=\log_{\text{x}^2}(\log\text{x}),$ the f(x) at x = e is:
  1. $0$
  2. $1$
  3. $\frac{1}{\text{e}}$
  4. $\frac{1}{2\text{e}}$
Answer
  1. $\frac{1}{2\text{e}}$
Solution:
We have, $\text{f(x)}=\log_{\text{x}^2}(\log\text{x})$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{\log\text{x}^2}$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{2\log\text{x}}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\frac{\text{d}}{\text{dx}}\bigg\{\frac{\log(\log\text{x})}{\log\text{x}}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}\times\log\text{x}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{x}}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{e}}-\frac{\log(\log)\text{e}}{\text{e}}}{(\log\text{e})^2}\Bigg\}$
[Putting x = e]
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\bigg\{\frac{\frac{1}{\text{e}}}{1}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2\text{e}}$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip