M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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26 questions · auto-graded multiple-choice test.

MCQ 11 Mark

If $\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}\ \text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}(\text{t})\cos\text{t},$ then $\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=$

A
$\text{f}(\text{t})-\text{f}(\text{t})$
B
$\{\text{f}(\text{t})-\text{f}(\text{t})\}^2$
✓
$\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$
D
$\text{None of these}$

Answer

Correct option: C.

$\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$

Here,
$\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}'(\text{t})\sin\text{t}$
$\text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}'(\text{t})\cos\text{t}$
$ \Rightarrow\frac{\text{dx}}{\text{dt}}=\text{f}'(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}-\text{f}'(\text{t})\cos\text{}\text{t}$
$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}'(\text{t})\sin\text)\cos\text{t}+\text{f}'{t}+\text{f}(\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}$
$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}$
Thus
$\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=\{-\text{f}(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\Big\}^2$
$=\{\text{f}(\text{t})\sin\text{t}+\text{f}''(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\}^2$
$=\sin^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2+\cos^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$
$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2(\sin^2\text{t}+\cos^2\text{t})$
$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

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MCQ 21 Mark

If $\text{y}=\text{a}+\text{bx}^2,\text{a,b}$ arbitrary constants, then

A
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{xy}$
✓
$\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$
C
$\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}+\text{y}=0$
D
$\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{xy}$

Answer

Correct option: B.

$\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

$\text{y}=\text{a}+\text{bx}^2$
$\Rightarrow\text{y}_1=2\text{bx}$
$\Rightarrow\text{y}_2=2\text{b}$
Multiply by x on both sides we get
$\text{xy}_2=2\text{bx}=\text{y}_1$
$\Rightarrow\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

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MCQ 31 Mark

If $\text{y}=\tan^{-1}\Big\{\frac{\log(\frac{\text{e}}{\text{x}})^2}{\log(\frac{\text{e}}{\text{x}})^2}\Big\}+\tan^{-1}\Big(\frac{3-2\log,\text{x}}{1-6\log,\text{x}}\Big)$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$

A
2
B
1
✓
0
D
-1

Answer

Correct option: C.

0

$\text{y}=\tan^{-1}\Big\{\frac{\log(\frac{\text{e}}{\text{x}})^2}{\log(\frac{\text{e}}{\text{x}})^2}\Big\}+\tan^{-1}\Big(\frac{3-2\log,\text{x}}{1-6\log,\text{x}}\Big)$
$=\tan^{-1}\Big\{\frac{1-2\log_\text{e}\text{x}}{1+2\log_\text{e}\text{x}}\Big\}+\tan^{-1}\Big\{\frac{3+2\log_\text{e}\text{x}}{1-6\log_\text{e}\text{x}}\Big\}$
$=\tan^{-1}1-\tan^{-1}(2\log_\text{e}\text{x})+\tan^{-1}(3)+\tan^{-1}(2\log_\text{e}\text{x})$
$=\tan^{-1}+\tan^{-1}(3)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

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MCQ 41 Mark

If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equals to:

✓
$\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$
B
$\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^2}$
C
$\frac{\text{g}''}{\text{f}''}$
D
$\frac{\text{f}''\text{g}'-\text{g}''\text{f}'}{(\text{g}')^3}$

Answer

Correct option: A.

$\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$

$\text{x}=\text{f}(\text{t})$ $\text{y}=\text{g}(\text{t}),$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{g}'}{\text{f}'}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)\frac{\text{dt}}{\text{dx}}$
$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{\text{f}'^2}\frac{1}{\text{f}'}$
$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$

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MCQ 51 Mark

If $\text{y}=\text{x}^{\text{n}-1}\log\text{x}\ \text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1$ is equals to :

✓
$-(n - 1)^2y$
B
$(n - 1)^2y$
C
$-n^2y$
D
$n^2y$

Answer

Correct option: A.

$-(n - 1)^2y$

Here,
$\text{y}=\text{x}^{\text{n}-1}\log\text{x}$
$\Rightarrow\text{y}_1=(\text{n}-1)\text{x}^{\text{n-2}}\log\text{x}+\frac{\text{x}^{\text{n}-1}}{\text{x}}$
$\Rightarrow\text{y}_1=\frac{(\text{n}-1)\text{x}^{\text{n}-1}\log\text{x}+\text{x}^{\text{n}-1}}{\text {x}}$
$\Rightarrow\text{xy}_1=(\text{n}-1)\text{y}+\text{x}^{\text{n-1}}$
$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n-2}}$
$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+\frac{(\text{n}-1)\text{x}^{\text{n}-1}}{\text{x}}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n}-1}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\{\text{xy}_1-(\text{n}-1)\text{y}\}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{xy}_1-(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=2\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1(1-2\text{n}+2)=-(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1=-(\text{n}-1)^2\text{y}$

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MCQ 61 Mark

If $\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}},$ then $(2\text{xy}_1+\text{y})\text{y}_3=$

✓
$3(xy_2 + y_1) y_2$
B
$3(xy_1 + y_2) y_2$
C
$3(xy_1 + y_2) y_1$
D
None of these

Answer

Correct option: A.

$3(xy_2 + y_1) y_2$

$\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}}$
$\Rightarrow(\text{x}^2+\text{c})\text{y}=\text{ax}+\text{b}$
Differentiating $\text{w.r.t. x},$ we get
$2\text{xy}+(\text{x}^2+\text{c})\frac{\text{dy}}{\text{dx}}=\text{a}$
Differentiating $\text{w.r.t. x},$ we get
$2\text{y}+2\text{xy}_1+2\text{xy}+(\text{x}^2+\text{c})\text{y}_2=0$
$\Rightarrow2\text{y}+4\text{xy}_1+\text{x}^2+\text{cy}_2=0$
Differentiating $\text{w.r.t. x},$ we get
$2\text{y}_1+4\text{y}_1+4\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3+2\text{xy}_2=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+\Big(\frac{-2\text{y}-4\text{xy}_1}{\text{y}_2}\Big)\text{y}_3=0$
$[\because2\text{y}+4\text{xy}_1+(\text{x}^2+\text{c})\text{y}_2=0]$
$\Rightarrow6\text{y}_1\text{y}_2+6\text{x}(\text{y}_2)^2-2\text{y}-4\text{xy}_1\text{y}_3=0$
$\Rightarrow3\text{y}_1\text{y}_2+3\text{x}(\text{y}_2)^2-\text{y}-2\text{xy}_1\text{y}_3=0$
$\Rightarrow(\text{y}_1+\text{xy}_2)3\text{y}_2=(2\text{xy}_1+\text{y})\text{y}_3$

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MCQ 71 Mark

$\frac{\text{d}^{20}}{\text{dx}^{20}}(2\cos\text{x}\cos3\text{x})=$

A
$2^{20}(\cos2\text{x}-2^{20}\cos4\text{x})$
✓
$2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$
C
$2^{20}(\sin2\text{x}+2^{20}\sin4^\text{x})$
D
$2^{20}(\sin2\text{x}-2^{20}\sin4^\text{x})$

Answer

Correct option: B.

$2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

$\text{y}=2\cos\text{x}\cos3\text{x}=\cos(3\text{x}-\text{x})+\cos(3\text{x}+\text{x})=\cos2\text{x}+\cos4\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}-4\sin4\text{x}=-2(\sin2\text{x}+2\sin4\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\cos2\text{x}-16\cos4\text{x}=-2^2(\cos2\text{x}+2^2\cos4\text{x})$
$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}=2^3(\sin2\text{x}+2^3\sin4\text{x})$
$\Rightarrow\frac{\text{d}^4\text{y}}{\text{dx}^4}=2^3(2\cos2\text{x}+4\times2^3\cos4\text{x})=2^4(\cos2\text{x}+2^4\cos4\text{x})$
$\therefore\frac{\text{d}^{20}(\cos2\text{x}+\cos4\text{x})}{\text{dx}^{20}}=2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

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MCQ 81 Mark

If $\text{y}=(\sin^{-1}\text{x})^2,$ then $(1-\text{x}^2)\text{y}_2$ is equal to :

✓
$xy_1 + 2$
B
$xy_1 - 2$
C
$-xy_1 + 2$
D
None of these

Answer

Correct option: A.

$xy_1 + 2$

Here,
$\text{y}=(\sin^{-1}\text{x})\frac{1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2(1-\text{x}^2)=2+\text{xy}_1$

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MCQ 91 Mark

If $\text{x}=\text{t}^2,\text{y}=\text{t}^3$ Then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$

A
$\frac{3}{2}$
B
$\frac{3}{4\text{t}}$
C
$\frac{3}{2\text{t}}$
✓
$\frac{3\text{t}}{2}$

Answer

Correct option: D.

$\frac{3\text{t}}{2}$

$\text{x}=\text{t}^2\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}$
$\text{y}=\text{t}^3\Rightarrow\frac{\text{dy}}{\text{dt}}=3\text{t}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}^2}{2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}}{2}$

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MCQ 101 Mark

If $\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$ then find $(\text{x}^2-1)\text{y}_2+\text{xy}_1=$

✓
$-n^2y$
B
$n^2y$
C
$0$
D
None of these.

Answer

Correct option: A.

$-n^2y$

$\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$
Differentiating both sides we get
$\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}-1}-\text{y}^{\frac{1}{\text{n}}-1}\Big)=2$
$\Rightarrow\text{y}_1\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)=2\text{ny}$
Again differentiating both sides we get
$\text{y}_2\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)+\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}-1}\Big)=2\text{ny}_1$
$\Rightarrow\text{ny}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+\frac{\text{y}^2_1}{\text{y}}\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)=2\text{n}^2\text{y}_1$
$\Rightarrow\text{nyy}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\text{nyy}_2\frac{2\text{ny}}{\text{y}_1}+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\frac{\text{n}^2\text{y}^2\text{y}_2}{\text{y}_1^2}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{4\text{x}^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow(\text{x}^2-1)\text{y}_2+\text{xy}_1=\text{n}^2\text{y}$

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MCQ 111 Mark

If $\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$ then the value of a, 0 < r < is equals to:

A
$\frac{\text{n}!}{\text{r}!}$
B
$\frac{(\text{n}-\text{r})!}{\text{r}!}$
✓
$\frac{\text{n}!}{(\text{n}-\text{r})!}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{\text{n}!}{(\text{n}-\text{r})!}$

$\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$
$\Rightarrow\text{e}^\text{x}(\text{nx}^{\text{n-1}}-\text{a}_1(\text{n}-1)\text{x}^{\text{n-2}}+\text{a}_2(\text{n}-2)\text{x}^{\text{n}-3}+...+(-1)^{\text{n}-1}\text{a}_{\text{n}-1}+\text{x}^\text{a}-\text{a}_1\text{x}^{\text{n-2}}+...+(-1)^\text{n}\text{a}_\text{n})=\text{x}^\text{n}\text{e}^\text{x}$
$\Rightarrow\text{e}^\text{x}(\text{x}^\text{n}+(\text{n}-\text{a}_1)\text{x}^{\text{n}-1}-(\text{a}_1(\text{n-1})-\text{a}_2)\text{x}^{\text{n}-2}\\+(\text{a}_2(\text{n}-2)-\text{a}_3)\text{x}^{\text{n}-3}-...)=\text{x}^\text{n}\text{e}^\text{x}$
on comparing both sides we get
$\text{n}-\text{a}_1=0$
$\Rightarrow\text{a}_1=\text{n}$
$\text{a}_1(\text{n}-1)-\text{a}_2=0$
$\Rightarrow\text{a}_2=\text{a}_1(\text{n}-1)=\text{n}(\text{n}-1)$
$\text{a}_2(\text{n}-2)-\text{a}_3=0$
$\Rightarrow\text{a}_3=\text{a}_2(\text{n}-2)=\text{n}(\text{n}-1)(\text{n}-2)$
So,
$\text{a}_\text{r}=\text{n}(\text{n-1})(\text{n}-2)...(\text{n}-(\text{r}-1)=\frac{\text{n}!}{(\text{n}-\text{r})!}$

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MCQ 121 Mark

If $\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2},$ then $(1-\text{x})^2\text{f}''(\text{x})-\text{xf}(\text{x})=$

✓
1
B
-1
C
0
D
None of these

Answer

Correct option: A.

1

Here,
$\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2}$
$\Rightarrow\sqrt{1-\text{x}}^2\text{f}(\text{x})=\sin^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\sqrt{1-\text{x}^2}\text{f}'(\text{x})-\frac{\text{x f}{(\text{x})}}{\sqrt{1-\text{x}}^2}=\frac{1}{\sqrt{1-\text{x}}^2}$
$\Rightarrow(1-\text{x}^2)\text{f}'(\text{x})-\text{xf}(\text{x})=1$

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MCQ 131 Mark

If $\text{x}=2\text{ at},\text{y}=\text{at}^2,$ where a is a constant, then $\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ at}\ \text{x}=\frac{1}{2}$ is:

✓
$\frac{1}{2}\text{a}$
B
1
C
2a
D
None of these

Answer

Correct option: A.

$\frac{1}{2}\text{a}$

Here,
$\text{x}=2\text{ at},\text{y}=\text{at}^2,$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$
Now, $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$

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MCQ 141 Mark

If $\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})(\cos3\text{x}+\text{i}\sin3\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})\ \text{and}\ \text{f}(1)=1, $ then f1 is equals to:

A
$$$\frac{\text{n}(\text{n}+1)}{2}$
B
$\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
✓
$-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$
D
$\text{none of these}$

Answer

Correct option: C.

$-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$

$\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos\text{x}+\text{i}\sin\text{x})^2...(\cos\text{x}+\text{i}\sin\text{x})^\text{n}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{1+2+3........\text{n}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{\frac{\text{n}(\text{n}+1)}{2}}$
$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^\text{a}$
$\Rightarrow\text{f}(\text{ax})=(\cos\text{ax}+\text{i}\sin\text{ax})...1$
$\Rightarrow\text{f}(1)=(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow1=(\cos\text{a}+\text{i}\sin\text{a})...2\ [\because\text{f}(1)=1]$
Differentiating eqn.1, we get
$\text{f}'(\text{x})=\text{a}(-\sin\text{ax}+\text{i}\cos\text{ax})$
$\Rightarrow\text{f}''(\text{x})=\text-{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''\text{x}=\text{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''(\text{x})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{ax}+\text{i}\sin\text{ax})$
$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{a}+\text{i}\sin\text{a})$
$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}\ [\text{using }2]$

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MCQ 151 Mark

If $\text{x}=\text{a}\cos\ \text{nt}-\text{b}\sin\ \text{nt}$ then $\frac{\text{d}^2\text{x}}{\text{dt}^2}$ is :

A
$n^2x$
✓
$-n^2x$
C
$-nx$
D
$nx$

Answer

Correct option: B.

$-n^2x$

Here
$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
Differentiating $\text{w.r.t. t},$ we get
$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$
Differentiating $\text{w.r.t. t},$ we get
$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$
$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$
$=-\text{n}^2\text{x}$

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MCQ 161 Mark

If $\text{y}^2=\text{ax}^2+\text{bx}+\text{c},$ then $\text{y}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is:

A
a constant
B
a function of x only
✓
a function of y only
D
a function of x and only

Answer

Correct option: C.

a function of y only

$\text{y}^2=\text{ax}^2+\text{bx}+\text{c}$
$\frac{\text{dy}}{\text{dx}}=2\text{ax}+\text{b}$
$\frac{\text{d}^2\text{y}}{\text{d}^2}=2\text{a}$
$=\text{y}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ay}^3$
= A function of y only

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MCQ 171 Mark

if $\text{y}=\text{e}^{{\tan}\text{x}},$ then $(\cos^2\text{x})\text{y}_2=$

A
$(1-\sin2\text{x})\text{y}_1$
B
$-(1+\sin2\text{x})\text{y}_1$
✓
$(1+\sin2\text{x})\text{y}_1$
D
$\text{None of these}$

Answer

Correct option: C.

$(1+\sin2\text{x})\text{y}_1$

$\text{y}=\text{e}^{{\tan}\text{x}},$
$\text{y}_1=\text{sec}^2\text{xe}^{\tan\text{x}}$
$\Rightarrow\cos^2\text{xy}_1=\text{e}^{\tan\text{x}}$
again differentiating w.r.t.x, we get
$\cos^2\text{xy}_2-2\cos\text{x}\sin\text{xy}_1=\sec^2\text{xe}^{\tan\text{x}}$
$\Rightarrow\cos^2\text{xy}_2=\text{y}_1\sin2\text{x}+\text{y}_1$

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MCQ 181 Mark

If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a} > \text{b}>0,$ then :

A
$\text{y}_1=\frac{-1}{\text{a}+\text{b}\cos\text{x}}$
✓
$\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
C
$\text{y}_1=\frac{1}{\text{a}-\text{b}\cos\text{x}}$
D
$\text{y}_2=\frac{-\text{b}\sin\text{x}}{(\text{a}-\text{b}\cos\text{x})^2}$

Answer

Correct option: B.

$\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$

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MCQ 191 Mark

If $\text{x}=\text{at}^2,\text{y}=2\text{at}$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$=

A
$-\frac{1}{\text{t}^2}$
B
$\frac{1}{2\text{at}^3}$
C
$-\frac{1}{\text{t}^3}$
✓
$-\frac{1}{2\text{at}^3}$

Answer

Correct option: D.

$-\frac{1}{2\text{at}^3}$

$\text{x}=\text{at}^2,\text{y}=2\text{at}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$

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MCQ 201 Mark

If $\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{bx}}\Big)^\text{x}$ then $\text{x}^3\text{y}_2=$

✓
$(\text{xy}_1-\text{y})^2$
B
$(1+\text{y})^2$
C
$\Big(\frac{\text{y}-\text{xy}_1}{\text{y}_1}\Big)^2$
D
$\text{None of these}$

Answer

Correct option: A.

$(\text{xy}_1-\text{y})^2$

$\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{b}}\Big)^\text{x}$
$\text{y}=\text{x}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$
$\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\text{x}\Big(\frac{1}{\text{x}}-\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+1-\frac{\text{bx}}{\text{a}+\text{bx}}$
$(\text{a}+\text{bx})\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))(\text{a}+\text{bx})+\text{a}$
Again differentiating w.r.t.x, we get
$(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))\\+(\text{a}+\text{bx})\Big(\frac{1}{\text{x}}\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\frac{\text{a}}{\text{x}}$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\frac{\text{by}}{\text{x}}+\frac{\text{a}}{\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\Big(\frac{\text{y}}{\text{x}}+\frac{\text{a}}{\text{a}+\text{bx}}\Big)$
$=\frac{(\text{a}+\text{by})(\text{a}+\text{bx})-\text{b}(\text{ay}+\text{bxy}+\text{ax})}{\text{x}(\text{a}+\text{b})^2}$
$=\frac{\text{a}^2+\text{abx}+\text{aby}+\text{b}^2\text{xy}-\text{bay}-\text{b}^2\text{xy}-\text{abx}}{\text{x}(\text{a}+\text{bx})^2}$
$=\frac{\text{a}^2}{\text{x}(\text{a}+\text{bx})^2}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}$
$(\text{xy}_1-\text{y})=\frac{\text{ax}}{\text{a}+\text{bx}}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}=(\text{xy}_1-\text{y})^2$

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MCQ 211 Mark

If $\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$ then $(1-\text{x}^2)\text{y}_2-\text{xy}_1$ is equal to :

A
$m^2y$
B
$my$
✓
$-m^2y$
D
None of these

Answer

Correct option: C.

$-m^2y$

Here,
$\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{mx}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xm}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)\times\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\text{y}_2=-\text{ym}^2+\text{xy}_1$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1=-\text{m}^2\text{y}$

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MCQ 221 Mark

If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ Then $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2} =$

A
$n(n - 1) y$
✓
$n(n + 1) y$
C
$ny$
D
$n^2y$

Answer

Correct option: B.

$n(n + 1) y$

Here
$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$
$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$
$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$
$=\text{n}(\text{n}+1)\text{y}$

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MCQ 231 Mark

If $\text{y}=\text{a}\sin\text{mx}+\text{b}\cos\text{mx},$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equal to :

✓
$-m^2y$
B
$m^2y$
C
$-my$
D
$my$

Answer

Correct option: A.

$-m^2y$

$\text{y}=\text{a}\sin\text{mx}+\text{b}\cos\text{mx}$
$\frac{\text{dy}}{\text{dx}}=\text{am}\cos\text{mx}-\text{bm}\sin\text{mx}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{am}^2\sin\text{mx}-\text{bm}^2\cos\text{mx}=-\text{m}^2\text{y}$

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MCQ 241 Mark

If $\text{xy}-\log_\text{e}\text{y}=1$ satisfies the equation $\text{x}(\text{yy}_2+\text{y}_1^2)-\text{y}_2+\lambda\text{yy}_1=0,$ then $\lambda=$

A
-3
B
1
✓
3
D
None of these

Answer

Correct option: C.

3

$\text{xy}-\log_\text{e}\text{y}=1$
$\Rightarrow\text{xy}_1+\text{y}-\frac{\text{y}_1}{\text{y}}=0$
$\Rightarrow\text{xyy}_1+\text{y}^2-\text{y}_1=0$
$\Rightarrow\text{yy}_1+\text{xy}_1\text{y}_1+\text{xyy}_2+2\text{yy}_1-\text{y}_2=0$
$\Rightarrow\text{x}(\text{y}_1^2+\text{yy}_2)-\text{y}_2+3\text{yy}_2=0$
$\therefore\lambda=3$

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MCQ 251 Mark

Let $f(x)$ be a polynomial. Then the second order derivatives of $f(e^x)$ is :

A
$\text{f}(\text{e}^\text{x})\text{e}^{2\text{x}}+\text{f}(\text{e}^\text{x})\text{e}^\text{x}$
B
$\text{f}(\text{e}^\text{x})\text{e}^{\text{x}}+\text{f}(\text{e}^\text{x})$
C
$\text{f}(\text{e}^\text{x})\text{e}^{2\text{x}}+\text{f}(\text{e}^\text{x})\text{e}^\text{x}$
✓
$\text{f}(\text{e}^\text{x})$

Answer

Correct option: D.

$\text{f}(\text{e}^\text{x})$

Let $\text{y}=\text{f}(\text{e}^\text{x}),$ then
$\text{y}_1=\text{f}\ '(\text{e}^\text{x})\text{e}^\text{x}$
$\text{y}_2=\text{f}\ ''(\text{e}^\text{x})\text{e}^\text{x}\text{e}^\text{x}+\text{f}\ '(\text{e}^\text{x})\text{e}^\text{x}$
$=\text{e}^{2\text{x}}\text{f}\ ''(\text{e}^\text{x})+\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

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MCQ 261 Mark

If $\text{y}=\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})$ then $\text{x}^2\text{y}^2+\text{xy}_1=$

A
0
B
y
✓
-y
D
None of these

Answer

Correct option: C.

-y

$\text{y}=\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})$
$\Rightarrow\text{y}_1=-\text{a}\sin(\log_\text{e}\text{x})\frac{1}{\text{x}}+\text{b}\cos(\log_\text{e}\text{x})\frac{1}{\text{x}}$
$\Rightarrow\text{y}_2=\frac{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})}{\text{x}}$
$\Rightarrow\text{y}_2=\frac{-\text{a}(\log_\text{e}\text{x})-\text{b}\sin(\log_\text{e}\text{x})-\{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{x}^2\text{y}_2=-\{\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})\}\\-\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}$
$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=-\text{y}$

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