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1

MCQ (Single Correct Answer)

If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is

A

$${n^2}y$$

B

$$-{n^2}y$$

C

$$-y$$

D

$$2{x^2}y$$

$$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}$$

$$ = {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}$$

or $$\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny$$

or $$\sqrt {1 + {x^2}} {y_1} = ny$$

$$\left( {{y_1} = {{dy} \over {dx}}} \right)$$

Squaring, $$\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}$$

Differentiating, $$\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x$$

$$ = {n^2}.2y{y_1}$$

or $$\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$$

$${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}$$

$$ = {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}$$

or $$\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny$$

or $$\sqrt {1 + {x^2}} {y_1} = ny$$

$$\left( {{y_1} = {{dy} \over {dx}}} \right)$$

Squaring, $$\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}$$

Differentiating, $$\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x$$

$$ = {n^2}.2y{y_1}$$

or $$\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y$$

On those following papers in MCQ (Single Correct Answer)

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Complex Numbers

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Application of Derivatives

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