NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

AIEEE 2002

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
$$2{x^2}y$$

Explanation

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

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

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

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

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

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

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

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12