Let $f$ be a twice differentiable function defined on $R$, such that $f(0) = 1,\,\,\,\,f'(0) = 2$ and $f''(x) \ne 0$ for all $x \in R.$ If $\left| {\begin{array}{*{20}{c}} {f(x)}&{f'(x)}\\ {f'(x)}&{f''(x)} \end{array}} \right| = 0$ for all $x \in R$, then the value of $f(1)$ lies in the interval

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f

be a twice differentiable function defined on

R

, such that

f(0) = 1,\,\,\,\,f'(0) = 2

and

f''(x) \ne 0

for all

x \in R.

\left| {\begin{array}{cccccccccccccccccccc} {f(x)}&{f'(x)}\\ {f'(x)}&{f''(x)} \end{array}} \right| = 0

for all

x \in R

, then the value of

f(1)

lies in the interval

View answer and explanation

Correct option: B

(6, 9)

Option B ((6, 9)) is correct because solving the differential equation and using the initial conditions leads to f(1) = e², which lies in the interval (6, 9).

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs →

Let

y=y(x)

be the solution of the differential equation

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

such that

y(1)=1

. Then

\left|(y(2))^{3}-12 y(2)\right|

is equal to : The degree and order of the differential equation whose solution is a parabola whose axis is

x

-axis, are If

{x^2} + {y^2} = 1

, then

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

then

\frac{{{d^2}x}}{{d{t^2}}}

is equal to Let

y=y_{1}(x)

and

y=y_{2}(x)

be the solution curves of the differential equation

\frac{d y}{d x}=y+7

with initial conditions

y_{1}(0)=0

and

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

and

y=y_{2}(x)

intersect at Let

y=y(t)

be a solution of the differential equation

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

where,

\alpha>0, \beta>0

and

\gamma>0

. Then

\lim \limits_{{x \rightarrow \infty}} y(t)