JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

In the function

f

given by

f\left( x \right) = {x^3} - 3\left( {a - 2} \right){x^2} + 3ax + 7,

for some

a \in R

is increasing in

\left( {0,\,1} \right]

and decreasing in

\left[ {1,\,5} \right),

then a root of the equation,

\frac{{f\left( x \right) - 14}}{{{{\left( {x - 1} \right)}^2}}} = 0\,\,\,\left( {x \ne 1} \right)

View answer and explanation

Correct option: C

Option C (7) is correct because by finding the value of 'a' using the given increasing/decreasing conditions, we can determine the function f(x) and then solve for the root of the given equation.

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

then

\frac{{dy}}{{dx}}

is equal to

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}