Let f, g: [- 1, 2] → ℝ be continuous functions which are twice differentiable on the interval (- 1, 2) . Let the values of f and g at the points - 1, 0 and 2 be as given in the following table: x = - 1 x = 0 x = 2 f ( x

Correct answer: B. f ^ ′ (x) - 3 g ^ ′ (x) = 0 has exactly one solution in (-1, 0) h(x) = f (x) - 3g (x) h(- 1) = 3 - 0 = 3 h(0) = 6 - 3 = 3 h(2) = 0 + 3 = 3 h′ ( x ) has at least one root is (- 1,0) And h ^ ′ (x) has at least one in ( 0,2 ) Hence h ^ ′′ (x) atleast one root in ( - 1,2 ) but Also, (f - 3g) ^ ′′ ≠ 0in(- 1, 0)and ( 0, 2 ) h ^ ′′ (x) = 0onlyatx = 0, so f ^ ′ (x) - 3 g ^ ′ (x) has exactly one solution in (-1, 0) and (0, 2)

JEE Advanced 2015Paper 2

Exam: JEE Advanced
Year: 2015
Paper: Paper 2
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2015 · Paper 2 · Mathematics · Application of Derivatives

Let f, g: [- 1, 2] → ℝ be continuous functions which are twice...

Let

f, g: \left[- 1, 2\right] \to \mathbb{R}

be continuous functions which are twice differentiable on the interval \left(- 1, 2\right)

. Let the values of

f and g

at the points

- 1, 0 and 2

be as given in the following table:<br /> <table border="1"> <tbody> <tr> <td> </td> <td>

x = - 1

</td> <td>

x = 0

</td> <td>

x = 2

</td> </tr> <tr> <td>

f ( x )

</td> <td>3</td> <td>6</td> <td>0</td> </tr> <tr> <td>

g\left(x\right) 0 1 -1
In each

of

the intervals \left(- 1, 0\right) and \left(0, 2\right) the function

( f - 3g ) "

never vanishes. Then the correct statement(s) is(are)

Options

A{f}^{′}\left(x\right) - 3{g}^{′}\left(x\right) = 0 $has exactly three solutions in \left(- 1, 0\right) \cup ( 0, 2 )$
B ${f}^{′}\left(x\right) - 3{g}^{′}\left(x\right) = 0$ has exactly one solution in (-1, 0)
C ${f}^{′}\left(x\right) - 3{g}^{′}\left(x\right) = 0$ has exactly one solution in (0, 2)
D ${f}^{′}\left(x\right) - 3{g}^{′}\left(x\right) = 0$ has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)

Correct answer

B. ${f}^{′}\left(x\right) - 3{g}^{′}\left(x\right) = 0$ has exactly one solution in (-1, 0)

Step-by-step solution

h\left(x\right) = f

\left(x\right) - 3g \left(x\right) <br >

h\left(- 1\right)

= 3 - 0 = 3

h\left(0\right)

= 6 - 3 = 3

<br >

h\left(2\right)

= 0 + 3 = 3

h&prime; ( x )

has at least one root is \left(- 1,0\right)
And

{h}^{&prime;}\left(x\right)

has at least one in

( 0,2 )

Hence

{h}^{&prime;&prime;}\left(x\right) atleast one root

in ( - 1,2 )

but<br >Also,

{\left(f - 3g\right)}^{′′} \ne 0in\left(- 1, 0\right)and ( 0, 2 )

<br >

{h}^{′′}\left(x\right) = 0onlyatx = 0, so

<br >

{f}^{′}\left(x\right)

- 3{g}^{&prime;}\left(x\right)

has exactly one solution in (-1, 0) and (0, 2)

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