If f x is twice differentiable function in - c 1-1- c 2-1- and f - c 1- f - c 2-0- f - c 1 - f - c 20- text then k -text P 1Wline II If f - c - 1- f - c - 2-0- - text then k -WlineWhineWline lendarray Which of the following is only CORRECT combination-A- I - Q B- II - Q C- III - P D- IV - R

The correct option is A (I)→(Q)For f”(c_1) f”(c_2) gt; 0 nbsp; nbsp;k = 2, nbsp; nbsp; nbsp; nbsp; m = 4 For f”(c_1) f”(c_2) lt; ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

If f x is twi...

Question

If $f (x)$ is twice differentiable function in $[c_{1} - 1, c_{2} + 1]$ and $f^{'} (c_{1}) = f^{'} (c_{2}) = 0, f^{''} (c_{1}) \cdot f^{''} (c_{2}) < 0, f (c_{1}) = 9, f (c_{2}) = 0$ . Let $k$ and $m$ be the minimum number of the roots of $f (x) = 0$ and $f^{'} (x) = 0$ respectively, in $[c_{1} - 1, c_{2} + 1]$

$\begin{matrix} List - I & List - II (I) If f^{''} (c_{1}) - f^{''} (c_{2}) > 0, then k = & (P) 1 (II) If f^{''} (c_{1}) - f^{''} (c_{2}) < 0, then k = & (Q) 2 (III) If f^{''} (c_{1}) - f^{''} (c_{2}) > 0, then m = & (R) 3 (IV) If f^{''} (c_{1}) - f^{''} (c_{2}) < 0, then m = & (S) 4 \end{matrix}$

Which of the following is only CORRECT combination?

(I) \to (Q)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(I I) \to (Q)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(I I I) \to (P)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(I V) \to (R)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(I) \to (Q)$
For $f^{''} (c_{1}) - f^{''} (c_{2}) > 0$
$k = 2, m = 4$

For $f^{''} (c_{1}) - f^{''} (c_{2}) < 0$
$k = 1 m = 2$

Suggest Corrections

Similar questions

Q. If

f (x)

is twice differentiable function in

[c_{1} - 1, c_{2} + 1]

and

f^{'} (c_{1}) = f^{'} (c_{2}) = 0, f^{''} (c_{1}) \cdot f^{''} (c_{2}) < 0, f (c_{1}) = 9, f (c_{2}) = 0

. Let

k

and

m

be the minimum number of the roots of

f (x) = 0

and

f^{'} (x) = 0

respectively,in

[c_{1} - 1, c_{2} + 1]

\begin{matrix} List - I & List - II (I) If f^{''} (c_{1}) - f^{''} (c_{2}) > 0, then k = & (P) 1 (II) If f^{''} (c_{1}) - f^{''} (c_{2}) < 0, then k = & (Q) 2 (III) If f^{''} (c_{1}) - f^{''} (c_{2}) > 0, then m = & (R) 3 (IV) If f^{''} (c_{1}) - f^{''} (c_{2}) < 0, then m = & (S) 4 \end{matrix}

Which of the following is only CORRECT combination?

Q. If

f (x)

is invertible and twice differentiable function satisfying

f^{'} (x) = f (x) \int 0 f^{- 1} (t) d t, \forall x \in R

and

f^{'} (0) = 1

, then

f^{'} (1)

can be

Q. Let

y = f (x) and y = g (x)

be two differentiable function in

[0, 2]

such that

f (0) = 3, f (2) = 5, g (0) = 1 and g (2) = 2.

If there exists atleast one

c ϵ (0, 2)

such that

f^{'} (c) = k g^{'} (c)

, then

k

must be

Q. Let

\int \frac{(1 + x^{4}) d x}{(1 - x^{4})^{3 / 2}} = f (x) + C_{1}

with

f (0) = 0

and

\int f (x) d x = g (x) + C_{2}

with

g (0) = 0.

g (\frac{1}{\sqrt{2}}) = \frac{π}{k},

then the value of

k

Q. Let

f : [0, 2] \to R

be continuous on

[0, 2]

and twice differentiable in

(0, 2) .

f (0) = 0,

f (1) = 1

and

f (2) = 1,

then

Join BYJU'S Learning Program

The correct option is A (I)→(Q)For f′′(c1)−f′′(c2)>0 k=2, m=4 For f′′(c1)−f′′(c2)<0 k=1 m=2

The correct option is A $(I) \to (Q)$
For $f^{''} (c_{1}) - f^{''} (c_{2}) > 0$
$k = 2, m = 4$

For $f^{''} (c_{1}) - f^{''} (c_{2}) < 0$
$k = 1 m = 2$