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; ...

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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)

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(I I) \to (Q)

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(I I I) \to (P)

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(I V) \to (R)

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Solution

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

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

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\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}

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