Match the statements in Column I with the intervals in Column II. Column I Column II (A) f ( x ) = 4 x − x 3 + lo g e ( a 2 − 3 a + 3 ) , if 0 ≤ x ≤ 3 ; = x − 18 , if x ≥ 3 . Then, the complete set

Question

Match the statements in Column I with the intervals in Column II.

Column I	Column II
(A) $f (x) = 4 x - x^{3} + lo g_{e} (a^{2} - 3 a + 3)$ , if $0 \leq x \leq 3$ ; $= x - 18$ , if $x \geq 3$ . Then, the complete set of values of 'a' such that $f (x)$ has a local maxima at $x = 3$ lie in the interval	(P) $(- \infty, - 2) \cup (2, \infty)$
(B) The equation $x + cos x = a$ has exactly a positive root, then, complete set of values of $a$ lie in the interval	(Q) $[1, 2]$
(C) If $f (x) = 3 + ∣ x - k ∣$ for $x \leq k$ ; $= a^{2} - 2 + \frac{s i n ( x - k )}{x - k}$ for $x > k$ , has minimum at $x = k$ , then, the complete set of values of 'a' lie in the interval	(R) $(- 20, - 16)$
(D) If $f (x) = x^{3} - 9 x^{2} + 24 x + a$ has three real and distinct roots, then, possible values of $a$ lie in the interval	(S) $(1, \infty)$

TestHub · Accepted Answer

A. f is decreasing for x < 3 and increasing for x > 3. ∴ for x = 3 to be a point of local maximum, f ( 3 ) ≥ f ( 3 − 0 ) ⇒ − 15 ≥ 12 − 27 + lo g e ( a 2 − 3 a + 3 ) ⇒ 0 < a 2 − 3 a + 3 ≤ 1 ⇒ ( a − 1 ) ( a − 2 ) ≤ 0 ⇒ a ∈ [ 1 , 2 ] . B. f ( x ) = x + cos x − a ⇒ f ′ ( x ) = 1 − sin x ≥ 0∀ x ∈ R ⇒ f is increasing on R . For positive root, f ( 0 ) = 1 − a < 0 ⇒ a > 1 ⇒ a ∈ ( 1 , ∞ ) . C. 3 + ∣ x − k ∣ has minimum value 3 at x = k . For f to have minimum at x = k , lim x → k + ( a 2 − 2 + x − k s i n ( x − k ) ) = a 2 − 1 ≥ 3 ⇒ a 2 ≥ 4 ⇒ a ∈ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) . D. d x d y = 3 x 2 − 18 x + 24 = 3 ( x − 2 ) ( x − 4 ) . For three real and distinct roots, f ( 2 ) ⋅ f ( 4 ) < 0 ⇒ ( 20 + a ) ( 16 + a ) < 0 ⇒ a ∈ ( − 20 , − 16 ) .

Mathematics - Application of Derivative Question with Solution | TestHub

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