If the area bounded by the curves y - -k-x2- y -k-4 x2 and 2 - -x- - 3 is 19- then k lies in where - represent greatest integer function

The required area 19= 2∫_2^3( [k]x^2 [k]4x^2 ) dx nbsp;⇒ 19 = nbsp;2∫_2^33[k]4x^2 dx ⇒ 19 = nbsp;[k]2 (27 8) ⇒ [k] = 2 ∴ k ∈ [2, 3) ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

If the area b...

Question

If the area bounded by the curves $y = [k] x^{2}, y = \frac{[k]}{4} x^{2}$ and $2 \leq | x | \leq 3$ is $19$ , then $k$ lies in
(where [.] represent greatest integer function)

[2, 3)

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[1, 3)

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[0, 3)

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[3, 4)

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Solution

The correct option is A $[2, 3)$

The required area
$19 = 2 3 \int 2 ([k] x^{2} - \frac{[k]}{4} x^{2}) d x \Rightarrow 19 = 2 3 \int 2 \frac{3 [k]}{4} x^{2} d x \Rightarrow 19 = \frac{[k]}{2} (27 - 8) \Rightarrow [k] = 2 ∴ k \in [2, 3)$

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If the area bounded by the curves y=[k]x2,y=[k]4x2 and 2≤|x|≤3 is 19, then k lies in (where [.] represent greatest integer function)

The correct option is A [2,3) The required area 19=23∫2([k]x2−[k]4x2) dx⇒19=23∫23[k]4x2 dx⇒19=[k]2(27−8)⇒[k]=2∴k∈[2,3)

If the area bounded by the curves $y = [k] x^{2}, y = \frac{[k]}{4} x^{2}$ and $2 \leq | x | \leq 3$ is $19$ , then $k$ lies in
(where [.] represent greatest integer function)

The correct option is A $[2, 3)$

The required area
$19 = 2 3 \int 2 ([k] x^{2} - \frac{[k]}{4} x^{2}) d x \Rightarrow 19 = 2 3 \int 2 \frac{3 [k]}{4} x^{2} d x \Rightarrow 19 = \frac{[k]}{2} (27 - 8) \Rightarrow [k] = 2 ∴ k \in [2, 3)$