If the range of the parameter t in the interval 0-2 - satisfying -2 x2-5 x-10-sin t x2-21-sin t x-9 sin t-4-0 for all real values of x is a- b- and a - b - k - then the value of k is

2x^2+5x 10 lt;0, ∀ x∈ R Let sin t = a ax^2+2(1+a)x+(9a+4) lt;0, ∀ x∈ R It is only possible if a lt;0 and 4(1+a)^2 4a(9a+4) lt;0 ⇒ 8a^2+2a 1 gt;0 ⇒ ( ...

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Byju's Answer

Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

If the range ...

Question

If the range of the parameter $t$ in the interval $(0, 2 π)$ , satisfying $\frac{- 2 x^{2} + 5 x - 10}{(sin t) x^{2} + 2 (1 + sin t) x + (9 sin t + 4)} > 0$ for all real values of $x$ is $(a, b)$ , and $(a + b) = k π$ , then the value of $k$ is

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Solution

$- 2 x^{2} + 5 x - 10 < 0, \forall x \in R$
Let $sin t = a$

$a x^{2} + 2 (1 + a) x + (9 a + 4) < 0, \forall x \in R$
It is only possible if $a < 0$ and $4 (1 + a)^{2} - 4 a (9 a + 4) < 0 \Rightarrow 8 a^{2} + 2 a - 1 > 0 \Rightarrow (4 a - 1) (2 a + 1) > 0 a > \frac{1}{4}, which is not possible \Rightarrow a < - \frac{1}{2} \Rightarrow sin t < - \frac{1}{2} \Rightarrow t \in (\frac{7 π}{6}, \frac{11 π}{6}) \Rightarrow a + b = 3 π \Rightarrow k = 3$

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If the range of the parameter t in the interval (0,2π), satisfying −2x2+5x−10(sint)x2+2(1+sint)x+(9sint+4)>0 for all real values of x is (a,b), and (a+b)=kπ, then the value of k is

−2x2+5x−10<0,∀x∈R Let sint=a ax2+2(1+a)x+(9a+4)<0,∀x∈R It is only possible if a<0 and 4(1+a)2−4a(9a+4)<0⇒8a2+2a−1>0⇒(4a−1)(2a+1)>0a>14, which is not possible⇒a<−12⇒sint<−12⇒t∈(7π6,11π6)⇒a+b=3π⇒k=3

If the range of the parameter $t$ in the interval $(0, 2 π)$ , satisfying $\frac{- 2 x^{2} + 5 x - 10}{(sin t) x^{2} + 2 (1 + sin t) x + (9 sin t + 4)} > 0$ for all real values of $x$ is $(a, b)$ , and $(a + b) = k π$ , then the value of $k$ is