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JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1
If a,ba, b and cc are real numbers such that a2+b2+c2=1a^2+b^2+c^2=1, then ab+bc+caa b+b c+c a lies in the interval
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Correct option: B
[−12,1]\left[-\frac{1}{2}, 1\right]
The correct answer is B ([−12,1\frac{-1}{2}, 1]) because using (a+b+c)2≥0(a+b+c)^2 \geq 0 and (a−b)2+(b−c)2+(c−a)2≥0(a-b)^2+(b-c)^2+(c-a)^2 \geq 0 allows us to find the interval for ab+bc+caab+bc+ca.

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The solution of the equation log⁡7log⁡5(x+5+x)=0\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 is If α,β\alpha ,\beta are the roots of x2−3x+a=0,a∈R{x^2} - 3x + a = 0,a \in R and α<1<β,\alpha < 1 < \beta , then If α,  β\alpha ,\;\beta be the roots of the equation x2−px+q=0{x^2} - px + q = 0 and α1,β1{\alpha _1},{\beta _1} be the roots of the equation x2−qx+p=0,{x^2} - qx + p = 0, then the equation whose roots are 1α1β+1αβ1and  1αα1+1ββ1,\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}, is If x2+4ax+2>0{x^2} + 4ax + 2 > 0 for all values of xx, then aa lies in the interval If f(x)=ax2+bx+c  f\left( x \right) = a{x^2} + bx + c\; and   f(−1)<1,f(1)>−1,f(3)<−4  \;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\; and   a≠0,\;a \ne 0, then Root(s) of the equation 9x2−18∣x∣+5=09{x^2} - 18\left| x \right| + 5 = 0 belonging to the domain of definition of the function f(x)=log⁡(x2−x−2)f\left( x \right) = \log \left( {{x^2} - x - 2} \right) is (are)
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