Solution :
Thus if \(x \ne 1\)
Now the denominator has negative discriminant with positive leading coefficient – thus it is positive for all real \(x\). So \(2-x \ge 0 \Rightarrow x \le 2\) & \(x \ne 1\).
Q-2) The condition that the equation \(\frac{1}{x} + \frac{1}{x+b} = \frac{1}{m}+\frac{1}{m+b}\) has real roots that are equal in magnitude but opposite in sign is :
a) \(b^2=m^2\) b) \(b^2=2m^2\) c) \(2b^2=m^2\) d) The roots of the equation will always satisfy this criteria
Solution:
We can see trivially that \(x=m\) is a root of the equation, and since upon cross multiplication, this is going to land us up with a quadratic equation, we know that there can be only 2 roots. Thus the other root has to be \(-m\). Plugging in \(x=-m\) we have
\(\frac{1}{b-m} – \frac{1}{b+m} = \frac{2}{m} \Rightarrow \frac{b+m-b+m}{b^2-m^2} = \frac{2}{m}\).
Thus \(2m^2=b^2\).
Q-3) Find all real \(a\) such that \(\log_2{(ax^2+x+a)} \ge 1\) for all real \(x\).
Solution:
We need to have \(ax^2+x+a \ge 2\) for all real \(x\).
That implies \(ax^2+x+(a-2) \ge 0\) for all real \(x\), the condition for which is that \(a>0\) and the discriminant should never be negative so that the quadratic does not intersect with the X-Axis. If the Discriminant is 0, that means the X-Axis is tangent to the quadratic and the equation has equal roots – which means it’s a perfect square.
Thus we have \(4a^2-8a-1 \ge 0\), which implies \(a>0\) & \(a \in \left( -\infty, 1-\frac{\sqrt{5}}{2}\right] \bigcup \left[1+\frac{\sqrt{5}}{2}, \infty \right )\)
Q-4) If \(a,b,c \in \mathbb{R}\) and \((a+b+c)c<0\), then the quadratic equation \(p(x)=ax^2+bx+c=0\) has:
a) A negative root b) 2 real roots c) 2 imaginary roots d)None of these
Solution:
We observe that \(p(1)=a+b+c\) and \(p(0)=c\). Since \(p(1)p(0)<0\), we know that \(p\) crosses the X-Axis in \((0,1)\) – thus it has a real root, which also means both the roots are real since \(a,b,c \in \mathbb{R}\).
Q-5) Let \(a,b,c \in \mathbb{R}\) be distinct such that each of the expressions \(ax^x+bx+c\), \(bx^2+cx+a\) and \(cx^2+ax+b\) is positive for all real \(x\). Let \(\alpha = \frac{bc+ca+ab}{a^2+b^2+c^2}\), then
a) \(\alpha <4\) b) \(\alpha <1\) c) \(\alpha > \frac{1}{4}\) d) \(\alpha >1\)
Solution:
We know that \(a,b,c >0\). Also all the 3 discriminants must be negative for the 3 euations to stay positive for all real \(x\). Hence we can conclude \(a^2+b^2+c^2 < 4(ab+bc+ca) \Rightarrow \alpha >\frac{1}{4}\).
Also \(a^2+b^2+c^2-(ab+bc+ca)=\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]>0\Rightarrow ab+bc+ca<a^2+b^2+c^2 \Rightarrow \alpha <1\). Thus A,B,C are the correct options.
Q-6) Show that the following equation has real roots:
Solution:
(You may prefer to look up the standard definition of \(e\) before solving this)
Let \(f(x)\) denote the numerator of the given equation after cross multiplication, thus we have
Now \(f(e)=\pi^{e+1}(\pi -e)>0\), and we also have \(f(\pi)=-e^{\pi+1}(\pi -e)<0\). We should note that for a continuous function \(f\), if \(f(a)f(b)<0\), then \(f\) has an odd number of real roots in \((a,b)\). The conclusion thus follows from the above observations.
Q-7) Let \(a,b,c \in \mathbb{R}\) such that the equations \(ax^2+bx+c=0\) & \(x^2+x+1=0\) have a common root. Show that \(a=b=c\).
Solution :
We know that the roots of \(x^2+x+1=0\) are complex, and also imaginary roots of an equation with real coefficients occur in conjugate pairs. Thus if these 2 equations share 1 root in common, they must also share the second root – which means both roots of the equations are same.
Thus \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\Rightarrow a=b=c\).
Q-8) Let \(a,b,c \in \mathbb{R}\) with \(a \ne 0\) such that the equation \(ax^2+bx+c=0\) has no real roots. Which of the following options are correct:
a) \(a+b+c>0\) b) \(a(a+b+c)>0\) c) \(b(a+b+c)>0\) d) \(c(a+b+c)>0\)
Solution:
Since the equation does not have any real root we can conclude that the expression \(ax^2+bx+c\) never changes its sign. Thus \(f(1)f(0)>0\) and \(af(1)>0\). Thus options B & D are correct.
Q-9)
The graph above is of the expression \(ax^2+bx+c\). Which of the following is/are true about this graph:
a) \(a>0\) b) \(b<0\) c) \(c>0\) d) \(b^2-4ac \le 0\)
Solution:
Clearly the equation has 2 distinct roots, hence option D is ruled out. Also the shape of the parabola says that \(a<0\). We see that the extrema occurs at \(x-0\) where \(x_0<0 \Rightarrow \frac{-b}{2a}<0 \Rightarrow \frac{b}{2a}>0\). Since we already know that \(a<0\), this implies \(b<0\).
Also product of the roots is negative, thus \(\frac{c}{a}<0\), we know \(a<0\) which implies \(c>0\). Thus options B & C are correct.
Q-10) Solve \(x(x+2)^2(x-1)^5(2x-3)(x-3)^4 \ge 0\)
Solution:
Let’s take out the squares and fourth powers away since they are always positive. We apply the method of intervals to the remaining expressions. Thus we have \(x,(x-1),(2x-3)\) where the exponenets are odd. The sign thus changes at \(0,1,\frac{3}{2}\). Also for \(x>\frac{3}{2}\) all the factors of the expression are positive. For \(x<0\), we observe that \(x(x-1)^5(2x-3)<0\), thus \(x>0\).
So far, we know that \(x>0\) and \(x>\frac{3}{2}\) for the expression to be positive. So let’s analyse what happens in \(\left[ 0,\frac{3}{2}\right]\). We have already stated that the expression will change sign at \(x=1\). In the interval \((0,1)\), we have \((x-1)^5<0\) and \((2x-3)<0 \Rightarrow (x-1)^5(2x-3)>0\). Thus \(x \in [0,1]\). In the interval \(\left( 1,\frac{3}{2}\right)\) we have \(x(x-1)^5>0\) but \((2x-3)<0\). Thus the final expression is negative.
\(x \in [0,1] \bigcup \left[ \frac{3}{2}, \infty \right)\) is the final set of solutions.
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