Location of Roots MCQ Quiz in தமிழ் - Objective Question with Answer for Location of Roots - இலவச PDF ஐப் பதிவிறக்கவும்

\( \alpha, \beta = \dfrac{\lambda - 2 \pm \sqrt{4-4\lambda + \lambda^2 - 40 + 4\lambda}}{2} = \dfrac{\lambda - 2 \pm \sqrt{\lambda^2 - 36}}{2} \).

The magnitude of the difference of the roots is clearly \( |\sqrt{\lambda^2-36}| \).

We have, \( \alpha^3 + \beta^3 = \dfrac{(\lambda-2)^3}{4} + \dfrac{3(\lambda-2)(\lambda^2 - 36)}{4} = \dfrac{(\lambda-2)(4\lambda^2 - 4\lambda - 104)}{4} = (\lambda-2)(\lambda^2-\lambda-26) \).

This function attains its minimum value at \( \lambda = 4 \).

Thus, the magnitude of the difference of the roots is clearly \( |i\sqrt{20}|=2\sqrt{5} \).

So the correct answer is option B.

\(g(t)=5+\left| t-1 \right|\)

\(f(t)={ t }^{ 2 }-2t\)

From the graph

\(g(t)\) and \(f(t)\) intersect at two points, but one value of \(t\) is negative, and only one value of \(t\) is positive.

So, number of real roots is \(1\)

\(f(0)=0\)

Hence

\(f(x)\) has a real root \(\alpha_{0}>0\).

Now \(f(x)\) is continuous on \([0,\alpha_{0}]\) and differentiable on \((0,\alpha_{0})\).

Hence there exists a \(\alpha_{1}\epsilon[0,\alpha_{0}]\) such that

\(f'(\alpha_{1})=0\) ...(Rolle's Theorem).

Consider \(f'(x)\).

It is continuous on \([0,\alpha_{1}]\) and differentiable on \((0,\alpha_{1})\).

Hence there exists a \(\alpha_{2}\epsilon[0,\alpha_{2}]\) such that

\(f'(\alpha_{2})=0\) ...(Rolle's Theorem).

And this can be repeated till \(f^{n}=constant\).

Hence all of the above is true.

\((z^{6}-28)^{2}=1296\)

\(z^{6}-28=\pm36\)

\(z^{6}=64\) and \(z^{6}=-8\)

\(z^{3}=\pm8\)

\(z=2\) and \(z=-2\) ... (i)

\(z^{6}=2^{3}.e^{i(2k-1)\pi}\)

\(z=2^{\frac{1}{2}}(e^{i\frac{(2k-1)\pi}{6}})\) where \(k=1,2,3..6\)

if \(\alpha, \beta, \gamma\) are roots of a cubic equation then,

cubic equation is written as;

\(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\alpha\gamma)x-\alpha\beta\gamma=0\)

Given, \(2\) and \(3\) are roots of the equation \(2x^3+mx^2-13x+n=0\)

Let \(\alpha\) be the third root.

\(5+\alpha =-\frac{m}{2}\) ....(1)

Also, \(6+2\alpha+3\alpha=-\frac{13}{2}\)

\(\Rightarrow \alpha=-\frac{5}{2}\)

Also, \(6\alpha=-\frac{n}{2}\)

\(\Rightarrow n=30\)

Now, by (1), \(m=-5\)

\(f(x)=3(x-a)(x-b)\)

\(f(1)=3(1-a)(1-b)=\) -ve as \(a < 1 < b\)

on \(f(1)=3-3 \sin \alpha -2(1-\sin^{2}\alpha ) < 0\)

or \(2\sin^{2}\alpha -3\sin\alpha +1 < 0\)

\((2 \sin \alpha -1)(\sin \alpha -1) < 0\)

\(2(\sin \alpha -\dfrac{1}{2})(\sin \alpha -1) < 0\)

\(\therefore \dfrac{1}{2} < \sin \alpha < 1\). Now \(\sin 30^{\circ}= \sin 150^{\circ}=\dfrac{1}{2}\)

\(\therefore \alpha\) lies between \(\dfrac{\pi }{6} \text{ to } \dfrac{5\pi }{6}\) excluding \(\dfrac{\pi }{2}\) as \(\sin\alpha\) is not equal to 1 and hence (d) is correct.

\(\|x+2\|=-2\)

\(y=\|x+2\|\,,y=-2\)

From the figure, No. of intersection points\(=0\)

\(\therefore\) No. of solutions\(=0\)

Thus, \(x+1> 0\)

\(\therefore \left | x+1 \right |= x+1\) ...(1)

Also, \(3+2x-x^{2}> 0\)

\(\Rightarrow x^{2}-2x-3< 0\)

or \(\left ( x-3 \right )\left ( x+1 \right )< 0\)

\(\therefore -1< x< 3\)

or \(-1< x< 0\) and \(0< x< 3\) ...(2)

Hence, the given equation by (1) reduces to

\(3+2x-x^{2}= \left ( x-3 \right )\left | x \right |\)

[\(\because { a }^{ \log _{ a }{ x } }=x\)]

Now, \(\left | x \right |= -x\) if \(x<0\), \(\left | x \right |= x\) when \(x>0\)

When \(-1< x< 0\)

then \(3+2x-x^{2}= \left ( x-3 \right )\left ( -x \right )\)

or \(3+2x-x^{2}= -x^{2}+3x\)

\(\therefore x= 3 \notin \left [ -1,0 \right ]\)

When \(0< x< 3\), then by (2)

\(3+2x-x^{2}= \left ( x-3 \right )x= x^{2}-3x\)

or \(2x^{2}-5x-3= 0 \therefore x= -\frac{1}{2}, 3\)

These values do not lie in \(0< x< 3\).

Hence, the equation has no solution.

Ans: C

Given:

The equation 2x2 + 5x + 5 = 0

Concept Used:

The quadratic equation Ax² + Bx + C = 0 

If roots are imaginary, B² - 4ac < 0 

Calculation:

Comparing of given equation

Now, A = 2, B = 5 & C = 5 

⇒ (5)² - 4 × 5 × 5 < 0

⇒ 25 - 100 < 0 

⇒ -75 < 0 

∴ The roots are imaginary and distinct. 

Explanation:

(x2 + x + 2)2 - (a - 3)(x2 + x + 2)(x2 + x + 1) + (a - 4)(x2 + x + 1)2 = 0

Dividing the equation by (x2 + x + 1)2  and we get

⇒ \(\left ( \frac{x^{2} + x + 2}{x^{2}+x+1} \right )^{2} \) - (a - 3 )\(\left ( \frac{x^{2} + x + 2}{x^{2}+x+1} \right ) \) + (a - 4) = 0 

Let \(\left ( \frac{x^{2} + x + 2}{x^{2}+x+1} \right ) \) = p   (p >1) , the equation becomes -

p² - (a - 3)p + (a - 4) = 0

⇒ p² - (a - 4)p + p + (a - 4) = 0

⇒ (p - 1)(p - a + 4) = 0

⇒ p = 1 ( ignore it because p > 1) and p = a - 4

So  \(\left ( \frac{x^{2} + x + 2}{x^{2}+x+1} \right ) \) = a - 4

⇒ (a -4)(x² + x + 1) = x² + x + 2

⇒ ax² + ax + a - 4x² - 4x - 4 = x² + x + 2

⇒ (a -5 )x² + (a - 5)x + (a - 6) = 0    ---- (1)

For Real Roots,

D ≥ 0 

⇒ b² - 4ac ≥ 0

⇒ (a - 5)² - 4(a - 5)(a - 6) ≥ 0

⇒ (a -5)[a - 5 - 4(a - 6)] ≥ 0

⇒ (a - 5)( - 3a + 19) ≥ 0

⇒ - (a - 5)(3a - 19) ≥ 0

⇒ (a - 5)(3a - 19) ≤ 0     ( on multiplying or dividing the inequality by negative quantity then inequality sign changes )

With the help of a wavy curve, we get 

⇒ 5 ≤ a ≤ \(\frac{19}{3} \)   ( but we ignore a = 5 because at a = 5 equation (1) does not remain a quadratic equation)

⇒ 5 < a ≤ \(\frac{19}{3} \)

∴ The integral value of a is 6.

The answer is 6.