12ஆம் வகுப்பு கணிதவியல் — சமன்பாட்டியல்: Book Back MCQs with Answers & Explanations

A zero is any \(x\) for which \(x^{3}+64=0\), i.e. \(x^{3}=-64\). Because \(-64=(-4)^{3}\), the real cube root is \(x=-4\), and substituting back gives \((-4)^{3}+64=-64+64=0\). None of \(0,\,4,\,4i\) satisfy the equation. Factoring confirms it: \(x^{3}+64=(x+4)\left(x^{2}-4x+16\right)\), so the remaining two zeros are the non-real numbers \(2\pm 2\sqrt{3}\,i\).

The composition \(h(x)=f\big(g(x)\big)\) feeds the polynomial \(g(x)\) into \(f\). The leading behaviour of \(f\) is its top term \(x^{m}\), so the leading behaviour of \(f\big(g(x)\big)\) is \(\big(g(x)\big)^{m}\). Raising a degree-\(n\) polynomial to the power \(m\) multiplies the degree, giving \(n\times m=mn\). For instance, with \(f(x)=x^{2}\) and \(g(x)=x^{3}+1\), \(f\big(g(x)\big)=\left(x^{3}+1\right)^{2}=x^{6}+2x^{3}+1\), which has degree \(6=2\times 3\).

By the Fundamental Theorem of Algebra a degree-\(n\) polynomial equation has exactly \(n\) roots in the complex number system, as long as each root is counted as often as its multiplicity. Since every real number is also complex, the phrase “complex roots” covers real and non-real roots together. The roots need not all be distinct (a repeated root collapses the count of different values) and need not all be real (non-real roots can occur), so those options fail. “At most one root” is plainly false for \(n\ge 2\).

Combine the reciprocals over one denominator: \(\displaystyle\sum\frac{1}{\alpha}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta\gamma+\gamma\alpha+\alpha\beta}{\alpha\beta\gamma}\). For \(x^{3}+px^{2}+qx+r\), Vieta’s formulas give the sum of pairwise products \(\alpha\beta+\beta\gamma+\gamma\alpha=q\) and the product \(\alpha\beta\gamma=-r\) (the sign alternates with degree for a cubic). Hence \(\displaystyle\sum\frac{1}{\alpha}=\frac{q}{-r}=-\frac{q}{r}\).

The Rational Root Theorem says any rational zero \(\dfrac{p}{q}\) in lowest terms must have \(p\) dividing the constant term \(-5\) and \(q\) dividing the leading coefficient \(4\). So \(p\in\{\pm 1,\pm 5\}\) and \(q\in\{\pm 1,\pm 2,\pm 4\}\). Testing the choices: \(-1=\frac{-1}{1}\), \(\frac{5}{4}\), and \(5=\frac{5}{1}\) all fit. But \(\frac{4}{5}\) has numerator \(4\) (which does not divide \(5\)) and denominator \(5\) (which does not divide \(4\)), so it can never be a rational zero of this polynomial.

Factor out \(x\): \(x^{3}-kx^{2}+9x=x\left(x^{2}-kx+9\right)\). One zero is \(x=0\), which is always real. The other two come from \(x^{2}-kx+9=0\) and are real exactly when the discriminant is non-negative: \((-k)^{2}-4(1)(9)\ge 0\), i.e. \(k^{2}-36\ge 0\), giving \(k^{2}\ge 36\) or \(|k|\ge 6\). When \(|k|=6\) the quadratic has a repeated real root, so all three zeros are still real (counted with multiplicity). Therefore three real zeros occur precisely when \(|k|\ge 6\). Choice \(|k|\le 6\) wrongly includes values such as \(k=0\), where \(x^{2}+9=0\) has no real root.

Read the left side as a quadratic in \(\sin^{2}x\). Put \(u=\sin^{2}x\); then \(u^{2}-2u+1=(u-1)^{2}=0\), so \(u=1\), i.e. \(\sin^{2}x=1\) and \(\sin x=\pm 1\). On \([0,\,2\pi]\), \(\sin x=1\) only at \(x=\dfrac{\pi}{2}\) and \(\sin x=-1\) only at \(x=\dfrac{3\pi}{2}\). That is exactly two values of \(x\).

Apply Descartes’ Rule of Signs to \(P(x)=x^{3}+12x^{2}+10ax+1999\). Its coefficients are \(1,\ 12,\ 10a,\ 1999\). When \(a\ge 0\) every coefficient is non-negative, so there is no sign change and hence no positive zero at all. Only when \(a\lt 0\) do the signs read \(+,\,+,\,-,\,+\), producing two sign changes — the single situation in which a positive zero can appear. So \(a\lt 0\) is the condition under which the polynomial admits a positive zero.

Write \(P(x)=x^{3}+0\cdot x^{2}+2x+3\). For positive zeros the non-zero coefficients are \(+,\,+,\,+\) with no sign change, so there are no positive zeros. For negative zeros, \(P(-x)=-x^{3}-2x+3\) has signs \(-,\,-,\,+\): one sign change, hence exactly one negative zero. Indeed \(P(-1)=-1-2+3=0\), and factoring gives \((x+1)\left(x^{2}-x+3\right)\); the quadratic has discriminant \(1-12=-11\lt 0\), so its two zeros are imaginary. The polynomial therefore has one negative real zero and two imaginary zeros.

By the Binomial Theorem, \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}(-1)^{r}x^{r}=\sum_{r=0}^{n}\binom{n}{r}(-x)^{r}=(1-x)^{n}\). The only zero of \((1-x)^{n}\) is \(x=1\), repeated \(n\) times. Since \(x=1\) is positive, all \(n\) zeros (counted with multiplicity) are positive. This also agrees with Descartes’ Rule: the coefficients strictly alternate in sign, giving \(n\) sign changes, and here every one of them is realised as a positive zero.