NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 2: Polynomials

Exercise 2.1

Step 1: Recall the condition.

If \(-3\) is a zero of the polynomial \((k-1)x^2 + kx + 1\), then by definition:

\((k-1)(-3)^2 + k(-3) + 1 = 0\).

Step 2: Simplify the expression.

\((k-1)(9) - 3k + 1 = 0\)

\(9k - 9 - 3k + 1 = 0\)

Step 3: Combine like terms.

\(9k - 3k = 6k\)

\(-9 + 1 = -8\)

So the equation becomes \(6k - 8 = 0\).

Step 4: Solve for \(k\).

\(6k = 8\)

\(k = \dfrac{8}{6}\)

\(k = \dfrac{4}{3}\)

Final Answer: The required value of \(k\) is \(\dfrac{4}{3}\).

Step 1: Write the factor form of the polynomial.

If the zeroes are \(-3\) and \(4\), then the polynomial must be a multiple of

\[(x - (-3))(x - 4) = (x + 3)(x - 4)\]

Step 2: Expand the product.

\[(x + 3)(x - 4) = x^2 - 4x + 3x - 12\]

\[= x^2 - x - 12\]

Step 3: Compare with the given options.

The basic polynomial is \(x^2 - x - 12\).

Any non-zero constant multiple of this is also valid.

Step 4: Check the options.

• Option A: \(x^2 - x + 12\) → wrong (constant term differs).
• Option B: \(x^2 + x + 12\) → wrong (signs do not match).
• Option C: \(\dfrac{1}{2}(x^2 - x - 12) = \dfrac{x^2}{2} - \dfrac{x}{2} - 6\) → correct.
• Option D: \(2x^2 + 2x - 24\) → wrong, because it equals \(2(x^2 + x - 12)\), not a multiple of \(x^2 - x - 12\).

Conclusion: The correct polynomial is

\(\dfrac{x^2}{2} - \dfrac{x}{2} - 6\).

Step 1: Recall the relationships of zeroes with coefficients.

If \(\alpha\) and \(\beta\) are the zeroes of \(x^2 + px + q\), then

\(\alpha + \beta = -p\)

\(\alpha \cdot \beta = q\)

Step 2: Apply to our polynomial.

Here, polynomial is \(x^2 + (a+1)x + b\).

So, coefficient of \(x\) is \(a+1\) and constant term is \(b\).

Thus,

Sum of zeroes \(= - (a+1)\)

Product of zeroes \(= b\)

Step 3: Use the given zeroes.

The zeroes are \(2\) and \(-3\).

So,

\(\alpha + \beta = 2 + (-3) = -1\)

\(\alpha \cdot \beta = 2 \times (-3) = -6\)

Step 4: Form equations for \(a\) and \(b\).

From sum: \(- (a+1) = -1\)

\(a+1 = 1\)

\(a = 0\)

From product: \(b = -6\)

Step 5: Final answer.

So the correct values are \(a = 0\) and \(b = -6\).

This corresponds to option D.

Step 1: If \(-2\) and \(5\) are the zeroes of a polynomial, then the factorised form of such a polynomial must include \((x + 2)(x - 5)\).

Step 2: For example, one polynomial is

\(p(x) = (x + 2)(x - 5)\).

Step 3: But multiplying this expression by any non-zero constant also gives a valid polynomial with the same zeroes. For instance:

\(q(x) = 2(x + 2)(x - 5)\),

\(r(x) = -3(x + 2)(x - 5)\),

\(s(x) = 10(x + 2)(x - 5)\),

and so on.

Step 4: All of these have the same zeroes \(-2\) and \(5\), because the constant multiple does not affect the roots.

Conclusion: There are infinitely many such polynomials. Therefore, the correct choice is (D) more than 3.

Step 1: Write down the roots.

The polynomial is \(ax^3 + bx^2 + cx + d\).

It is given that one root is \(0\). Let the other two roots be \(\alpha\) and \(\beta\).

Step 2: Use relations between coefficients and roots.

For a cubic polynomial \(ax^3 + bx^2 + cx + d\) with roots \(r_1, r_2, r_3\):

\(r_1 + r_2 + r_3 = -\dfrac{b}{a}\)

\(r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{c}{a}\)

\(r_1r_2r_3 = -\dfrac{d}{a}\)

Step 3: Substitute the known root.

Take \(r_1 = 0\), \(r_2 = \alpha\), \(r_3 = \beta\).

Then

\(r_1r_2 + r_2r_3 + r_3r_1 = 0 \cdot \alpha + \alpha\beta + 0 \cdot \beta\)

\(= \alpha\beta\).

Step 4: Compare with the formula.

We know \(r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{c}{a}\).

So \(\alpha\beta = \dfrac{c}{a}\).

Answer: The product of the other two zeroes is \(\dfrac{c}{a}\).

Step 1: Use the fact that \(-1\) is a root.

Substitute \(x=-1\) in the polynomial:

\((-1)^3 + a(-1)^2 + b(-1) + c = 0\)

\(-1 + a - b + c = 0\)

So, \(c = b - a + 1\).

Step 2: Recall the relationship of roots with coefficients.

For a cubic polynomial \(x^3 + ax^2 + bx + c\), if the roots are \(\alpha, \beta, \gamma\), then:

\(\alpha + \beta + \gamma = -a\)

\(\alpha\beta + \beta\gamma + \gamma\alpha = b\)

\(\alpha\beta\gamma = -c\)

Step 3: Express product of the other two zeroes.

Let the roots be \(-1, r, s\).

Then \((-1)rs = -c\)

So, \(rs = c\).

Step 4: Substitute the value of \(c\).

Since \(c = b - a + 1\), we get

\(rs = b - a + 1\).

Final Answer: Option (A) \(b - a + 1\).

Step 1: Recall the relations between coefficients and zeroes.

For quadratic polynomial \(ax^2 + bx + c\):

Sum of zeroes = \(-\dfrac{b}{a}\),

Product of zeroes = \(\dfrac{c}{a}\).

Step 2: Apply to the given polynomial.

Here, \(a = 1\), \(b = 99\), \(c = 127\).

So,

Sum = \(-\dfrac{99}{1} = -99\).

Product = \(\dfrac{127}{1} = 127\).

Step 3: Interpret the signs.

Since product \(127 > 0\), both zeroes must have the same sign.

Since sum \(-99 < 0\), the common sign must be negative.

Conclusion: Both zeroes are negative.

Step 1: Recall relations between coefficients and zeroes.

For a quadratic \(x^2 + kx + k\):

Sum of zeroes = \(-k\).

Product of zeroes = \(k\).

Step 2: Assume both zeroes are positive.

If both are positive, then:

Sum > 0 ⇒ \(-k > 0\) ⇒ \(k < 0\).

Product > 0 ⇒ \(k > 0\).

This is a contradiction (\(k\) cannot be both positive and negative at the same time).

Step 3: Conclusion.

It is impossible for both zeroes to be positive.

Correct Option: (A) cannot both be positive

Step 1: Condition for equal zeroes.

A quadratic equation \(ax^2 + bx + c = 0\) has equal roots when its discriminant is zero.

That is,

\(D = b^2 - 4ac = 0\).

So, \(b^2 = 4ac\).

Step 2: Interpret the relation.

Since \(b^2 \geq 0\), we see that \(4ac = b^2 \geq 0\).

This implies that \(ac \geq 0\).

Step 3: Exclude the zero case.

We are told \(c \ne 0\) and clearly \(a \ne 0\) (otherwise it is not quadratic).

So, \(ac \gt 0\).

Step 4: Final conclusion.

Thus, \(a\) and \(c\) must have the same sign.

Correct Option: (C)

Step 1: Assume the zeroes.

Let the two zeroes be \(r\) and \(-r\).

Step 2: Form the quadratic polynomial.

If \(r\) and \(-r\) are the zeroes, then the polynomial is

\((x - r)(x + r) = x^2 - r^2\).

Step 3: Compare with the given form.

The polynomial becomes \(x^2 + ax + b\).

Here, the coefficient of \(x\) is \(a = 0\), so there is no linear term.

The constant term is \(b = -r^2\).

Step 4: Sign of the constant term.

Since \(r^2 \geq 0\) for all real \(r\),

\(b = -r^2 \leq 0\).

Therefore, the constant term is negative (unless \(r = 0\), in which case the polynomial is just \(x^2\)).

Conclusion: The polynomial has no linear term and the constant term is negative. So the correct option is (A).

Step 1: Recall the property of quadratic graphs.

The graph of a quadratic polynomial \(ax^2 + bx + c\) is always a parabola.

A parabola is a smooth U-shaped curve that opens either upwards (if \(a > 0\)) or downwards (if \(a < 0\)).

Important: A quadratic polynomial graph can have only one turning point, called its vertex.

Step 2: Check each option.

Option (A): The graph is a parabola opening upwards. ✅ Valid quadratic graph.

Option (B): The graph is a parabola opening downwards. ✅ Valid quadratic graph.

Option (C): The graph is also a parabola opening downwards. ✅ Valid quadratic graph.

Option (D): The curve bends twice, showing two turning points. ❌ This is not possible for a quadratic polynomial, but it is possible for a cubic polynomial.

Conclusion: The graph in option (D) is not the graph of a quadratic polynomial.

Correct answer: (D).