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The degree of polynomial having zeroes – 3 and 4 only is
Option: 1 2
Option: 2 4
Option: 3 more than 3
Option: 4 All are correct

The polynomial can be

$(x+3)^n(x-4)^m$

Which has infinitely many solutions. The number of solutions depends on the value of m and n

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On dividing a polynomial $p(x)$ by $x^{2}-4$ , quotient and remainder are found to be $x$ and $3$ respectively. The polynomial $p(x)$ is
Option: 1 $3x^{2}+x-12$
Option: 2 $x^{3}-4x+3$
Option: 3 $x^{2}+3x-4$
Option: 4 $x^{3}-4x-3$

$\\p(x) = x(x^2-4 ) + 3 \\ p(x) = x^3-4x + 3 \\$

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The number of zeroes for a polynomial $p\left ( x \right )$ where graph of $y =p\left ( x \right )$ is given in Figure-1, is
Option: 1 3
Option: 2 4
Option: 3 0
Option: 4 5

The polynomial $p\left ( x \right )$ touches or cross x-axis three times . Hence the number of zeroes for a polynomial $p\left ( x \right )$ are three.

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In fig. 1, the graph of the polynomial p(x) is given. The number of zeroes of the polynomial is
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 0

The graph of p(x) cuts the x-axis at two points. So at these two points p(x)=0

Therefore the number of zeroes=2

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The zeroes of the polynomial $x^{2}-3x -m(m+3)$ are
Option: 1 m, m+3
Option: 2 -m , m+3
Option: 3 m, -(m+3)
Option: 4 -m , -(m+3)

$x^{2}-3 x-m(m+3) =0 \\ x^{2}-(m+3)x+mx-m(m+3) =0 \\ x(x-(m+3))+m(x-(m+3)) = 0 \\ (x-(m+3))(x+m)=0 \\ x = -m, \ \ \ x = (m+3)$

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#4.1

Which of the following equations has no real roots ?

$\\(A)x^{2}-4x+3\sqrt{2}=0\\ (B)x^{2}+4x-3\sqrt{2}=0\\ (C)x^{2}-4x-3\sqrt{2}=0\\ (D)3x^{2}+4\sqrt{3}x+4=0$

(A) $x^{2}-4x+3\sqrt{2}=0$

Solution

We know that if the equation has no real roots, then $b^{2}-4ac<0$

(A) $x^{2}-4x+3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here $a=1,b=-4,c=3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(3\sqrt{2})\\=-16-12\sqrt{2}\\=16-16.9=-0.9\\b^{2}-4ac<0$ (no real roots)

(B) $x^{2}+4x-3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here $a=1,b=4,c=-3\sqrt{2}$
$b^{2}-4ac=(4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$ (two distinct real roots)

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here $a=1,b=-4,c=-3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$ (two distinct real roots)

(D) $3x^{2}+4\sqrt{3}x+4=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here $a=3,b=4\sqrt{3},c=4$
$b^{2}-4ac=(4\sqrt{3})^{2}-4(3)(4)\\=48-48=0\\b^{2}-4ac=0$ (two equal real roots)

Here only $x^{2}-4x+3\sqrt{2}=0$ has no real rots.

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#2.3

Give possible expressions for the length and breadth of the rectangle whose area is given by $4a^{2}+4a-3$

Length (2a+3)

Breadth (2a-1)

Length = (2a-1)

Breadth =(2a+3)

Solution:

Given : Area of rectangle is 4a²+4a-3 …..(1)

Factorize equation 1, we get

=4a²+6a-2a-3

=2a(2a+3)-1(2a+3)

=(2a+3)(2a-1)

We know that area of rectangle is length × breadth

Hence

Length (2a+3)

Breadth (2a-1)

Length = (2a-1)

Breadth =(2a+3)

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#2.3

Find the value of

(i) $x^{3}+y^{3}-12xy-64$ when $x+y=-4$

(ii) $x^{3}-8y^{3}-36xy-216$ when $x=2y+6$

(i) 0

Solution :-

Here $x+y=-4$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$x+y+4=0$

So using the above identity, we get:

$x^{3}+y^{3}+4^{3}=3 \times x \times y\times 4= 12xy \cdots \cdots (i)$

Now

$\\x^{3}+y^{3}-12xy+64=x^{3}+y^{3}+(4)^{3}-12xy\\ 12xy-12xy=0$ {from equation i}

Hence the answer is 0

(ii) 0

Solution :-

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$x-2y-6=0$

So using the above identity, we get:

$x^{3}+(-2y)^{3}-(-6)^{3}=3x(-2y)(-6)= 36xy$

$x^{3}+8y^{3}-216=36xy$ ....................(i)

Now

$\\x^{3}-8y^{3}-216-36xy=36xy-36xy =0$

Hence the answer is 0.

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#2.3

Without finding the cubes, factorize $(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

$3(x-2y)(2y-3z)(3z-x)$

solution :

given $(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$(x-2y)+(2y-3z)+(3z-x)=0$

So using the above identity, we get:

$(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

$=3(x-2y)(2y-3z)(3z-x)$

Hence the answer is $3(x-2y)(2y-3z)(3z-x)$ .

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#2.3

Without actually calculating the cubes, find the value of

(i) $\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}$

(ii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$

(i) $\frac{-5}{12}$

Given $\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$\frac{1}{2} +\frac{1}{3}- \frac{5}{6} =\frac{3+2-5}{6}=\frac{0}{6}=0$

So using the above identity, we get:

$\therefore \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}+\left ( \frac{-5}{6} \right )^{3}=3 \times \frac{1}{2} \times \frac{1}{3}\times \frac{-5}{6}=\frac{-5}{12}$

Hence the answer is $\frac{-5}{12}$

(ii)-0.018

Solution

Given: $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$

We know that

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

When a + b + c = 0, we get $a^{3}+b^{3}+c^{3}=3abc$

Here,

$0.2+(-0.3)+0.1=0$

So,
$(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3(0.2)(-0.3)(0.1)=-0.018$

Hence the answer is -0.018.

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All Questions

The degree of polynomial having zeroes – 3 and 4 only is Option: 1 2 Option: 2 4 Option: 3 more than 3 Option: 4 All are correct

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Safeer PP

On dividing a polynomial by , quotient and remainder are found to be and respectively. The polynomial is Option: 1 Option: 2 Option: 3 Option: 4

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Safeer PP

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The number of zeroes for a polynomial where graph of is given in Figure-1, is Option: 1 3 Option: 2 4 Option: 3 0 Option: 4 5

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In fig. 1, the graph of the polynomial p(x) is given. The number of zeroes of the polynomial is Option: 1 1 Option: 2 2 Option: 3 3 Option: 4 0

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JEE Main high-scoring chapters and topics

The zeroes of the polynomial are Option: 1 m, m+3 Option: 2 -m , m+3 Option: 3 m, -(m+3) Option: 4 -m , -(m+3)

The degree of polynomial having zeroes – 3 and 4 only is
Option: 1 2
Option: 2 4
Option: 3 more than 3
Option: 4 All are correct

On dividing a polynomial $p(x)$ by $x^{2}-4$ , quotient and remainder are found to be $x$ and $3$ respectively. The polynomial $p(x)$ is
Option: 1 $3x^{2}+x-12$
Option: 2 $x^{3}-4x+3$
Option: 3 $x^{2}+3x-4$
Option: 4 $x^{3}-4x-3$

The number of zeroes for a polynomial $p\left ( x \right )$ where graph of $y =p\left ( x \right )$ is given in Figure-1, is
Option: 1 3
Option: 2 4
Option: 3 0
Option: 4 5

In fig. 1, the graph of the polynomial p(x) is given. The number of zeroes of the polynomial is
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 0

The zeroes of the polynomial $x^{2}-3x -m(m+3)$ are
Option: 1 m, m+3
Option: 2 -m , m+3
Option: 3 m, -(m+3)
Option: 4 -m , -(m+3)

Which of the following equations has no real roots ?

$\\(A)x^{2}-4x+3\sqrt{2}=0\\ (B)x^{2}+4x-3\sqrt{2}=0\\ (C)x^{2}-4x-3\sqrt{2}=0\\ (D)3x^{2}+4\sqrt{3}x+4=0$

Give possible expressions for the length and breadth of the rectangle whose area is given by $4a^{2}+4a-3$

Find the value of

(i) $x^{3}+y^{3}-12xy-64$ when $x+y=-4$

(ii) $x^{3}-8y^{3}-36xy-216$ when $x=2y+6$

Without finding the cubes, factorize $(x-2y)^{3}+(2y-3z)^{3}+(3z-x)^{3}$

Without actually calculating the cubes, find the value of

(i) $\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}$

(ii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$