Chapter 8

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry · 132 exercises

Problem 1

If both the roots of \(a x^{2}+b x+c\) are zero, then find the value of \(a, b, c\).

5 step solution

Problem 3

If the product of the roots of the equation \(m x^{2}+6 x+(2 m-1)=0\) is \(-1\), then find \(m\).

4 step solution

Problem 4

If the sum of the roots of the equation \((a+1) x^{2}+(2 a+3) x+(3 a+4)=0\) is \(-1\), then find the product of the roots.

3 step solution

Problem 5

If \(a, b\) are the roots of the equation \(x^{2}+x+1=0\), then find the value of \(a^{2}+b^{2}\).

4 step solution

Problem 6

If \(\alpha, \beta\) are the roots of the equation \(4 x^{2}+3 x+7=0\), then find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\).

4 step solution

Problem 7

If \(\alpha\) and \(\beta\) are the roots of \(a x^{2}+b x+c=0\), find the values of the following:- i. \(\frac{1}{a \alpha+b}+\frac{1}{a \beta+b} .\left\\{\right.\) ii. \(\frac{\beta}{a \alpha+b}+\frac{\alpha}{a \beta+b} .\) iii. \((a \alpha+b)^{-3}+(a \beta+b)^{-3}\). \(\left\\{\right.\) iv. \((a \alpha+b)^{-2}+(a \beta+b)^{-2}\). \(\left\\{\right.\)

6 step solution

Problem 8

If \(\alpha, \beta\) are roots of the equation \(6 x^{2}-6 x+1=0\), then find the value of \(\frac{1}{2}\left(a+b \alpha+c \alpha^{2}+d \alpha^{3}\right)+\frac{1}{2}\left(a+b \beta+c \beta^{2}+d \beta^{3}\right)\) in terms of \(a, b, c, d\).

5 step solution

Problem 9

If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+a x+b=0\), then prove that \(\frac{\alpha}{\beta}\) is a root of the equation \(b x^{2}+\left(2 b-a^{2}\right) x+b=0 .\)

5 step solution

Problem 10

If \(\alpha\) and \(\beta\) are the roots of \(x^{2}-p(x+1)-c=0\), show that \((\alpha+1)(\beta+1)=1-c .\) Hence prove that \(\frac{\alpha^{2}+2 \alpha+1}{\alpha^{2}+2 \alpha+c}+\frac{\beta^{2}+2 \beta+1}{\beta^{2}+2 \beta+c}=1\)

3 step solution

Problem 11

If the roots of the equation \((x-a)(x-b)-k=0\) be \(c\) and \(d\), then prove that the roots of the equation \((x-c)(x-d)+k=0\) are \(a\) and \(b\).

6 step solution

Problem 12

If the roots of the equation \(x^{2}+p x+q=0\) are obtained \(-2\) and \(-15\) when the coefficient of \(x\) was misread 17 in place of 13 , then find the correct roots of the equation.

3 step solution

Problem 13

Two candidates attempt to solve a quadratic of the form \(x^{2}+p x+q=0 .\) One starts with a wrong value of \(p\) and finds the roots to be 2 and 6 . The other starts with a wrong value of \(q\) and finds the roots to be 2 and \(-9 .\) Find the correct roots.

4 step solution

Problem 14

If \(\alpha\) be a root of \(4 x^{2}+2 x-1=0\), prove that \(4 \alpha^{3}-3 \alpha\) is the other root.

3 step solution

Problem 15

Let \(a, b, c\) be real numbers with \(a \neq 0\) and let \(\alpha, \beta\) be the roots of the equation \(a x^{2}+b x+c=0 .\) Express the roots of \(a^{3} x^{2}+a b c x+c^{3}=0\) in terms of \(\alpha, \beta\).

3 step solution

Problem 16

If \(\alpha+\beta=3\) and \(\alpha^{3}+\beta^{3}=7\), then show that \(\alpha\) and \(\beta\) are the roots of \(9 x^{2}-27 x+20=0\).

3 step solution

Problem 17

If \(\alpha, \beta\) be the roots of \(a x^{2}+2 b x+c=0\) and \(\alpha+\delta, \beta+\delta\) be those of \(A x^{2}+2 B x+C=0\), then prove that \(\frac{b^{2}-a c}{B^{2}-A C}=\left(\frac{a}{A}\right)^{2}\)

7 step solution

Problem 18

The ratio of the roots of the equation \(a x^{2}+b x+c=0\) is same as the ratio of the roots of the equation \(A x^{2}+B x+C=0 .\) If \(D_{1}\) and \(D_{2}\) are the discriminants of \(a x^{2}+b x+c=0\) and \(A x^{2}+B x+C=0\) respectively, then show that \(D_{1}: D_{2}=b^{2}: B^{2}\).

4 step solution

Problem 19

If \(\alpha\) and \(\beta\) be the roots of \(x^{2}+p x-q=0\) and \(\gamma, \delta\) the roots of \(x^{2}+p x+r=0\), prove that \((\alpha-\gamma)(\alpha-\delta)=(\beta-\gamma)(\beta-\delta)=q+r\)

5 step solution

Problem 20

If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+p x+1=0\) and \(\gamma, \delta\) the roots of \(x^{2}+q x+1=0\), show that \(q^{2}-p^{2}=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta) .\)

2 step solution

Problem 21

If \(\sin \alpha\) and \(\cos \alpha\) are roots of the equation \(p x^{2}+q x+r=0\), then show that \(p^{2}-q^{2}+2 p r=0\).

6 step solution

Problem 22

If one root of the equation \(5 x^{2}+13 x+k=0\) is reciprocal of other, then find the value of \(k\).

5 step solution

Problem 23

If the roots of \(p x^{2}+q x+2=0\) are reciprocal of each other, then find \(p\).

3 step solution

Problem 24

Find the condition that the roots of the equation \(a x^{2}+b x+c=0\) be such that i. One root is \(n\) times the other. ii. One root is three times the other.

4 step solution

Problem 25

If the roots of the equation \(a x^{2}+b x+c=0\) are of the form \(\frac{k+1}{k}\) and \(\frac{k+2}{k+1}\), prove that \((a+b+c)^{2}=b^{2}-4 a c\)

5 step solution

Problem 26

If one root of the equation \(a x^{2}+b x+c=0\) be the square of the other, then prove that \(b^{3}+a c^{2}+a^{2} c=3 a b c\).

4 step solution

Problem 27

Find the relation between \(a, b, c\) such that one root of the equation \(a x^{2}+b x+c=0\) may be double of the other.

4 step solution

Problem 28

If one root of the quadratic equation \(a x^{2}+b x+c=0\) is equal to the \(n\) th power of the other root, then show that \(\left(a c^{n}\right)^{\frac{1}{n+1}}+\left(a^{n} c\right)^{\frac{1}{n+1}}+b=0\).

5 step solution

Problem 29

If the equation \(\frac{a}{x-a}+\frac{b}{x-b}=1\) has roots equal in magnitude but opposite in sign, then find the value of \(a+b\).

3 step solution

Problem 30

If the roots of the equation \(\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}\) are equal in magnitude but opposite in sign, show that \(p+q=2 r\) and the product of the roots is equal to \(-\frac{1}{2}\left(p^{2}+q^{2}\right)\).

4 step solution

Problem 31

If the equation \(\left(k^{2}-5 k+6\right) x^{2}+\left(k^{2}-3 k+2\right) x+\left(k^{2}-4\right)=0\) is satisfied by more than two values of \(x\), then determine the value of \(k\).

3 step solution

Problem 32

If the sum of the roots of \(a x^{2}+b x+c=0\) be equal to sum of their squares, prove that \(2 a c=a b+b^{2}\).

4 step solution

Problem 33

\(\alpha, \beta\) are the roots of the equation \(\lambda\left(x^{2}-x\right)+x+5=0\). If \(\lambda_{1}\) and \(\lambda_{2}\) be the two values of \(\lambda\) for which \(\alpha\) and \(\beta\) are connected by the relation \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{5}\), then find the value of \(\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}\).

5 step solution

Problem 35

Find the value of \(p\) for which \(x+1\) is a factor of \(x^{4}+(p-3) x^{3}-(3 p-5) x^{2}+(2 p-9) x+6\). Find the remaining factors for this value of \(p\).

3 step solution

Problem 36

If \(x^{2}-3 x+2\) is a factor of \(x^{4}-p x^{2}+q\), prove \(p=5, q=4\).

8 step solution

Problem 37

If the difference of the roots of the equation \(x^{2}+p x+12=0\) is 1 , find the value of \(p\).

3 step solution

Problem 38

Find the value of \(p\) for which the difference between the roots of the equation \(x^{2}+p x+8=0\) is 2 .

3 step solution

Problem 39

If the roots of the equation \(x^{2}-p x+q=0\) differ by unity, then prove that \(p^{2}=4 q+1\).

7 step solution

Problem 40

If \(a, b\) are roots of the equation \(x^{2}+a b=(a+1) x\), then find the value of \(b\).

3 step solution

Problem 41

Knowing that 2 and 3 are the roots of the equation \(2 x^{3}+m x^{2}-13 x+n=0\), determine \(m\) and \(n\) and find the third root of the equation.

6 step solution

Problem 42

Find the value of \(m\) for which the equation \(\frac{a}{x+a+m}+\frac{b}{x+b+m}=1\) has roots equal in magnitude but opposite in sign. \

3 step solution

Problem 43

If \(p\) and \(q\) are roots of the quadratic equation \(x^{2}+m x+m^{2}+a=0\), then find the value of \(p^{2}+q^{2}+p q\).

6 step solution

Problem 44

If \(\alpha, \beta\) are the roots of the equation \(x^{2}+x+1=0\) and \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha}\) are roots of the equation \(x^{2}+p x+q=0\), then find the value of \(p\).

4 step solution

Problem 45

If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+p x+q=0\) and also of \(x^{2 n}+p^{n} x^{n}+q^{n}=0\) and if \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha}\) are the roots of \(x^{n}+1+(x+1)^{n}=0\), then prove that \(n\) must be a even integer.

5 step solution

Problem 46

Find the polynomial whose roots are 1,2 and 3 .

4 step solution

Problem 47

Form a quadratic equation whose roots are \(\frac{a}{\sqrt{a} \pm \sqrt{a-b}}\).

5 step solution

Problem 48

If \(\alpha\) and \(\beta\) are the roots of \(2 x^{2}-3 x-6=0\), find the equation whose roots are \(\alpha^{2}+2, \beta^{2}+2\).

4 step solution

Problem 49

Find the equation whose roots are \((\alpha+\beta)^{2}\) and \((\alpha-\beta)^{2}\), where \(\alpha, \beta\) are the roots of \(2 x^{2}+2(m+n) x+\left(m^{2}+n^{2}\right)=0 .\) \\{Ans. \(\left.x^{2}-4 m n x-\left(m^{2}-n^{2}\right)^{2}=0\right\\}\)

5 step solution

Problem 50

If \(\alpha\) and \(\beta\) are the roots of \(x^{2}-2 x+3=0\), find the equation whose roots are:i. \(\quad \alpha+2, \beta+2 .\left\\{\right.\) ii. \(\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}\). iii. \(\alpha^{3}-3 \alpha^{2}+5 \alpha-2, \beta^{3}-\beta^{2}+\beta+5\).

7 step solution

Problem 51

If \(\alpha\) and \(\beta\) are the roots of \(a x^{2}+b x+c=0\), find the equation whose roots are given below:- i. \(\quad \frac{1}{\alpha+\beta}, \frac{1}{\alpha}+\frac{1}{\beta} .\left\\{\right.\) ii. \(\frac{\alpha}{\beta}, \frac{\beta}{\alpha} .\left\\{\right.\) iii. \(\alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} .\left\\{\right.\) iv. \(\alpha^{2}+\beta^{2}, \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} .\)c\right)\left(a^{2}+c^{2}\right) x+\left(b^{2}-2 a c\right)^{2}=0\right\\}\( v. \)\frac{1}{a \alpha+b}, \frac{1}{a \beta+b}$.

5 step solution

Problem 52

If \(\alpha, \beta\) are the roots of \(a x^{2}+b x+c=0, \alpha_{1},-\beta\) are the roots of \(a_{1} x^{2}+b_{1} x+c_{1}=0\), show that \(\alpha, \alpha_{1}\) are the roots of \(\frac{x^{2}}{\frac{b}{a}+\frac{b_{1}}{a_{1}}}+x+\frac{1}{\frac{b}{c}+\frac{b_{1}}{c_{1}}}=0\).

4 step solution

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