Q: If α and β are the roots of the equation x2+4x−7=0, the equation whose roots are 1+αα and 1+ββ is

Let y=1+xx⇒xy+y=x⇒x(y−1)=−y ⇒x=1−yySubstituting for x in the given equation⇒(1−y)2y2+1−y4y−7=0⇒10y2−18y+7=0Or the required equation is 10x2−18x+7=0

Q: The ratio of the roots of the equation x2−7x+1=0 is the same as the ratio of the roots of the equation λx2+4x−5 =0. The value of λ equals

Let kα and α be the roots of x2−7x+1=0 kα+α=7,kα2=1⇒(k+17)2=k1⇒k(k+1)2=49Similarly, for the second equation, we get[λ(k+1)−4]2=λk−5⇒k(k+1)2=5λ2−16λ=5λ−16We have 5λ−16=49⇒λ=245−16

Q: If x2+px−5=0 and x2+qx+5=0,p=q have a common root, the value of (p2−q2) is

Let α denote the common root Then,5p+5qα2=−10α=q−p1α=(p−q)10⇒α2=(p−q)−5(p+q)⇒(p−q)−5(p+q)=(p−q)2100⇒−(p+q)=p−q20⇒(p2−q2)=−20

Q: Statement 1 :The roots of the equation x2+7x+12=0 are both negative.Statement 2 : If a,b,c are of the same sign then both the roots of ax2+bx+c=0 are negative.

The roots of x2+7x+12=0 are −3 and −4. So Statement 1 is true.Consider the equation x2+x+4=0Although a,b,c are positive the roots are complex.Statement 2 is false.

Q: Statement 1 Number of real roots of the equation x2−∣x∣−2=0 is 2 .Statement 2 A quadratic equation ax2+bx+c=0 has two and only two roots.

Statement 2 is trueStatement 1Let x>0⇒x2−x−2=0⇒x=2 is a solution.Let x<0⇒x2+x−2=0⇒x=−2 is a solution x=0 does not satisfy the given equation.Equation has 2 roots. However, it does not follow from statement 2Choice (b)

Q: Statement 1 If the roots of the quadratic equation ax2+bx+c=0 lie between −2 and 2 , then both (4a+2b+c) and (4a−2b+c) must be positive.Statement 2 If α and β are the real roots of the equation ax2+bx+c=0, then (ax2+bx+c) will have the same sign as that of a if x is chosen as a number lying beyond α and β.

Statement 2 is trueStatement 1 is false since the sign of (4a+2b+c) and (4a−2b+c) depends on the sign of a. Also, other conditions have to be satisfied for the roots to lie between −2 and 2Choice (d)

Q: Statement 1 Minimum value of the quadratic expression (x2+x+4) is 415. andStatement 2 Minimum or maximum value of the quadratic expression (ax2+bx+c) where a,b,c are real occurs at x=2α+β where α and β are the roots of the equation ax2+bx+c=0

Statement 2 is trueStatement 1α+β=−1Hence, using statement 2 , the minimum value of x2+x+4 is(2−1)2−21+4=415Choice (a)

Question 1

If α and β are the roots of the equation x2+4x−7=0, the equation whose roots are 1+αα​ and 1+ββ​ is

Accepted Answer

Let y=1+xx​⇒xy+y=x⇒x(y−1)=−y ⇒x=1−yy​Substituting for x in the given equation⇒(1−y)2y2​+1−y4y​−7=0⇒10y2−18y+7=0Or the required equation is 10x2−18x+7=0

Question 2

The ratio of the roots of the equation x2−7x+1=0 is the same as the ratio of the roots of the equation λx2+4x−5 =0. The value of λ equals

Accepted Answer

Let kα and α be the roots of x2−7x+1=0 kα+α=7,kα2=1⇒(k+17​)2=k1​⇒k(k+1)2​=49Similarly, for the second equation, we get[λ(k+1)−4​]2=λk−5​⇒k(k+1)2​=5λ2−16λ​=5λ−16​We have 5λ−16​=49⇒λ=245−16​

Question 3

Number of positive roots of the equation (x−1)(x−2)(x−3)(x−4)=15 is

Accepted Answer

The equation may be rewritten as (x2−5x+4)(x2−5x+ 6) =15Let y=x2−5x+4.Then we have y(y+2)−15=0⇒y=−5,3y=−5⇒x2−5x+9=0⇒ No real rootsy=3⇒x2−5x+1=0⇒ Two positive roots

Question 4

If x2+px−5=0 and x2+qx+5=0,p=q have a common root, the value of (p2−q2) is

Accepted Answer

Let α denote the common root Then,5p+5qα2​=−10α​=q−p1​α=(p−q)10​⇒α2​=(p−q)−5(p+q)​⇒(p−q)−5(p+q)​=(p−q)2100​⇒−(p+q)=p−q20​⇒(p2−q2)=−20

Question 5

The integer k for which the inequality x2−2(4k−1)x+15k2−2k−7>0 is valid for real x is

Accepted Answer

Let f(x)=x2−2(4k−1)x+15k2−2k−7>0Then f(x)>0⇒ Discriminant <04(4k−1)2−4(15k2−2k−7)<016k2−8k−15k2+2k+1+7<0k2−6k+8<0⇒(k−4)(k−2)<02

Question 6

Statement 1 :The roots of the equation x2+7x+12=0 are both negative.Statement 2 : If a,b,c are of the same sign then both the roots of ax2+bx+c=0 are negative.

Accepted Answer

The roots of x2+7x+12=0 are −3 and −4. So Statement 1 is true.Consider the equation x2+x+4=0Although a,b,c are positive the roots are complex.Statement 2 is false.

Question 7

Statement 1  Number of real roots of the equation x2−∣x∣−2=0 is 2 .Statement 2 A quadratic equation ax2+bx+c=0 has two and only two roots.

Accepted Answer

Statement 2 is trueStatement 1Let x>0⇒x2−x−2=0⇒x=2 is a solution.Let x<0⇒x2+x−2=0⇒x=−2 is a solution x=0 does not satisfy the given equation.Equation has 2 roots. However, it does not follow from statement 2Choice (b)

Question 8

Statement 1 If the roots of the quadratic equation ax2+bx+c=0 lie between −2 and 2 , then both (4a+2b+c) and (4a−2b+c) must be positive.Statement 2 If α and β are the real roots of the equation ax2+bx+c=0, then (ax2+bx+c) will have the same sign as that of a if x is chosen as a number lying beyond α and β.

Accepted Answer

Statement 2 is trueStatement 1 is false since the sign of (4a+2b+c) and (4a−2b+c) depends on the sign of a. Also, other conditions have to be satisfied for the roots to lie between −2 and 2Choice (d)

Question 9

Statement 1 Minimum value of the quadratic expression (x2+x+4) is 415​. andStatement 2 Minimum or maximum value of the quadratic expression (ax2+bx+c) where a,b,c are real occurs at x=2α+β​ where α and β are the roots of the equation ax2+bx+c=0

Accepted Answer

Statement 2 is trueStatement 1α+β=−1Hence, using statement 2 , the minimum value of x2+x+4 is(2−1​)2−21​+4=415​Choice (a)

Question 10

Statement 1 If 7x2−8x+105loge​(x−e)2​>0 then x∈[e−1,e+1] andStatement 2 7x2−8x+105>0 for all real x

Accepted Answer

Statement 2For 7x2−8x+105=0, its discriminant64−7×4×105<0∴7x2−8x+105>0∀x∈R∴ Statement 2 is true.Statement 17x2−8x+105loge(x−e)2>0⇒(x−e)2>1using R⇒x−e>1 or x−e<−1⇒x>e+1 or x

Question 11

Given that (3−10​) is a root of the equation x4−8x3+16x2−28x−5=0, the sum of the reciprocals of the squares of the roots of the equation is equal to

Accepted Answer

Since the coefficients of the equation are rational, (3+10​) is also a root of the equation. Factors corresponding to the two roots above are (x−3−10​) and (x−3−10​)Their product is (x−3)2−10 or (x2−6x−1)Dividing (x4−8x3+16x2−28x−5) by (x2−6x−1) we get the quotient as x2−2x+5x2−2x+5=0gives x=1±2iRoots of the equation are3±10​ and 1±2iSum of the squares of the reciprocals of the roots=(3−10​)21​+(3+10​)21​+(1+2i)21​+(1−2i)21​=(3+10​)2+(3−10​)2+25(1−2i)2​+25(1+2i)2​=2(9+10)+252​{1−4}=38−256​=25944​

Question 12

Given that (2+i3​) is a root of the equation 2x4−5x3−3x2+41x−35=0, the other three roots are such that

Accepted Answer

Since the coefficients of the equation are real, (2−i3​) is also a root of the equation. Product of the factors corresponding to the roots 2±i3​ is (x−2)2+3 or x2−4x+7Dividing (2x4−5x3−3x2+41x−35) by (x2−4x+7), the quotient is (2x2+3x−5)2x2+3x−5=0 has 2 rational roots, since the discriminant is a perfect square.

Question 13

Two roots of the equation x4−6x3+18x2−30x+25=0 are of the form (a+ib) and (b+ia) where, a and b are real. Then, (a3+b3) equals

Accepted Answer

Since the coefficients of the equation are real, (a−ib) and (b−ia) are the other two roots.sum of the roots =2a+2b=6⇒a+b=3Product of the roots =(a2+b2)2=25⇒a2+b2=5⇒(a+b)2−2ab=5⇒9−2ab=52ab=4⇒ab=2a=2,b=1ora=1,b=2a3+b3=8+1=9

Question 14

The equation (1−p2)x2+2px−1=0 has both its roots in the interval (0,2), if

Accepted Answer

Let f(x)=(1−p2)x2+2px−1Both the roots of f(x)=0 are in (0,2)⇒D≥0 and coefficient of x2,f(0) and f(2) have the same sign.⇒4p2+4(1−p2)≥0,(1−p2)f(0)>0 and (1−p2)f(2)>0⇒4≥0 (which is true always), (1−p2)(−1)>0 and (1−p2)[4(1−p2)+4p−1]>0⇒(p2−1)>0 and (1−p2)(−4p2+4p+3)>0(2p−3)(2p+1)>0⇒(p<−1, or p>1) and (p<−1,or (p<−1, or −2123)⇒p<−1 or p>23−(A)Also, x=2α+β=(1−p2)−pMust lie between 0 and 2i.e., 0<1−p2−p<2⇒01 or −10⇒p41+17OR41−1741+17 or 41−1723Choices (c) and (d) are true

Question 15

The equations px2+qx+r=0 and qx2+rx+p=0(r=0) have a common root, if

Accepted Answer

The equations, px2+qx+r=0 and qx2+rx+p=0 have a common root.⇒(qr−p2)2=(rp−q2)(pq−r2)−(1)⇒p(p3+q3+r3)=3p2qr⇒p3+q3+r3=3pqr(∵p=0)−(2)⇒pqrp3+q3+r3​=3(∵p,q,r=0)∴(d) is true and (c) is not.(2) can be written as0=p3+q3+r3−3pqr=2(p+q+r)​[(p−q)2+(q−r)2+(r−p)2]⇒p+q+r=0 or p=q=r=4p{p3+q3+r3−3pqr}=0(b) is trueChoices (a), (b), (d)

Question 16

If the equation 2(log3​x)2−∣log3​x∣+k=0 has four solutions, if

Accepted Answer

t=∣log3x∣2t2−t+k=0Discriminant >0⇒1−8k>0⇒k<81t=∣log3x∣>0The roots of (1) are positive∴ product of the roots is >02k>0⇒k>0Hence 0

Question 17

The roots of the equation t+t1​=6 are

Accepted Answer

t2−6t+1=0⇒t=26±36−4​​=26±42​​=3±22​.

Question 18

Roots of the equation 7x+1+71−x=50 are

Accepted Answer

By trial methods −1 and +1 satisfy the equation.

Question 19

If the roots of x2+(k−1)x=2k+1 are equal, value of k is

Accepted Answer

b2−4ac=0⇒(k−1)2+4(2k+1)=0k2−2k+1+8k+4⇒k2+6k+5=0k=−5,−1

Filters

Filters

book

topic

difficulty

Solutions for IIT-JEE Super Course in Mathematics -Algebra I by Trishna Knowledge Systems

Found the solution, but did not understand the concept?

Ask my first question