Q 1 :

Let the sum of the maximum and the minimum values of the function f(x)=2x2-3x+82x2+3x+8  be mn, where gcd(m,n)=1. Then m+n is equal to                  [2024]

  • 182

     

  • 217

     

  • 201

     

  • 195

     

(3)

Let f(x)=2x2-3x+82x2+3x+8=y,2x2+3x+8>0 ∀x∈R

⇒x2(2y-2)+x(3y+3)+8y-8=0

For real roots, D≥0

⇒(3y+3)2-4(2y-2)(8y-8)≥0

⇒(3y+3)2-(8y-8)2≥0

⇒(11y-5)(-5y+11)≥0

⇒(y-511)(y-115)≤0⇒y∈[511,115]

Sum of maximum and minimum values

ymax+ymin=511+115=14655=mn⇒m+n=201



Q 2 :

The coefficients a,b,c in the quadratic equation ax2+bx+c=0 are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8}. The probability of this equation having repeated roots is:             [2024]

  • 3256  

     

  • 1128

     

  • 164

     

  • 3128

     

(3)

For equation ax2+bx+c=0 to have repeated roots we have

D=b2-4ac=0

⇒b2=4ac=22(ac)

⇒ac must be a perfect square.

⇒(a,c)={(1,1),(1,4),(2,2),(2,8),(3,3),(4,1),(4,4),(5,5),(6,6),(7,7),(8,2),(8,8)}

∴(a,b,c)={(1,2,1),(1,4,4),(2,4,2),(2,8,8),(3,6,3),(4,4,1),(4,8,4),(8,8,2)}

Required probability =883=164



Q 3 :

Let α,β be the distinct roots of the equation x2-(t2-5t+6)x+1=0, t∈R and an=αn+βn. Then the minimum value of a2023+a2025a2024 is            [2024]

  • -1/2

     

  • 1/2

     

  • 1/4

     

  • -1/4

(4)

Newton's Theorem says that for a quadratic equation ax2+bx+c=0 if a and b are its roots and an=αn±βn, then aan+1+ban+can-1=0.

So, by Newton's theorem, we have

         a2025-(t2-5t+6)a2024+a2023=0

⇒t2-5t+6=a2025+a2023a2024

Now, t2-5t+6=(t-52)2-14

Minimum value =-14

 



Q 4 :

Let S be the set of positive integral values of a for which ax2+2(a+1)x+9a+4x2-8x+32<0, ∀ x∈R. Then, the number of elements is S is                    [2024]

  • ∞

     

  • 0

     

  • 1

     

  • 3

     

(2)

Given ax2+2(a+1)x+9a+4x2-8x+32<0,∀x∈R

For x2-8x+32,

D=b2-4ac,D=(-8)2-4(1)(32),D=64-128,

D=-64<0,a>0

So, x2-8x+32>0, when

∴   ax2+2(a+1)x+9a+4<0,∀x∈R

       a<0 and D<0

Here, S is the set of positive integral values of x, which is not possible.

So, the number of elements in S=0

 



Q 5 :

The coefficients a,b,c in the quadratic equation ax2+bx+c=0 are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is p, then 216p equals:                    [2024]

  • 76

     

  • 38

     

  • 57

     

  • 19

     

(2)

We need to find the probability that the given equation ax2+bx+c=0 has real and distinct roots.

       [∵ One root will be bigger if roots are distinct]

∴    D>0

⇒b2-4ac>0

b≠1,2          ∵ if b=1or 2

then ac<14 or ac<1 which is not possible as

a,b,c∈{1,2,3,4,5,6}.

If b=3

⇒ac<94⇒(a,c)={(1,1),(1,2),(2,1)}

i.e., 3 ways

If b=4

⇒ac<4⇒(a,c)={(1,1),(1,2),(2,1),(3,1),(1,3)}.

i.e., 5 ways

If b=5

⇒ac<254⇒(a,c)={(1,1),(1,2),(2,1),(3,1),(1,3),(2,2),(1,4),(1,5),(2,3),(3,2),(4,1),(5,1),(6,1),(1,6)}.

i.e., 14 ways

If b=6

⇒ac<9⇒(a,c)={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(4,1),(4,2),(5,1),(6,1)}

i.e., 16 ways

∴    Total number of ways = 3 + 5 + 14 + 16 = 38

⇒ Required probability =3863=p

So, 216p=38216×216=38



Q 6 :

The number of distinct real roots of the equation |x| |x+2|-5|x+1|-1=0 is _________              [2024]



(3)

f(x)=|x||x+2|-5|x+1|-1=0

Case 1 : x≥0

∴    f(x)=x(x+2)-5(x+1)-1=0

⇒x2+2x-5x-6=0⇒x2-3x-6=0

⇒x=3±9+242

⇒x=3±332, One positive root [∵ x≥0]

Case 2: -1≤x<0

⇒f(x)=-x(x+2)-5(x+1)-1=0

⇒-x2-7x-6=0⇒x2+7x+6=0

⇒(x+6)(x+1)=0

⇒x=-6,-1  One root i.e., x=-1

Case 3 : -2≤x<-1

f(x)=-x(x+2)+5(x+1)-1=0

⇒-x2-2x+5x+4=0

⇒-x2+3x+4=0⇒x2-3x-4=0

⇒(x+1)(x-4)=0⇒x=-1,4

No root possible in given range of x.

Case 4 : x<-2

f(x)=-x(-x-2)+5(x+1)-1=0

⇒x2+2x+5x+4=0

⇒x2+7x+4=0

⇒x=-7±49-162=-7±332

One root in the range.

∴       We have 3 distinct real roots



Q 7 :

The number of real solutions of the equation x|x+5|+2|x+7|-2=0 is _______.       [2024]



(3)

Let f(x)=x|x+5|+2|x+7|-2 and f(x)=0

Case 1 : x≥-5

⇒x2+5x+2x+12=0

⇒x2+7x+12=0⇒x=-3 or x=-4

Case 2 : -7<x<-5

⇒-x2-5x+2x+12=0

⇒-x2-3x+12=0⇒x2+3x-12=0

⇒x=-3±572

⇒x≠-3+572                               [∵ -7<x<-5]

Case 3 : x≤-7

⇒-x2-7x-16=0⇒x2+7x+16=0

No real solution

So Number of real solutions = 3

 



Q 8 :

The number of distinct real roots of the equation |x+1||x+3|-4|x+2|+5=0 is ________.         [2024]



(2)

|x+1||x+3|-4|x+2|+5=0

(I) If x<-3,  x2+4x+3+4x+8+5=0

⇒x2+8x+16=0⇒x=-4 (one solution)

(II) If -3≤x<-2,  -x2-4x-3+4x+8+5=0

⇒x2-10=0⇒x=±10

which do not satisfy -3≤x<-2

(III) If -2≤x<-1,  -x2-4x-3-4x-8+5=0

⇒x2+8x+6=0⇒(x+4)2=10

⇒x=-4±10 which do not satisfy -2≤x<-1

(IV) If x≥-1,  x2+4x+3-4x-8+5=0

⇒x2=0⇒x=0 (one solution)

Hence, the number of distinct real roots are two.



Q 9 :

Let the set of all values of p∈R, for which both the roots of the equation x2–(p+2)x+(2p+9)=0 are negative real numbers, be the interval (α,β]. Then β–2α is equal to          [2025]

  • 5

     

  • 0

     

  • 20

     

  • 9

     

(1)

Given, x2–(p+2)x+(2p+9)=0

D≥0

⇒ (p+2)2–4(2p+9)≥0

⇒ p2+4p+4–8p–36≥0

⇒ p2–4p–32≥0

⇒ (p–8)(p+4)≥0

∴  p∈(–∞,–4]∪[8,∞)           ... (i)

So, sum of roots : p + 2 < 0

⇒  p < –2           ... (ii)

and product of roots : 2p + 9 > 0

⇒p>–92           ... (iii)

Using (i), (ii) and (iii), we get

p∈(–92,–4]

∴  β–2α=–4–2(–92)=5.



Q 10 :

The number of real roots of the equation x|x–2|+3|x–3|+1=0          [2025]

  • 3

     

  • 2

     

  • 4

     

  • 1

     

(4)

(I) When x < 2, we have

–x2+2x–3x+9+1=0

⇒ x2+x–10=0

⇒ x=–1+412, –1–412; x=–1–412           (∵  x < 2)

(II) When 2≤x<3, we have

x2–2x–3x+9+1=0

⇒ x2–5x+10=0

As D < 0 ⇒ No real roots.

(III) When x≥3, we have

x2–2x+3x–9+1=0

⇒ x2+x–8=0

⇒ x=–1+332, –1–332  (rejected)

Thus, only one real root exists.



Q 11 :

The sum of the squares of the roots of |x–2|2+|x–2|–2=0 and the squares of the roots of x2–2|x–3|–5=0 is          [2025]

  • 24

     

  • 36

     

  • 30

     

  • 26

     

(2)

We have, |x–2|2+|x–2|–2=0

⇒ |x–2|2+2|x–2|–|x–2|–2=0

⇒ (|x–2|+2)(|x–2|–1)=0

⇒ |(x–2)|=1          [∵  |x–2|≠–2]

⇒ x=2±1 ⇒ x=3,1

Sum of square of roots = 9 + 1 = 10

Now, we have x2–2|x–3|–5=0

Case I : When x – 3 > 0

⇒ x2–2x+1=0 ⇒ (x–1)2=0 ⇒ x=1

but x > 3 ⇒ x≠1

Case II : When x – 3 < 0

⇒ x2+2x–11=0

Discriminant, D = 4 + 44 = 48 > 0

x = –2±482=–2±432=–1±23

SInce, x < 3, so both roots are valid.

Sum of squares of roots = (–1+23)2+(–1–23)2

     =1+12–43+1+12+43=26

∴   Required sum = 10 + 26 = 36.



Q 12 :

If the set of all a∈R–{1}, for which the roots of the equation (1–a)x2+2(a–3)x+9=0 are positive is (–∞,–α]∪[β,γ), then 2α+β+γ is equal to __________.          [2025]



(7)

Let x1 and x2 be the roots of (1–a)x2+2(a–3)x+9=0.

Since, x1,x2>0

∴  x1+x2>0 and x1x2>0

⇒ –2(a–3)1–a>0 ⇒ a–31–a<0

⇒ a∈(–∞,1)∪(3,∞)

Also, x1x2>0 ⇒ 91–a>0 ⇒ 1–a>0 ⇒ a<1

On combining both conditions, we get a∈(–∞,1)

Now, for real roots, discriminant must be non negative.

i.e., (2(a–3))2–4(1–a)9≥0

⇒ 4(a2–6a+9)–36+36a≥0

⇒ 4a2–24a+36–36+36a≥0

⇒ 4a2+12a≥0 ⇒ a(a+3)≥0

⇒ a∈(–∞,–3]∪[0,∞)

Combining all the conditions, we get a∈(–∞,–3]∪[0,1)

∴  α=3, β=0 and γ=1

∴  2α+β+γ=2×3+0+1=7



Q 13 :

The set of all a∈ℝ for which the equation x|x-1|+|x+2|+a=0 has exactly one real root, is          [2023]

  • (-∞,∞)

     

  • (-6,∞)

     

  • (-∞,-3)

     

  • (-6,-3)

     

(1)

Let f(x)=x|x-1|+|x+2|

Given, x|x-1|+|x+2|+a=0

⇒ f(x)+a=0⇒-a=f(x)

f(x)={-x2-2,-∞<x<-2 -x2+2x+2,-2≤x<1x2+2,1≤x<∞

All values are increasing.

 



Q 14 :

The number of real roots of the equation x|x|-5|x+2|+6=0, is             [2023]

  • 5

     

  • 3

     

  • 4

     

  • 6

     

(2)

We have, x|x|-5|x+2|+6=0

Case I: x≤-2

 -x2+5x+10+6=0

⇒-x2+5x+16=0⇒x2-5x-16=0

⇒x=5±(-5)2+16×42=5±25+642=5±892

Only x=5-892 belongs to (-∞,-2]

So, one solution exists in this case.

Case II: -2<x≤0

-x2-5x-10+6=0

⇒-x2-5x-4=0⇒x2+5x+4=0

⇒(x+1)(x+4)=0⇒x=-1 or x=-4

Since -2<x≤0, x=-4 is not a solution. 

So, x=-1 is a solution in this case.

Case III: x>0

x2-5x-4=0⇒x=5±25+162=5±412

But x>0. So, x=5+412 is the only solution in this case.

Therefore, the equation has three solutions.



Q 15 :

The number of integral values of k, for which one root of the equation 2x2-8x+k=0 lies in the interval (1, 2) and its other root lies in the interval (2, 3), is      [2023]

  • 2

     

  • 0

     

  • 1

     

  • 3

     

(3)

2x2-8x+k=0 represents an upward parabola.

f(1)·f(2)<0  and  f(2)·f(3)<0

(2(1)2-8(1)+k)·(2(2)2-8(2)+k)<0

and  (2(2)2-8(2)+k)(2(3)2-8(3)+k)<0

(k-6)(k-8)<0  and  (k-8)(k-6)<0

k∈(6,8)  and  k∈(6,8)

Integral value of k=7



Q 16 :

Let m and n be the numbers of real roots of the quadratic equations x2-12x+[x]+31=0 and x2-5|x+2|-4=0 respectively, where [x] denotes the greatest integer ≤x. Then m2+mn+n2 is equal to __________ .         [2023]



(9)

x2-12x+[x]+31=0 

⇒  {x}=x2-11x+31⇒0≤x2-11x+31<1 

⇒  x2-11x+30<0⇒x∈(5,6) 

So,  [x]=5 

Now,  x2-12x+5+31=0 

⇒  x2-12x+36=0⇒x=6 but x∈(5,6)

m=0

Now, for  x2-5|x+2|-4=0

When

x≥-2;                     x<-2 

x2-5x-14=0;     x2+5x+6=0

(x-7)(x+2)=0;    (x+3)(x+2)=0

x=7,-2;                  x=-3,-2

Now,  x2-5|x+2|-4=0⇒x={7,-2,-3}      ∴ n=3

So,  m2+mn+n2=9



Q 17 :

Let S={α:log2(92α-4+13)-log2(53·32α-4+1)=2}. Then the maximum value of β for which the equation

 x2-2(∑α∈Sα)2x+∑α∈S(α+1)2β=0 has real roots, is ________ .               [2023]



(25)

Given log2(92α-4+13)-log2(52·32α-4+1)=2 
 
⇒log2(92α-4+1352·32α-4+1)=2⇒92α-4+1352·32α-4+1=4 

⇒  92α-4+13=10·32α-4+4⇒10·32α-4-92α-4=9  

Put α=1,  10·3-2-9-2=109-181=8981≠9 (not satisfy)

Put α=2,  10·30-90=10-1=9 (satisfy) 

Put α=3,  10·32-92=90-81=9 (satisfy)

Hence,  α=2 or 3.

Now,

∑α∈Sα=2+3=5 and ∑α∈S(α+1)2=∑α∈Sα2+2∑α∈Sα+∑α∈S1

=4+9+2(5)+2=25

So, x2-2(5)2x+25β=0=x2-50x+25β=0

For real roots, D≥0 

⇒2500-4(25β)≥0⇒100β≤2500⇒β≤25

Hence, βmax=25



Q 18 :

Let f(x)=x2+bx+c, minimum value of f(x) is -5, then absolute value of the difference of the roots of f(x) is:

  • 5

     

  • 20

     

  • 15

     

  • Can't be determined

     

(2)

Minimum value -D4=-5⇒D=20        

|α-β|=D1=20



Q 19 :

Which of the following is always true

  • One root of the equation ax2+bx+c=0, in the form of p+q, then other root is p-q (p∈z, q∈z+)

     

  • If the equation ax2+bx+c=0 is satisfied by more than two distinct numbers (real or complex), then it becomes an identity.

     

  • If a=2 and b,c are integers and the roots of the quadratic equation ax2+bx+c=0 are rational, then the roots must be integers.

     

  • a,b,c are in A.P and G.P then (a,b,c) can be (k,k,k) for all k∈R

     

(2)

a, b can be irrational



Q 20 :

If the quadratic equation (λ+2)x2–3λx+4λx=0, λ≠–2, has two positive roots, then the number of possible integral values of λ is :          [2026]

  • 1

     

  • 2

     

  • 3

     

  • 4

     

(2)

Figure

Let α,β be the roots of (λ+2)x2–3λx+4x=0

Since, both roots are positive

∴  Sum > 0, Product > 0, Discriminant ≥ 0

Now, Sum > 0

⇒ 3λλ2+2>0

f(x)=x2–3λxλ+2+4λλ+2

Product > 0 ⇒ 4λλ+2>0

Both gives λ∈(–∞,–2)∪(0,∞)          ... (i)

Also, discriminant ≥ 0, i.e.,D≥0

9λ2(λ+2)2–16λλ+2≥0

⇒ 9λ2–16λ(λ+2)≥0

⇒ –7λ2–32λ≥0 ⇒ 7λ2+32λ≤0

⇒ λ∈[–327,0]              ... (ii)

From (i) and (ii), we get

λ∈[–327,–2)

∵ Integral values are –4, –3.

So, number of possible integral values of λ is 2.



Q 21 :

Statement – I: If α=cos(2π7)+isin(2π7), p=α+α2+α4, q=α3+α5+α6, then the equation whose roots are p and q is x2+x+2.

Statement – II: If α is a root of z7=1, then 1+α+α2+…+α6=0

  • Statement 1 is true, Statement – 2 is true

     

  • Statement 1 is false, Statement – 2 is false

     

  • Statement 1 is true, Statement – 2 is false

     

  • Statement 1 is false, Statement – 2 is true

     

(1)

α is 7th root of unity    ⇒    1+α+α2+⋯+α6=0,  p+q=-1

pq=α4+α6+α5+α7+α8+α7+α9+α10

=3+(α+α2+α3+⋯+α6)=3+(-1)=2

⇒x2+x+2=0

Both I and II are true and II is the correct explanation.



Q 22 :

The number of real solutions of the equation e4x+4e3x-58e2x+4ex+1=0 is

  • 4

     

  • 6

     

  • 2

     

  • 8

     

(3)

Given equation is

e4x+4e3x-58e2x+4ex+1=0

Take, f(x)=(e2x+1e2x+4(ex+1ex)-58)

Let ex+1ex=p (>0)    …(i)

p2+4p-60=0

p=6  or  p=-10

Only p=6 is allowed

ex+1ex=6

Two real and distinct values of x



Q 23 :

If x be real, then the minimum value of x2−8x+17 is

  • -1

     

  • 0

     

  • 1

     

  • 2

     

(3)

x2-8x+17=(x-4)2+1 or differentiate or use formula for minimum of quadratic function.



Q 24 :

Statement I: If a1x2+b1x+c1=0 and a2x2+b2x+c2=0 (a1≠0,a2≠0) have a common root.

Then (c1a2-c2a1)2=(b1c2-b2c1)(a1b2-a2b1)

Statement II: The quadratic equations x2-6x+a=0 and x2-cx+6=0 have one root in common. The other roots of the first and second equations are integers in the ratio 4:3, if the common root is α then α3 is 8

  • Statement I is true and Statement II is true

     

  • Statement I is false and Statement II is true

     

  • Statement I is true and Statement II is false

     

  • Statement I is false and Statement II is false

     

(1)

Let root of x2-6x+a=0 are α, 4β=4αβ=a

x2-cx+6=0 are α, 3β=3α, β=6⇒a=8

⇒ x2-6x+8=0⇒x2-cx+6=0

Has a root common ⇒c=5 or c=112

Integral roots are 2, 4 and 2, 3. Common root is 2