Mathematics is one of the most interesting subjects of all time. For a few students, algebra is one of the trickiest chapters for JEE. However, it’s easy to learn and understand. You need to make sure you practice a lot of problems and keep the standard tricks right up your sleeves.

We are here to help you with the top tough questions with the solutions to help you brush up on your concepts. Make sure to practise them thoroughly and regularly to score more in the upcoming JEE Mains 2022 exams.

### Question 1

The value of a for which one root of the quadratic equation (a2 – 5a + 3) x2 + (3a – 1) x + 2 = 0 is twice as large as other, is

(A) –2/3

(B) ⅓

(C) –1/3

(D) 2/3

Answer:

(D) Let α and 2 be the roots of the given equation ,

then (a^2 – 5a + 3)α^2 + (3a – 1)α + 2 = 0 … (i)

and (a^2 – 5a + 3) (4 α ^2 ) + (3a – 1) (2(A) + 2 = 0 … (ii)

Solving (i) and (ii), we get α = -3/(3a-1)

put this value in (ii), we get a = 2/3

### Question 2

Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively are given by _________.

Answer:

Let r = λb + μc and c = ± (xi + yj).

Since c and b are perpendicular, we have 4x + 3y = 0

⇒ c = ±x (i − 43j), {Because, y = [−4 / 3]x}

Now, projection of r on b = [r . b] / [|b|] = 1

⇒ [(λb + μc) . b] / [|b|]

= [λb . B] / [|b|] = 1

⇒ λ = 1 / 5

Again, projection of r on c = [r . c] / [|c|] = 2

This gives μx = [6 / 5]

⇒ r = [1 / 5] (4i + 3j) + [6 / 5] (i − [4 / 3]j)

= 2i−j or

r = [1 / 5] (4i + 3j) − [6 / 5] (i − [4 / 3]j)

= [−2 / 5] i + [11 / 5] j

### Question 3

A unit vector makes an angle π / 4 with a z-axis. If a + i + j is a unit vector, then a is equal to _________.

Answer:

Let a = li + mj + nk, where l2 + m2 + n2 = 1. a makes an angle π / 4 with z−axis.

Hence, n = 1 / √2, l2 + m2 = 1 / 2 …..(i)

Therefore, a = li + mj + k / √2

a + i + j = (l + 1) i + (m + 1) j + k / √2

Its magnitude is 1, hence (l + 1)2 + (m + 1)2 = 1 / 2 …..(ii)

From (i) and (ii),

2lm = 1 / 2

⇒ l = m = −1 / 2

Hence, a = [−i / 2] − [j / 2] + [k / √2].

### Question 4

Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation p × {(x − q) × p} + q × {(x − r) × q} + r × {(x − p) × r} = 0, then x is given by ____________.

Answer:

|p| = |q| = |r| = c, (say) and

p . q = 0 = p . r = q . r

p × |( x − q) × p |+ q × |(x − r) × q| + r × |( x − p) × r| = 0

⇒ (p . p) (x − q) − {p . (x − q)} p + . . . . . . . . . = 0

⇒ c2 (x − q + x − r + x − p) − (p . x) p − (q . x) q − (r . x) r = 0

⇒ c2 {3x − (p + q + r)} − [(p . x) p + (q . x) q + (r . x) r] = 0

which is satisfied by x = [1 / 2] (p + q+ r).

### Question 5

If a,b,c are real and x3 − 3b2x + 2c3 is divisible by x − a and x − b, then what is a?

Answer:

As f(x) = x3 − 3b2x + 2c3 is divisible by x − a and x − b,

Therefore, f(a) = 0 ⇒ a3 − 3b2a + 2c3 = 0 …..(i) and

f(b)=0

b3 − 3b3 + 2c3 = 0 …..(ii)

From (ii), b = c

From (i), a3 − 3ab2 +2b3 = 0 (Putting b=c) ⇒ (a−b) (a2 + ab − 2b2) = 0 ⇒ a = b or a2 + ab = 2b

Thus a = b = c or a2 + ab = 2b2 and b = ca2 + ab = 2b2 is satisfied by a = −2b.

But b = c , a2 + ab − 2b2 and b = c is equivalent to a = −2b = −2c

### Question 6

Find the number of real values of x for which the equality ∣3x2 + 12x + 6∣ = 5x + 16 holds good.

Answer:

Equation is |3x2 + 12x + 6|= 5x + 16 …….(i)

when 3x2 + 12x + 6 ≥ 0

⇔ x2+ 4x ≥ −2

⇔ |x + 2|2 ≥ 4 − 2

⇔ |x + 2| ≥ (√2)2

⇔ x + 2 ≤ √−2 or x + 2 ≥ √2 …….(ii)

Then (i) becomes 3x2 + 12x + 6 = 5x + 16

⇔3x2 + 7x −10 = 0

⇒x = 1, −10 / 3

But x = −10 / 3 does not satisfy (ii).

When 3x2 + 12x + 6 < 0 ⇒ x2+ 4x < −2

|x + 2| ≤ √2 ⇒ [−√2 −2] ≤ x ≤ [−2 + √2] ? …….(iii)

Then (i) becomes ⇒ 3×2 + 12x + 6 = −(5 x + 16)

⇒3x2 + 17x + 22 = 0

⇒x =−2,−11 / 3

But x = −11 / 3 does not satisfy (iii).

So, 1 and – 2 are the only solutions.

### Question 7

What is the set of all real numbers x for which x2−|x + 2|+ x > 0?

Answer:

Case I:

When x + 2 ≥ 0 i.e. x ≥ −2, then given inequality becomes x2−|x + 2|+ x > 0

x2−2 > 0 ⇒|x| > √2

x<√−2 or x > √2

As x ≥ −2, therefore, in this case the part of the solution set is [−2 ,√−2) ∪ (√2 , ∞).

Case II: When x + 2 ≤ 0 i.e. x ≤ −2, then given inequality becomes x2 + (x + 2) + x > 0

⇒ x2 + 2x + 2 > 0 ⇒ (x + 1)2 + 1 > 0, which is true for all real x.

Hence, the part of the solution set in this case is (−∞, −2].

Combining the two cases, the solution set is (−∞, −2) ∪ ([−2, √−2] ∪ (√2, ∞) = (−∞, √−2) ∪ (√2, ∞).

### Question 8

If a, b are the roots of x2 + px + 1 = 0 and c, d are the roots of x2 + qx + 1 = 0, then the value of (a – (C) (b – (C) (a + (D) (b + (D) is

(A) p 2 – q 2

(B) q 2 – p 2

(C) q 2 + p2

(D) none of these

Answer:

(B)

Given, x^2+ px + 1 = (x – (A) (x – (B) => a + b = -p

and ab = 1 and x^2 + qx + 1 = (x – (C) (x – (D) => c + d = -q and cd = 1

Therefore, (a – (C) (b – (C) (a + (D) (b + (D) = (c – (A) (c – (B) (-d – (A) (-d – (B)

= (c^2 + pc + 1) (d2 – pd + 1)

But c^2 + qc + 1 = 0 and d^2 + qd + 1 = 0.

Hence,

(a – (C) (b – (C) (a + (D) (b + (D) = (-qc + p(C) (-qd – p(D) = cd (q – p) (q + p) = q^2 – p^2

### Question 9

If the coefficient of x7 in (ax2 + [1/bx])11 is equal to the coefficient of x−7 in (ax − 1/bx2)11, then ab = __________.

Answer:

In the expansion of (ax2 + [1/bx])11, the general term is

Tr + 1 = 11Cr (ax2)11 − r (1/bx)r

= 11Cr * a11 − r * [ 1/br ] * x22 − 3r

For x7, we must have 22 – 3r = 7

r = 5, and the coefficient of x7 = 11C5 * a11−5 * [1/b5] = 11C5 * a6 * b5

Similarly, in the expansion of (ax−1bx2)11, the general term is

Tr + 1 = 11Cr * (−1)r * [a11 − r/br ] * x11 − 3r

For x-7 we must have, 11 – 3r = -7 , r = 6, and the coefficient of x−7 is 11C6 * [a5/b6] = 11C5 * a5 * b6.

As given, 11C5 [a6/b5] = 11C5 * [a5/b6]

⇒ ab = 1

### Question 10

Let R = (5√5 + 11)2n + 1 and f = R − [R], where [.] denotes the greatest integer function. The value of R * f is

Answer:

Since (5√5 − 11) (5√5 + 11) = 4

5√5 − 11 = 4 / (5√5 + 11), Because 0 < 5√5 − 11 < 1 ⇒ 0 < (5√5 − 11)2n + 1 < 1, for positive integer n.

Again, (5√5 + 11)2n + 1 − (5√5 − 11)2n + 1 = 2 {2n+1C1 (5√5)2n * 11 + 2n + 1C3(5√5)2n − 2 × 113 + . . .. +

2n + 1C2n+1 112n+1}

= 2 {2n+1C1(125)n * 11 + 2n+1C3(125)n−1 113 +. . . . . . + 2n+1C2n+1 112n+1}

= 2k, (for some positive integer k)

Let f′= (5√5 − 11)2n + 1 , then [R] + f − f′=2k

f − f′ = 2k − [R]

⇒ f − f′ is an integer.

But, 0 ≤ f < 1; 0 < f′<1

⇒ −1 < f − f′ < 1

f − f′ = 0 (integer)

f = f′

Therefore, Rf = Rf′ = (5√5 + 11)2n + 1 * (5√5 − 11)2n + 1

= ([5√5]2 + 112)2n + 1

= 42n+1

### Question 11

If the coefficients of pth, (p + 1)th and (p + 2)th terms in the expansion of (1 + x )n are in A.P., then find the equation in terms of n.

Answer:

Coefficient of pth, (p + 1)th and (p + 2)th terms in expansion of (1 + x )n are nCp−1, nCp ,nCp+1. Then 2nCp = nCp−1 + nCp+1

⇒n2 − n (4p + 1) + 4p2 − 2 = 0

Trick: Let p = 1, hence nC0, nC1 and nC2 are in A.P.

⇒ 2 * nC1 = nC0 + nC2

⇒ 2n = 1 + [n (n − 1)] / [2]

⇒ 4n = 2 + n2 − n

⇒ n2 − 5n + 2 = 0

### Question 12

If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].

Answer:

f (x) = cos (log x)

Now let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

y = 0

### Question 13

If log10 5 = a and log10 3 = b then

(a) log30 8 = 3(1-a)/(b+1)

(b) log 40 15 = (a+b)/(3-2a)

(c) log 243 32 = (1-a)/b

(d) none of these

Solution:

Given log10 5 = a and log10 3 = b

We check all the given options.

log30 8 = log 23/log (3×10)

= 3 log 2/( log 3 + log 10)

= 3 log (10/5)/(1 + log 3)

= 3 (1 – log 5)/(1 + log 3)

= 3(1 – a)/(1+b)

Hence option a is correct.

log 40 15 = log 15/log 40

= log(3×5)/log(10×4)

= (log 3 + log 5)/(1 + log 22)

= (log 3 + log 5)/(1 +2 log 2)

= (log 3 + log 5)/(1 +2 log (10/5))

= (log 3 + log 5)/(1 +2 (1 – log 5))

= (log 3 + log 5)/(1 +2 – 2 log 5)

= (b+a)/(3 – 2a)

Hence option b is correct.

log 32-243 = log 32/log 243

= log 25/log 35

= 5 log 2/5 log 3

= log 2/log 3

= (1 – log 5)/log 3 (since log 2 = log (10/5) = log 10 – log 5 = 1 – log 5)

= (1-a)/b

So option c is correct.

Hence option a, b and c are correct.

These above-mentioned questions are from important chapters in algebra which includes Vector Algebra, Quadratic Equations, Logarithm, Binomial Theorem and Complex Numbers.

This subject requires you to be quick with solving the problems and this can be achieved if you practice more and more problems. Because there is a limited number of formulas in algebra, a pro tip will be to make a note of all the important formulas in one place. Revise this sheet on a daily basis. Keep updating your list with new formulas you come across. Algebra tests how fluent you are with the application of the concepts. All the best.