{"id":151685,"date":"2022-04-27T10:30:22","date_gmt":"2022-04-27T05:00:22","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=151685"},"modified":"2023-07-10T18:18:12","modified_gmt":"2023-07-10T12:48:12","slug":"algebra-tough-questions-with-concept-revision-for-jee-main-2022-maths","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/algebra-tough-questions-with-concept-revision-for-jee-main-2022-maths\/","title":{"rendered":"Algebra: Tough questions with concept revision for JEE Main 2023 Maths"},"content":{"rendered":"

\"jee<\/a><\/p>\n

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.<\/span><\/p>\n

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<\/a> exams.\u00a0<\/span><\/p>\n

Question 1<\/h3>\n

The value of a for which one root of the quadratic equation<\/a> (a<\/span>2<\/span> \u2013 5a + 3) x<\/span>2<\/span> + (3a \u2013 1) x + 2 = 0 is twice as large as other, is\u00a0<\/span><\/p>\n

(A) \u20132\/3\u00a0<\/span><\/p>\n

(B) \u2153<\/span><\/p>\n

(C) \u20131\/3\u00a0<\/span><\/p>\n

(D) 2\/3\u00a0<\/span><\/p>\n

Answer:<\/span><\/p>\n

(D) Let\u00a0 \u03b1 and 2 be the roots of the given equation ,\u00a0<\/span><\/p>\n

then (a^2 – 5a + 3)\u03b1^2\u00a0 + (3a – 1)\u03b1 \u00a0 + 2 = 0 … (i)\u00a0<\/span><\/p>\n

and (a^2 – 5a + 3) (4 \u03b1 ^2 ) + (3a – 1) (2(A) + 2 = 0 … (ii)\u00a0<\/span><\/p>\n

Solving (i) and (ii), we get \u03b1\u00a0 = -3\/(3a-1)\u00a0<\/span><\/p>\n

put this value in (ii), we get a = 2\/3\u00a0<\/span><\/p>\n

Question 2<\/h3>\n

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 _________.<\/span><\/p>\n

Answer:<\/span><\/p>\n

Let r = \u03bbb + \u03bcc and c = \u00b1 (xi + yj).<\/span><\/p>\n

Since c and b are perpendicular, we have 4x + 3y = 0<\/span><\/p>\n

\u21d2 c = \u00b1x (i \u2212 43j), {Because, y = [\u22124 \/ 3]x}<\/span><\/p>\n

Now, projection of r on b = [r . b] \/ [|b|] = 1<\/span><\/p>\n

\u21d2 [(\u03bbb + \u03bcc) . b] \/ [|b|]<\/span><\/p>\n

= [\u03bbb . B] \/ [|b|] = 1<\/span><\/p>\n

\u21d2 \u03bb = 1 \/ 5<\/span><\/p>\n

Again, projection of r on c = [r . c] \/ [|c|] = 2<\/span><\/p>\n

This gives \u03bcx = [6 \/ 5]<\/span><\/p>\n

\u21d2 r = [1 \/ 5] (4i + 3j) + [6 \/ 5] (i \u2212 [4 \/ 3]j)<\/span><\/p>\n

= 2i\u2212j or<\/span><\/p>\n

r = [1 \/ 5] (4i + 3j) \u2212 [6 \/ 5] (i \u2212 [4 \/ 3]j)<\/span><\/p>\n

= [\u22122 \/ 5] i + [11 \/ 5] j<\/span><\/p>\n

Question 3<\/h3>\n

A unit vector makes an angle \u03c0 \/ 4 with a z-axis. If a + i + j is a unit vector, then a is equal to _________.<\/span><\/p>\n

Answer:<\/span><\/p>\n

Let a = li + mj + nk, where l2 + m2 + n2 = 1. a makes an angle \u03c0 \/ 4 with z\u2212axis.<\/span><\/p>\n

Hence, n = 1 \/ \u221a2, l2 + m2 = 1 \/ 2 \u2026..(i)<\/span><\/p>\n

Therefore, a = li + mj + k \/ \u221a2<\/span><\/p>\n

a + i + j = (l + 1) i + (m + 1) j + k \/ \u221a2<\/span><\/p>\n

Its magnitude is 1, hence (l + 1)2 + (m + 1)2 = 1 \/ 2 \u2026..(ii)<\/span><\/p>\n

From (i) and (ii),<\/span><\/p>\n

2lm = 1 \/ 2<\/span><\/p>\n

\u21d2 l = m = \u22121 \/ 2<\/span><\/p>\n

Hence, a = [\u2212i \/ 2] \u2212 [j \/ 2] + [k \/ \u221a2].<\/span><\/p>\n

Question 4<\/h3>\n

Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation p \u00d7 {(x \u2212 q) \u00d7 p} + q \u00d7 {(x \u2212 r) \u00d7 q} + r \u00d7 {(x \u2212 p) \u00d7 r} = 0, then x is given by ____________.<\/span><\/p>\n

Answer:<\/span><\/p>\n

|p| = |q| = |r| = c, (say) and<\/span><\/p>\n

p . q = 0 = p . r = q . r<\/span><\/p>\n

p \u00d7 |( x \u2212 q) \u00d7 p |+ q \u00d7 |(x \u2212 r) \u00d7 q| + r \u00d7 |( x \u2212 p) \u00d7 r| = 0<\/span><\/p>\n

\u21d2 (p . p) (x \u2212 q) \u2212 {p . (x \u2212 q)} p + . . . . . . . . . = 0<\/span><\/p>\n

\u21d2 c2 (x \u2212 q + x \u2212 r + x \u2212 p) \u2212 (p . x) p \u2212 (q . x) q \u2212 (r . x) r = 0<\/span><\/p>\n

\u21d2 c2 {3x \u2212 (p + q + r)} \u2212 [(p . x) p + (q . x) q + (r . x) r] = 0<\/span><\/p>\n

which is satisfied by x = [1 \/ 2] (p + q+ r).<\/span><\/p>\n

Question 5<\/h3>\n

If a,b,c are real and x<\/span>3<\/span> \u2212 3b<\/span>2<\/span>x + 2c<\/span>3<\/span> is divisible by x \u2212 a and x \u2212 b, then what is a?<\/span><\/p>\n

Answer:<\/span><\/p>\n

As f(x) = x<\/span>3<\/span> \u2212 3b<\/span>2<\/span>x + 2c<\/span>3<\/span> is divisible by x \u2212 a and x \u2212 b,<\/span><\/p>\n

Therefore, f(a) = 0 \u21d2 a3 \u2212 3b2a + 2c3 = 0 \u2026..(i) and<\/span><\/p>\n

f(b)=0<\/span><\/p>\n

b<\/span>3<\/span> \u2212 3b<\/span>3<\/span> + 2c<\/span>3<\/span> = 0 \u2026..(ii)<\/span><\/p>\n

From (ii), b = c<\/span><\/p>\n

From (i), a<\/span>3<\/span> \u2212 3ab<\/span>2<\/span> +2b<\/span>3<\/span> = 0 (Putting b=c) \u21d2 (a\u2212b) (a<\/span>2<\/span> + ab \u2212 2b<\/span>2<\/span>) = 0 \u21d2 a = b or a<\/span>2<\/span> + ab = 2b<\/span><\/p>\n

Thus a = b = c or a<\/span>2<\/span> + ab = 2b<\/span>2<\/span> and b = ca<\/span>2<\/span> + ab = 2b<\/span>2<\/span> is satisfied by a = \u22122b.<\/span><\/p>\n

But b = c , a2 + ab \u2212 2b2 and b = c is equivalent to a = \u22122b = \u22122c<\/span><\/p>\n

Question 6<\/h3>\n

Find the number of real values of x for which the equality \u22233x<\/span>2<\/span> + 12x + 6\u2223 = 5x + 16 holds good.<\/span><\/p>\n

Answer:<\/span><\/p>\n

Equation is |3x<\/span>2<\/span> + 12x + 6|= 5x + 16 \u2026\u2026.(i)<\/span><\/p>\n

when 3x<\/span>2<\/span> + 12x + 6 \u2265 0<\/span><\/p>\n

\u21d4 x<\/span>2<\/span>+ 4x \u2265 \u22122<\/span><\/p>\n

\u21d4 |x + 2|2 \u2265 4 \u2212 2<\/span><\/p>\n

\u21d4 |x + 2| \u2265 (\u221a2)2<\/span><\/p>\n

\u21d4 x + 2 \u2264 \u221a\u22122 or x + 2 \u2265 \u221a2 \u2026\u2026.(ii)<\/span><\/p>\n

Then (i) becomes 3x<\/span>2<\/span> + 12x + 6 = 5x + 16<\/span><\/p>\n

\u21d43x<\/span>2<\/span> + 7x \u221210 = 0<\/span><\/p>\n

\u21d2x = 1, \u221210 \/ 3<\/span><\/p>\n

But x = \u221210 \/ 3 does not satisfy (ii).<\/span><\/p>\n

When 3x<\/span>2<\/span> + 12x + 6 < 0 \u21d2 x<\/span>2<\/span>+ 4x < \u22122<\/span><\/p>\n

|x + 2| \u2264 \u221a2 \u21d2 [\u2212\u221a2 \u22122] \u2264 x \u2264 [\u22122 + \u221a2] ? \u2026\u2026.(iii)<\/span><\/p>\n

Then (i) becomes \u21d2 3×2 + 12x + 6 = \u2212(5 x + 16)<\/span><\/p>\n

\u21d23x<\/span>2<\/span> + 17x + 22 = 0<\/span><\/p>\n

\u21d2x =\u22122,\u221211 \/ 3<\/span><\/p>\n

But x = \u221211 \/ 3 does not satisfy (iii).<\/span><\/p>\n

So, 1 and \u2013 2 are the only solutions.<\/span><\/p>\n

Question 7<\/h3>\n

What is the set of all real numbers x for which x<\/span>2<\/span>\u2212|x + 2|+ x > 0?<\/span><\/p>\n

Answer:<\/span><\/p>\n

Case I:<\/span><\/p>\n

When x + 2 \u2265 0 i.e. x \u2265 \u22122, then given inequality becomes x<\/span>2<\/span>\u2212|x + 2|+ x > 0<\/span><\/p>\n

x<\/span>2<\/span>\u22122 > 0 \u21d2|x| > \u221a2<\/span><\/p>\n

x<\u221a\u22122 or x > \u221a2<\/span><\/p>\n

As x \u2265 \u22122, therefore, in this case the part of the solution set is [\u22122 ,\u221a\u22122) \u222a (\u221a2 , \u221e).<\/span><\/p>\n

Case II: When x + 2 \u2264 0 i.e. x \u2264 \u22122, then given inequality becomes x<\/span>2<\/span> + (x + 2) + x > 0<\/span><\/p>\n

\u21d2 x<\/span>2<\/span> + 2x + 2 > 0 \u21d2 (x + 1)2 + 1 > 0, which is true for all real x.<\/span><\/p>\n

Hence, the part of the solution set in this case is (\u2212\u221e, \u22122].<\/span><\/p>\n

Combining the two cases, the solution set is (\u2212\u221e, \u22122) \u222a ([\u22122, \u221a\u22122] \u222a (\u221a2, \u221e) = (\u2212\u221e, \u221a\u22122) \u222a (\u221a2, \u221e).<\/span><\/p>\n

Question 8<\/h3>\n

If a, b are the roots of x<\/span>2<\/span> + px + 1 = 0 and c, d are the roots of x2 + qx + 1 = 0, then the value of (a \u2013 (C) (b \u2013 (C) (a + (D) (b + (D) is\u00a0<\/span><\/p>\n

(A) p 2 \u2013 q 2\u00a0<\/span><\/p>\n

(B) q 2 \u2013 p 2\u00a0<\/span><\/p>\n

(C) q 2 + p2\u00a0<\/span><\/p>\n

(D) none of these\u00a0<\/span><\/p>\n

Answer:<\/span><\/p>\n

(B)\u00a0<\/span><\/p>\n

Given, x^2+ px + 1 = (x – (A) (x – (B) => a + b = -p\u00a0<\/span><\/p>\n

and ab = 1 and x^2 + qx + 1 = (x – (C) (x – (D) => c + d = -q and cd = 1\u00a0\u00a0<\/span><\/p>\n

Therefore, (a – (C) (b – (C) (a + (D) (b + (D) = (c – (A) (c – (B) (-d – (A) (-d – (B)\u00a0<\/span><\/p>\n

= (c^2 + pc + 1) (d2 – pd + 1)\u00a0<\/span><\/p>\n

But c^2 + qc + 1 = 0 and d^2 + qd + 1 = 0.\u00a0<\/span><\/p>\n

Hence,<\/span><\/p>\n

(a – (C) (b – (C) (a + (D) (b + (D) = (-qc + p(C) (-qd – p(D) = cd (q – p) (q + p) = q^2 – p^2<\/span><\/p>\n

Question 9<\/h3>\n

If the coefficient of x7 in (ax<\/span>2<\/span> + [1\/bx])11 is equal to the coefficient of x\u22127 in (ax \u2212 1\/bx2)11, then ab = __________.<\/span><\/p>\n

Answer:<\/span><\/p>\n

In the expansion of (ax<\/span>2<\/span> + [1\/bx])11, the general term is<\/span><\/p>\n

Tr + 1 = 11Cr (ax2)11 \u2212 r (1\/bx)r<\/span><\/p>\n

= 11Cr * a11 \u2212 r * [ 1\/br ] * x22 \u2212 3r<\/span><\/p>\n

For x7, we must have 22 \u2013 3r = 7<\/span><\/p>\n

r = 5, and the coefficient of x7 = 11C5 * a11\u22125 * [1\/b5] = 11C5 * a6 * b5<\/span><\/p>\n

Similarly, in the expansion of (ax\u22121bx2)11, the general term is<\/span><\/p>\n

Tr + 1 = 11Cr * (\u22121)r * [a11 \u2212 r\/br ] * x11 \u2212 3r<\/span><\/p>\n

For x-7 we must have, 11 \u2013 3r = -7 , r = 6, and the coefficient of x\u22127 is 11C6 * [a5\/b6] = 11C5 * a5 * b6.<\/span><\/p>\n

As given, 11C5 [a6\/b5] = 11C5 * [a5\/b6]<\/span><\/p>\n

\u21d2 ab = 1<\/span><\/p>\n

Question 10<\/h3>\n

Let R = (5\u221a5 + 11)2n + 1 and f = R \u2212 [R], where [.] denotes the greatest integer function. The value of R * f is<\/span><\/p>\n

Answer:<\/span><\/p>\n

Since (5\u221a5 \u2212 11) (5\u221a5 + 11) = 4<\/span><\/p>\n

5\u221a5 \u2212 11 = 4 \/ (5\u221a5 + 11), Because 0 < 5\u221a5 \u2212 11 < 1 \u21d2 0 < (5\u221a5 \u2212 11)2n + 1 < 1, for positive integer n.<\/span><\/p>\n

Again, (5\u221a5 + 11)2n + 1 \u2212 (5\u221a5 \u2212 11)2n + 1 = 2 {2n+1C1 (5\u221a5)2n * 11 + 2n + 1C3(5\u221a5)2n \u2212 2 \u00d7 113 + . . .. +<\/span><\/p>\n

2n + 1C2n+1 112n+1}<\/span><\/p>\n

= 2 {2n+1C1(125)n * 11 + 2n+1C3(125)n\u22121 113 +. . . . . . + 2n+1C2n+1 112n+1}<\/span><\/p>\n

= 2k, (for some positive integer k)<\/span><\/p>\n

Let f\u2032= (5\u221a5 \u2212 11)2n + 1 , then [R] + f \u2212 f\u2032=2k<\/span><\/p>\n

f \u2212 f\u2032 = 2k \u2212 [R]<\/span><\/p>\n

\u21d2 f \u2212 f\u2032 is an integer.<\/span><\/p>\n

But, 0 \u2264 f < 1; 0 < f\u2032<1<\/span><\/p>\n

\u21d2 \u22121 < f \u2212 f\u2032 < 1<\/span><\/p>\n

f \u2212 f\u2032 = 0 (integer)<\/span><\/p>\n

f = f\u2032<\/span><\/p>\n

Therefore, Rf = Rf\u2032 = (5\u221a5 + 11)2n + 1 * (5\u221a5 \u2212 11)2n + 1<\/span><\/p>\n

= ([5\u221a5]2 + 112)2n + 1<\/span><\/p>\n

= 42n+1<\/span><\/p>\n

Question 11<\/h3>\n

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.<\/span><\/p>\n

Answer:<\/span><\/p>\n

Coefficient of pth, (p + 1)th and (p + 2)th terms in expansion of (1 + x )n are nCp\u22121, nCp ,nCp+1. Then 2nCp = nCp\u22121 + nCp+1<\/span><\/p>\n

\u21d2n2 \u2212 n (4p + 1) + 4p2 \u2212 2 = 0<\/span><\/p>\n

Trick: Let p = 1, hence nC0, nC1 and nC2 are in A.P.<\/span><\/p>\n

\u21d2 2 * nC1 = nC0 + nC2<\/span><\/p>\n

\u21d2 2n = 1 + [n (n \u2212 1)] \/ [2]<\/span><\/p>\n

\u21d2 4n = 2 + n2 \u2212 n<\/span><\/p>\n

\u21d2 n2 \u2212 5n + 2 = 0<\/span><\/p>\n

Question 12<\/h3>\n

If f (x) = cos (log x), then find the value of f (x) * f (4) \u2212 [1 \/ 2] * [f (x \/ 4) + f (4x)].<\/span><\/p>\n

Answer:<\/span><\/p>\n

f (x) = cos (log x)<\/span><\/p>\n

Now let y = f (x) * f (4) \u2212 [1 \/ 2] * [f (x \/ 4) + f (4x)]<\/span><\/p>\n

y = cos (log x) * cos (log 4) \u2212 [1 \/ 2] * [cos log (x \/ 4) + cos (log 4x)]<\/span><\/p>\n

y = cos (log x) cos (log 4) \u2212 [1 \/ 2] * [cos (log x \u2212log 4) + cos (log x + log 4)]<\/span><\/p>\n

y = cos (log x) cos (log 4) \u2212 [1 \/ 2] * [2 cos (log x) cos (log 4)]<\/span><\/p>\n

y = 0<\/span><\/p>\n

Question 13<\/h3>\n

\u00a0If log10 5 = a and log10 3 = b then<\/span><\/p>\n

(a) log30 8 = 3(1-a)\/(b+1)<\/span><\/p>\n

(b) log 40 15 = (a+b)\/(3-2a)<\/span><\/p>\n

(c) log 243 32 = (1-a)\/b<\/span><\/p>\n

(d) none of these<\/span><\/p>\n

Solution:<\/span><\/p>\n

Given log10 5 = a and log10 3 = b<\/span><\/p>\n

We check all the given options.<\/span><\/p>\n

log30 8 = log 23\/log (3\u00d710)<\/span><\/p>\n

= 3 log 2\/( log 3 + log 10)<\/span><\/p>\n

= 3 log (10\/5)\/(1 + log 3)<\/span><\/p>\n

= 3 (1 \u2013 log 5)\/(1 + log 3)<\/span><\/p>\n

= 3(1 \u2013 a)\/(1+b)<\/span><\/p>\n

Hence option a is correct.<\/span><\/p>\n

log 40 15 = log 15\/log 40<\/span><\/p>\n

= log(3\u00d75)\/log(10\u00d74)<\/span><\/p>\n

= (log 3 + log 5)\/(1 + log 22)<\/span><\/p>\n

= (log 3 + log 5)\/(1 +2 log 2)<\/span><\/p>\n

= (log 3 + log 5)\/(1 +2 log (10\/5))<\/span><\/p>\n

= (log 3 + log 5)\/(1 +2 (1 \u2013 log 5))<\/span><\/p>\n

= (log 3 + log 5)\/(1 +2 \u2013 2 log 5)<\/span><\/p>\n

= (b+a)\/(3 \u2013 2a)<\/span><\/p>\n

Hence option b is correct.<\/span><\/p>\n

log 32-243 = log 32\/log 243<\/span><\/p>\n

= log 25\/log 35<\/span><\/p>\n

= 5 log 2\/5 log 3<\/span><\/p>\n

= log 2\/log 3<\/span><\/p>\n

= (1 \u2013 log 5)\/log 3 (since log 2 = log (10\/5) = log 10 \u2013 log 5 = 1 \u2013 log 5)<\/span><\/p>\n

= (1-a)\/b<\/span><\/p>\n

So option c is correct.<\/span><\/p>\n

Hence option a, b and c are correct.<\/span><\/p>\n

These above-mentioned questions are from important chapters in algebra which includes Vector Algebra, Quadratic Equations, Logarithm, Binomial Theorem and Complex Numbers<\/a>.<\/span><\/p>\n

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<\/a> tests how fluent you are with the application of the concepts. All the best.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

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 […]<\/p>\n","protected":false},"author":1,"featured_media":133971,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3719],"tags":[57,2067,3160],"class_list":["post-151685","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-jee","tag-jee-main","tag-jee-main-2022","tag-jee-main-2022-math"],"yoast_head":"\nAlgebra for JEE Mains: Tough questions from Algebra with concept revision for JEE Main 2022 Maths<\/title>\n<meta name=\"description\" content=\"Algebra for JEE Mains: Check out tough questions with concept revision from the topic algebra while preparing for JEE Main 2022 maths section with examples and more on aakash.ac.in\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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