Mathematics Unsolved Sample Paper Solutions ISC Class 12

Section-A

Answer 1.

(i) (b) (9, 6) ∈ R

 Explanation :

Given, R = {(x, y) : x = y + 3, y > 5}

For y = 6, x = 6 + 3 = 9

(9, 6) ∈ R.

(ii) (c) 8

 Explanation :

Here, n = n(A) = 3, m = n(B) = 2

No. of functions = mn = 23 = 8.

(iii) (a) 1/2

 Explanation :
172

(iv) (c) Determinant is a number associated to a square matrix

 Explanation :
174

(vi) (b) 3

 Explanation :

(vii) (b) 5/2

 Explanation :
176

(viii) (c) x + y = 0

 Explanation :
178

(ix) (d) ab = 1

 Explanation :
179

(x) (d) 7/8

 Explanation :
  • (xi) Consider,
181
182

Answer 2.

  • (i) Here, R = {(a, b) : b = a + 1}
  • R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6)}
  • ⇒ R = {(1,2,) (2,3), (3,4), (4,5), (5,6)}
  • (a) R is not reflexive as (a, a) ∉ R
  • (b) R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R
  • (c) R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R

OR

  • (ii) Given, f(x) = x2 + 4
  • Let f(x1) = f(x2)
  • ⇒ x21 + 4 = x22 + 4
  • ⇒ x1 = x2
  • Thus, f(x) is one-one.
  • Since, x2 + 4 is a real number. Thus, for every y in the co-domain of f, there exists a number x in R+ such that f(x) = y = x2 + 4
  • Thus, we can say that f(x) is onto.
  • Now, f(x) is one-one and onto. Hence, f(x) is invertible.
  • Let f(x) = y ⇒ x2 + 4 = y
  • ⇒ x2 = y − 4
  • i.e. x = √(y − 4)
  • Also, x = f −1(y)
  • f –1(y) = √(y − 4) . Hence Proved.

Answer 3.

184
185

Answer 4.

186

Answer 5.

187

OR

Answer 6.

  • Given, y = 2 cos (log x) + 3 sin (log x)
  • On differentiating both sides w.r.t. x, we get

Answer 7.

190

Answer 8.

191
192

Answer 9.

193

OR

194

Answer 10.

195
196
197
198

Answer 11.

199

Answer 12.

200

OR

201

Answer 13.

202
203

OR

204
205
206
207

Answer 14.

209

Section-B

Answer 15.

210
211
213
213
214
215

Answer 16.

216
217

OR

218

Answer 17.

219

OR

220

Answer 18.

222

Section-B

Answer 19.

224
226
228
228
229

Answer 20.

231

OR

232

Answer 21.

Marks in
Physics x
Marks in
Chemistry y
dx = x – 40 dx dy = y – 36 dy (dx)2 (dy)2 dxdy
46 40 6 4 36 16 24
42 38 2 2 4 4 4
44 36 4 0 16 0 0
40 35 0 -1 0 1 0
43 39 3 3 9 9 9
41 37 1 1 1 1 1
45 41 5 5 25 25 25
Σdx = 21 Σdy = 14 Σ(dx)2 = 91 Σ(dy)2 = 56 Σdxdy = 63
234

Answer 22.

235

The feasible region is the shaded region.

Corner Points Z = x + y
A (25, 0), Z = 25 + 0 = 25
B (20, 10), Z = 20 + 10 = 30
C (0, 20), Z = 0 + 20 = 20

Hence, maximum number of cakes = 20 + 10 = 30.

OR

236
237

ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2023 

ISC 36 Sample Question Papers

All Subjects Combined for Class 12 Exam 2023

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