Find X if [ 1 1 1 1 1 − 1 2 1 − 1 ] · X = [ 6 0 1 ] where X = [ x 1 x 2 x 3 ] .

The given system of equation is A X = B where A = [ 1 1 1 1 1 − 1 2 1 − 1 ] and B = 6 0 1 . A = 1 ( - 1 + 1 ) - 1 ( - 1 + 2 ) + 1 ( 1 - 2 ) = - 2 So the given system has a unique solution. A i j is the cofactor of the a i j A 11 = ( − 1 ) 1 + 1 | 1 − 1 1 − 1 | = 0 A 12 = ( − 1 ) 1 + 2 | 1 − 1 2 − 1 | = − 1 A 13 = ( − 1 ) 1 + 3 | 1 1 2 1 | = − 1 A 21 = ( − 1 ) 2 + 1 | 1 1 1 − 1 | = 2 A 22 = ( − 1 ) 2 + 2 | 1 1 2 − 1 | = − 3 A 23 = ( − 1 ) 2 + 3 | 1 1 2 1 | = 1 A 31 = ( − 1 ) 3 + 1 | 1 1 1 − 1 | = − 2 A 32 = ( − 1 ) 3 + 2 | 1 1 1 − 1 | = 2 A 33 = ( − 1 ) 3 + 3 | 1 1 1 1 | = 0 ∴ A d j A = [ 0 2 − 2 − 1 − 3 2 − 1 1 0 ] The solution of A X = B is X = A - 1 B . X = A d j A A B X = 1 ( - 2 ) [ 0 2 − 2 − 1 − 3 2 − 1 1 0 ] 6 0 1 x y z = 1 2 0 × 6 + 2 × 0 + ( - 2 ) × 1 ( - 1 ) × 6 + ( - 3 ) × 0 + 2 × 1 ( - 1 ) × 6 + 1 × 0 + 0 × 1 x y z = 1 2 3 ⇒ x = 1 , y = 2 , z = 3

If D j = j a n ( n + 1 ) / 2 j 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 j 3 c n 2 ( n + 1 ) 2 / 4 , then prove that ∑ j = 1 n D j = 0 .

Given that D j = j a n ( n + 1 ) / 2 j 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 j 3 c n 2 ( n + 1 ) 2 / 4 So, we have D 1 = 1 a n ( n + 1 ) / 2 1 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 c n 2 ( n + 1 ) 2 / 4 , D 2 = 2 a n ( n + 1 ) / 2 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 2 3 c n 2 ( n + 1 ) 2 / 4 .... D n = n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 ∴ ∑ j = 1 n D j = 1 a n ( n + 1 ) / 2 1 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 c n 2 ( n + 1 ) 2 / 4 + 2 a n ( n + 1 ) / 2 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 2 3 c n 2 ( n + 1 ) 2 / 4 + . . . n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = 1 + 2 a n ( n + 1 ) / 2 1 2 + 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 + 2 3 c n 2 ( n + 1 ) 2 / 4 + 3 a n ( n + 1 ) / 2 3 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 3 3 c n 2 ( n + 1 ) 2 / 4 + . . . n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 Thus, using the properties of sum of determinants, we get ∑ j = 1 n D j = 1 + 2 + 3 + . . n a n ( n + 1 ) / 2 1 2 + 2 2 + 3 2 + . . n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 + 2 3 + 3 3 + . . n 3 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = n ( n + 1 ) / 2 a n ( n + 1 ) / 2 n ( n + 1 ) ( 2 n + 1 ) / 6 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 2 ( n + 1 ) 2 / 4 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = 0

If f ( x ) = 1 + sin 2 x cos 2 x 4 sin 2 x sin 2 x 1 + cos 2 x 4 sin 2 x sin 2 x cos 2 x 1 + 4 sin 2 x , what is the least value of f ( x ) ?

Given that f ( x ) = 1 + sin 2 x cos 2 x 4 sin 2 x sin 2 x 1 + cos 2 x 4 sin 2 x sin 2 x cos 2 x 1 + 4 sin 2 x = 1 - 1 0 sin 2 x 1 + cos 2 x 4 sin 2 x sin 2 x cos 2 x 1 + 4 sin 2 x (applying R 1 → R 1 - R 2 ) = 1 - 1 0 0 1 - 1 sin 2 x cos 2 x 1 + 4 sin 2 x (applying R 2 → R 2 - R 3 ) = 1 0 0 0 1 - 1 sin 2 x 1 1 + 4 sin 2 x (applying C 2 → C 2 + C 1 ) = 1 × [ 1 × ( 1 + 4 sin 2 x ) - 1 × ( - 1 ) ] + 0 + 0 = 2 + 4 sin 2 x We know that - 1 ≤ sin 2 x ≤ 1 . ⇒ - 4 ≤ 4 sin 2 x ≤ 4 ⇒ - 2 ≤ 4 sin 2 x + 2 ≤ 6 Thus, the least value of f ( x ) = 2 + 4 sin 2 x is - 2 .

If f r ( x ) , g r ( x ) , h r ( x ) , r = 1 , 2 , 3 are polynomials in x such that f r ( a ) = g r ( a ) = h r ( a ) and F ( x ) = f 1 ( x ) f 2 ( x ) f 3 ( x ) g 1 ( x ) g 2 ( x ) g 3 ( x ) h 1 ( x ) h 2 ( x ) h 3 ( x ) , find F ′ ( x ) at x = a .

f r ( x ) , g r ( x ) , h r ( x ) , r = 1 , 2 , 3 are polynomials in x such that f r ( a ) = g r ( a ) = h r ( a ) and F ( x ) = f 1 ( x ) f 2 ( x ) f 3 ( x ) g 1 ( x ) g 2 ( x ) g 3 ( x ) h 1 ( x ) h 2 ( x ) h 3 ( x ) . So, f r ( x ) , g r ( x ) , h r ( x ) where r = 1 , 2 , 3 are differentiable functions. We have to find F ' ( x ) at x = a . We know that F ′ ( x ) = | f 1 ′ ( x ) f 2 ′ ( x ) f 3 ′ ( x ) g 1 ( x ) g 2 ( x ) g 3 ( x ) h 1 ( x ) h 2 ( x ) h 3 ( x ) | + | f 1 ( x ) f 2 ( x ) f 3 ( x ) g 1 ′ ( x ) g 2 ′ ( x ) g 3 ′ ( x ) h 1 ( x ) h 2 ( x ) h 3 ( x ) | + | f 1 ( x ) f 2 ( x ) f 3 ( x ) g 1 ( x ) g 2 ( x ) g 3 ( x ) h 1 ′ ( x ) h 2 ′ ( x ) h 3 ′ ( x ) | ∴ F ′ ( a ) = | f 1 ′ ( a ) f 2 ′ ( a ) f 3 ′ ( a ) g 1 ( a ) g 2 ( a ) g 3 ( a ) h 1 ( a ) h 2 ( a ) h 3 ( a ) | + | f 1 ( a ) f 2 ( a ) f 3 ( a ) g 1 ′ ( a ) g 2 ′ ( a ) g 3 ′ ( a ) h 1 ( a ) h 2 ( a ) h 3 ( a ) | + | f 1 ( a ) f 2 ( a ) f 3 ( a ) g 1 ( a ) g 2 ( a ) g 3 ( a ) h 1 ′ ( a ) h 2 ′ ( a ) h 3 ′ ( a ) | We know that f r ( a ) = g r ( a ) = h r ( a ) for r = 1 , 2 , 3 . Also, the value of determinant with two identical rows is zero. Hence, F ' ( a ) = 0 + 0 + 0 = 0

E M B I B E

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)

Odisha Board Solutions for Chapter: Determinants, Exercise 2: EXERCISES 5 (b)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Determinants, Exercise 2: EXERCISES 5 (b)

Attempt the practice questions on Chapter 5: Determinants, Exercise 2: EXERCISES 5 (b) with hints and solutions to strengthen your understanding. Elements of Mathematics Class 12 solutions are prepared by Experienced Embibe Experts.

Questions from Odisha Board Solutions for Chapter: Determinants, Exercise 2: EXERCISES 5 (b) with Hints & Solutions

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 1

Write the number of solutions of the following system of equation: $x - 2 y = 0$

MEDIUM

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 10

Find $X$ if $[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & - 1 \\ 2 & 1 & - 1 \end{array}] \cdot X = [\begin{array}{l} 6 \\ 0 \\ 1 \end{array}]$ where $X = [\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}]$ .

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 21

For what value of $λ$ , the system of equations $x + y + z = 6, 4 x + λ y - λ z = 0, 3 x + 2 y - 4 z = - 5$ does not possess a solution?

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 29

If $D_{j} = |\begin{matrix} j & a & n (n + 1) / 2 \\ j^{2} & b & n (n + 1) (2 n + 1) / 6 \\ j^{3} & c & n^{2} (n + 1)^{2} / 4 \end{matrix}|$ , then prove that $\sum_{j = 1}^{n} D_{j} = 0$ .

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 30

If $a_{1}, a_{2}, \dots \dots a_{n}$ are in G.P., and $a_{i} > 0$ for every $i$ , then find the value of $|\begin{matrix} \log a_{n} & \log a_{n + 1} & \log a_{n + 2} \\ \log a_{n + 1} & \log a_{n + 2} & \log a_{n + 3} \\ \log a_{n + 2} & \log a_{n + 3} & \log a_{n + 4} \end{matrix}|$

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 31

If $f (x) =$ $|\begin{matrix} 1 + \sin^{2} x & \cos^{2} x & 4 \sin 2 x \\ \sin^{2} x & 1 + \cos^{2} x & 4 \sin 2 x \\ \sin^{2} x & \cos^{2} x & 1 + 4 \sin 2 x \end{matrix}|$ , what is the least value of $f (x)$ ?

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 32

If $f_{r} (x), g_{r} (x), h_{r} (x), r = 1, 2, 3$ are polynomials in $x$ such that $f_{r} (a) = g_{r} (a) = h_{r} (a)$ and $F (x)$ $= |\begin{array}{l} f_{1} (x) & f_{2} (x) & f_{3} (x) \\ g_{1} (x) & g_{2} (x) & g_{3} (x) \\ h_{1} (x) & h_{2} (x) & h_{3} (x) \end{array}|$ , find $F^{'} (x) at x = a$ .

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (b)>Q 33

If $f (x) =$ $| \begin{array}{ccc} \cos x & \sin x & \cos x \\ \cos 2 x & \sin 2 x & 2 \cos 2 x \\ \cos 3 x & \sin 3 x & 3 \cos 3 x \end{array} |$ , find $f^{'} (\frac{π}{2}) .$

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