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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

The maximum value of = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix}]$ is ( $θ$ is real number)

(A) $\frac{1}{2}$

(B) √3/2

(C) √2

(D) $\frac{2√3}{4}$

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow = | \begin{matrix} 0 & 0 & 1 \\ - s i n θ & s i n θ & 1 \\ c o s θ & 0 & 1 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow = 1 | \begin{matrix} - s i n θ & s i n θ \\ c o s θ & 0 \end{matrix} | = - s i n θ c o s θ \\ \Rightarrow = - \frac{1}{2} . 2 s i n θ c o s θ = - \frac{1}{2} s i n 2 θ \\ b u t maximum value of sin2 θ =1 \Rightarrow | - \frac{1}{2} . 1 | = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Ncert Solutions Maths class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

$f (x) = {\begin{matrix} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{matrix}$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 2^{+}}{l i m} f (x) = 3 x + 5 \\ = \underset{h \to 0}{l i m} 3 (2 + h) + 5 = 11 \\ \underset{x \to 2}{l i m} f (x) = 3 x + 5 = 3 (2) + 5 = 11 \\ \underset{x \to 2^{-}}{l i m} f (x) = x^{2} = \underset{h \to 0}{l i m} {(2 - h)}^{2} \\ = \underset{h \to 0}{l i m} {(2)}^{2} + h^{2} - 4 h = {(2)}^{2} = 4 \\ Since \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \neq \underset{x \to 2^{-}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 2 . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Let f(t|) $[\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}],$ then $\underset{t \to 0}{l i m} \frac{f (t)}{t^{2}}$ is equal to

(A) 0

(B) –1

(C) 2

(D) 3

10 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (t) = [\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}] \\ E x p a n d i n g a l o n g R_{1} \\ = c o s t | \begin{matrix} t & 2 t \\ t & t \end{matrix} | - t | \begin{matrix} 2 s i n t & 2 t \\ s i n t & t \end{matrix} | + 1 | \begin{matrix} 2 s i n t & t \\ s i n t & t \end{matrix} | \\ = c o s t (t^{2} - 2 t^{2}) - t (2 t s i n t - 2 t s i n t) + (2 t s i n t - t s i n t) \\ = t^{2} c o s t + t s i n t \\ ∴ \frac{f (t)}{t^{2}} = \frac{- t^{2} c o s t + t s i n t}{t^{2}} \\ \Rightarrow \frac{f (t)}{t^{2}} = - c o s t + \frac{s i n t}{t^{2}} \\ \Rightarrow \underset{t \to 0}{l i m} \frac{f (t)}{t^{2}} = \underset{t \to 0}{l i m} (- c o s t) + \underset{t \to 0}{l i m} \frac{s i n t}{t^{2}} = - 1 + 1 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $A, B a n d C$ are angles of a triangle, then the determinant $[\begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix}]$ is equal to

(A) 0

(B) –1

(C) 1

(D) None of these

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix} | \\ C_{1} \to a C_{1} + b C_{2} + c C_{3} \\ \Rightarrow | \begin{matrix} - a + b c o s C + c c o s B & c o s C & c o s B \\ a c o s C - b + c c o s A & - 1 & c o s A \\ a c o s B + b c o s A - C & c o s A & - 1 \end{matrix} | \\ \Rightarrow | \begin{matrix} - a + a & c o s C & c o s B \\ - b + b & - 1 & c o s A \\ - c + c & c o s A & - 1 \end{matrix} | [\begin{array}{l} ? F r o m p r o t e c t i o n f o r m u l a \\ a = b c o s C + c c o s B \\ b = a c o s C + c c o s A \\ c = b c o s A + a c o s B \end{array}] \\ \Rightarrow | \begin{matrix} 0 & c o s C & c o s B \\ 0 & - 1 & c o s A \\ 0 & c o s A & - 1 \end{matrix} | = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Ncert Solutions Maths class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Examine the continuity of the function
$f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t y = f (x) w i l l b e c o n t i n u o u s a t x = a i f \\ \underset{x \to a^{-}}{l i m} f (x) = \underset{x \to a}{l i m} f (x) = \underset{x \to a^{+}}{l i m} f (x) \\ G i v e n f (x) = x^{3} + 2 x^{2} - 1 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1}{l i m} f (x) = {(1)}^{3} + 2 {(1)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{+}}{l i m} f (x) = \underset{h \to 0}{l i m} {(1 + h)}^{3} + 2 {(1 + h)}^{2} - 1 = 1 + 2 - 1 = 2 \\ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = 2 . \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

The number of distinct real roots of $[\begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix}] = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(A) 0

(B) 2

(C) 1

(D) 3

7 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix} | = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} 2 c o s x + s i n x & c o s x & c o s x \\ 2 c o s x + s i n x & s i n x & c o s x \\ 2 c o s x + s i n x & c o s x & s i n x \end{matrix} | = 0 (T a k i n g 2 c o s x + s i n x c o m m o n f r o m C_{1}) \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 1 & s i n x & c o s x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 0 & c o s x - s i n x & 0 \\ 0 & s i n x - c o s x & c o s x - s i n x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ \Rightarrow (2 c o s x + s i n x) [1 | \begin{matrix} c o s x - s i n x & 0 \\ s i n x - c o s x & c o s x - s i n x \end{matrix} |] \\ \Rightarrow (2 c o s x + s i n x) {(c o s x - s i n x)}^{2} = 0 \\ 2 c o s x + s i n x = 0 a n d {(c o s x - s i n x)}^{2} = 0 \\ 2 + t a n x = 0 a n d (c o s x - s i n x) = 0 \\ ∴ t a n x = - 2 a n d \Rightarrow t a n x = 1 \\ B u t - \frac{π}{4} \leq x \leq \frac{π}{4} a n d \Rightarrow t a n x = t a n \frac{π}{4} \\ ∴ x = \frac{π}{4} \in [- \frac{π}{4}, \frac{π}{4}] \\ S o, x &thins \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

The determinant $[\begin{matrix} b^{2} - a b & b - c & b c - a c \\ b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix}]$ equals

(A) $a b c (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(C) $(a + b + c) (b - c) (c - a) (a - b)$

(D) None of these

9 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix} | \\ \Rightarrow = | \begin{matrix} b (b - a) & b - c & c (b - a) \\ a (b - a) & a - b & b (b - a) \\ c (b - a) & c - a & a (b - a) \end{matrix} | (T a k i n g (b - a) c o m m o n f r o m C_{1} a n d C_{3}) \\ \Rightarrow = {(b - a)}^{2} | \begin{matrix} b & b - c & c \\ a & a - b & b \\ c & c - a & a \end{matrix} | \\ C_{1} \to C_{1} - C_{3} \\ \Rightarrow = {(a - b)}^{2} | \begin{matrix} b - c & b - c & c \\ a - b & a - b & b \\ c - a & c - a & a \end{matrix} | (C_{1} a n d C_{2} a r e i d e n t i c a l c o l u m n s .) \\ \Rightarrow = {(a - b)}^{2} . 0 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Ncert Solutions Maths class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

The area of a triangle with vertices $(- 3, 0)$ , $(3, 0)$ , and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(C) –9

(D) 6

21 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A r e a o f t r i a n g l e w i t h v e r t i c e s (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) w i l l b e : \\ Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | \Rightarrow Δ = \frac{1}{2} | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} | \\ \Rightarrow = \frac{1}{2} [- 3 | \begin{matrix} 0 & 1 \\ k & 1 \end{matrix} | - 0 | \begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} | + 1 | \begin{matrix} 3 & 0 \\ 0 & k \end{matrix} |] \\ \Rightarrow = \frac{1}{2} [- 3 (- k) - 0 + 1 (3 k)] \\ \Rightarrow = \frac{1}{2} (3 k + 3 k) \Rightarrow = \frac{1}{2} (6 k) = 3 k \\ 3 k = 9 \Rightarrow k = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

The value of determinant $[\begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix}]$ is

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 b c$

(C) $a^{3} + b^{3} + c^{3} - 3 a b c$

(D) None of these

66 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, w e h a v e | \begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix} | \\ C_{2} \to C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} a - b & a + b + c & a \\ b - a & a + b + c & b \\ c - a & a + b + c & c \end{matrix} | \\ \Rightarrow (a + b + c) | \begin{matrix} a - b & 1 & a \\ b - a & 1 & b \\ c - a & 1 & c \end{matrix} | (T a k i n g a + b + c c o m m o n f r o m C_{2}) \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (a + b + c) | \begin{matrix} 2 (a - b) & 0 & a - b \\ b - c & 0 & b - c \\ c - a & 1 & c \end{matrix} | \\ T a k i n g (a - b) a n d (b - c) c o m m o n f r o m R_{1} a n d R_{2} r e s p e c t i v e l y \\ \Rightarrow (a + b + c) (a - b) (b - c) | \begin{matrix} 2 & 0 & 1 \\ 1 & 0 & 1 \\ c - a & 1 & c \end{matrix} | \\ E x p a n d i n g a l o n g C_{2} \\ \Rightarrow (a + b + c) (a - b) (b - c) [- 1 | \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} |] \\ \Rightarrow (a + b + c) (a - b) (b - c) (- 1) \\ \Rightarrow (a + b + c) (a - b) (c - b) \\ H e n c e, t h e c o r r e c t x^{2} o p t i o n i s (d) . \end{array}$

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The maximum value of = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix}]$ is ( $θ$ is real number)

(A) $\frac{1}{2}$

(B) √3/2

(C) √2

(D) $\frac{2√3}{4}$

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

$f (x) = {\begin{matrix} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{matrix}$

Let f(t|) $[\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}],$ then $\underset{t \to 0}{l i m} \frac{f (t)}{t^{2}}$ is equal to

(A) 0

(B) –1

(C) 2

(D) 3

If $A, B a n d C$ are angles of a triangle, then the determinant $[\begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix}]$ is equal to

(A) 0

(B) –1

(C) 1

(D) None of these

Examine the continuity of the function
$f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

The number of distinct real roots of $[\begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix}] = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(A) 0

(B) 2

(C) 1

(D) 3

The determinant $[\begin{matrix} b^{2} - a b & b - c & b c - a c \\ b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix}]$ equals

(A) $a b c (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(C) $(a + b + c) (b - c) (c - a) (a - b)$

(D) None of these

Kindly consider the following:

The area of a triangle with vertices $(- 3, 0)$ , $(3, 0)$ , and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(C) –9

(D) 6

The value of determinant $[\begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix}]$ is

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 b c$

(C) $a^{3} + b^{3} + c^{3} - 3 a b c$

(D) None of these