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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write down the negation of the following compound statements:

i. All rational numbers are real and complex.

ii. All real numbers are rational or irrationals.

iii. x = 2 and x = 3 are roots of the quadratic equation x² – 5x + 6 = 0.

iv. A triangle has either 3 sides or 4 sides.

v. 35 is a prime number or a composite number.

vi. All prime integers are either even or odd.

vii. x is equal to either x or –x.

viii. 6 is divisible by 2 and 3.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) p : A l l r a t i o n a l n u m b e r s a r e r e a l . \\ \sim p : A l l r a t i o n a l n u m b e r s a r e n o t r e a l . \\ q : A l l r a t i o n a l n u m b e r s a r e c o m p l e x . \\ \sim q : A l l r a t i o n a l n u m b e r s a r e n o t c o m p l e x . \\ \sim (p \land q) = (\sim p \lor \sim q) : A l l r a t i o n a l n u m b e r s a r e n e i t h e r r e a l n o r c o m p l e x . \\ (i i) p : A l l r e a l n u m b e r s a r e r a t i o n a l s . \\ q : A l l r e a l n u m b e r s a r e i r r a t i o n a l s . \\ T h e n e g a t i o n o f t h e a b o v e s t a t e m e n t i s \\ \sim (p \lor q) = (\sim p \land \sim q) : A l l r e a l n u m b e r s a r e n o t r a t i o n a l s a n d a l l r e a l n u m b e r s a r e \\ n o t i r r a t i o n a l s . \\ (i i i) p : x = 2 i s r o o t o f t h e e q u a t i o n x^{2} - 5 x + 6 = 0 . \\ q : x = 3 i s r o o t o f t h e e q u a t i o n x^{2} - 5 x + 6 = 0 . \\ T h e n e g a t i o n o f t h e a b o v e s t a t e m e n t i s \\ \sim (p \land q) = (\sim p \lor \sim q) : x = 2 i s n o t t h e r o o t o f t h e e q u a t i o n x^{2} - 5 x + 6 = 0 o r x = 3 i s \\ n o t t h e r o o t o f t h e e q u a t i o n x^{2} - 5 x + 6 = 0 . \\ (i v) p : A t r i a n g l e h a s 3 - s i d e s . \\ q : A t r i a n g l e h a s 4 - s i d e s . \\ T h e n e g a t i o n o f t h e a b o v e s t a t e m e n t i s \\ \sim (p \lor q) = (\sim p \land \sim q) : A t r i a n g l e h a s n e i t h e r 3 - s i d e s n o r 4 - s i d e s . \\ (v) p : 35 i s a p r i m e n u m b e r . \\ q : 35 i s a c o m p o s i t e n u m b e r . \\ T h e n e g a t i o n o f t h e a b o v e s t a t e m e n t i s \\ \sim (p \lor q) = (\sim p \land \sim q) : 35 i s n o t a p r i m e n u m b e r a n d i t &thins \end{array}$

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New question posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write down the negation of the following compound statements:

i. All rational numbers are real and complex.

ii. All real numbers are rational or irrationals.

iii. x = 2 and x = 3 are roots of the quadratic equation x² – 5x + 6 = 0.

iv. A triangle has either 3 sides or 4 sides.

v. 35 is a prime number or a composite number.

vi. All prime integers are either even or odd.

vii. x is equal to either x or –x.

viii. 6 is divisible by 2 and 3.

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Translate the following statements into symbolic form:

i. Rahul passed in Hindi and English.

ii. x and y are even integers.

iii. 2, 3, and 6 are factors of 12.

iv. Either x or x + 1 is an odd integer.

v. A number is either divisible by 2 or 3.

vi. Either x = 2 or x = 3 is a root of 3x² – x – 10 = 0.

vii. Students can take Hindi or English as an optional paper.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) p : R a h u l p a s s e d i n H i n d i . \\ q : R a h u l p a s s e d i n E n g l i s h . \\ p \land q : R a h u l p a s s e d i n H i n d i a n d E n g l i s h . \\ (i i) p : x i s a n e v e n i n t e g e r . \\ q : y i s a n e v e n i n t e g e r . \\ p \land q : x a n d y a r e e v e n i n t e g e r . \\ (i i i) p : 2 i s a f a c t o r o f 12 . \\ q : 3 i s a f a c t o r o f 12 . \\ r : 6 i s a f a c t o r o f 12 . \\ p \land q \land r : 2, 3 a n d 6 a r e f a c t o r s o f 12 . \\ (i v) p : x i s a n o d d i n t e g e r . \\ q : x + 1 i s a n o d d i n t e g e r . \\ p \lor q : E i t h e r x o r x + 1 i s a n o d d i n t e g e r . \\ (v) p : A n u m b e r i s d i v i s i b l e b y 2 . \\ q : A n u m b e r i s d i v i s i b l e b y 3 . \\ p \lor q : A n u m b e r i s d i v i s i b l e b y 2 o r 3 . \\ (v i) p : x = 2 i s a r o o t o f t h e e q u a t i o n 3 x^{2} - x - 10 = 0 . \\ q : x = 3 i s a r o o t o f t h e e q u a t i o n 3 x^{2} - x - 10 = 0 . \\ p \lor q : E i t h e r x = 2 o r x = 3 i s t h e r o o t o f t h e e q u a t i o n 3 x^{2} - x - 10 = 0 . \\ (v i i) p : H i n d i i s t h e o p t i o n a l p a p e r . \\ q : E n g l i s h i s t h e o p t i o n a l p a p e r . \\ p \lor q : E i t h e r H i n d i o r E n g l i s h i s o p t i o n a l p a p e r . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write the negation of the following simple statements:

i. The number 17 is prime.

ii. 2 + 7 = 6.

iii. Violets are blue.

iv. $\sqrt$ 5 is a rational number.

v. 2 is not a prime number.

vi. Every real number is an irrational number.

vii. Cow has four legs.

viii. A leap year has 366 days.

ix. All similar triangles are congruent.

x. Area of a circle is the same as the perimeter of the circle.

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) T h e n u m b e r 17 i s n o t p r i m e . \\ (i i) 2 + 7 \neq 6 . \\ (i i i) V i o l e t s a r e n o t b l u e . \\ (i v) \sqrt[]{5} i s n o t a r a t i o n a l n u m b e r . \\ (v) 2 i s a p r i m e n u m b e r . \\ (v i) E v e r y r e a l n u m b e r i s n o t a n i r r a t i o n a l n u m b e r . \\ (v i i) C o w d o e s n o t h a v e f o u r l e g s . \\ (v i i i) A l e a p y e a r d o e s n o t h a v e 366 d a y s . \\ (i x) T h e r e e x i s t s i m i l a r t r i a n g l e s w h i c h a r e n o t c o n g r u e n t . \\ (x) A r e a o f a c i r c l e i s n o t s a m e a s t h e p e r i m e t e r o f t h e c i r c l e . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Write the component statements of the following compound statements and check whether the compound statement is true or false:

i. 57 is divisible by 2 or 3.

ii. 24 is a multiple of 4 and 6.

iii. All living things have two eyes and two legs.

iv. 2 is an even number and a prime number.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) H e r e t h e g i v e n s t a t e m e n t i s t h e f o r m p \lor q w h i c h h a s t h e t r u t h v a l u e T w h e n e v e r e i t h e r \\ p o r q b o t h h a v e t h e t r u t h v a l u e T . \\ H e n c e, I t i s a t r u e s t a t e m e n t a n d i t s c o m p o n e n t s t a t e m e n t s a r e \\ p : 57 i s d i v i s i b l e b y 2 . (F a l s e) \\ q : 57 i s d i v i s i b l e b y 3 . (T r u e) \\ (i i) H e r e t h e g i v e n s t a t e m e n t i s t h e f o r m p \land q w h i c h h a s t h e t r u t h v a l u e T w h e n e v e r b o t h \\ p o r q h a v e t h e t r u t h v a l u e T . \\ H e n c e, I t i s a t r u e s t a t e m e n t a n d i t s c o m p o n e n t s t a t e m e n t s a r e \\ p : 24 i s a m u l t i p l e o f 4 . (T r u e) \\ q : 24 i s a m u l t i p l e o f 6 . (T r u e) \\ (i i i) I t i s a f a l s e s t a t e m e n t . S ince p \land q w h i c h h a s t h e t r u t h v a l u e F w h e n e v e r e i t h e r \\ p o r q o r b o t h h a v e t h e t r u t h v a l u e F . I t s c o m p o n e n t s t a t e m e n t s a r e \\ p : A l l l i v i n g t h i n g s h a v e t w o e y e s . (F a l s e) \\ q : A l l l i v i n g t h i n g s &thins \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(x + 2)}^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x} \\ P u t x + 2 = y \Rightarrow x = y - 2 \\ = \underset{y - 2 \to 0}{l i m} \frac{y^{\frac{1}{3}} - 2^{\frac{1}{3}}}{y - 2} = \underset{y \to 2}{l i m} \frac{y^{\frac{1}{3}} - 2^{\frac{1}{3}}}{y - 2} \\ = \frac{1}{3} . {(2)}^{\frac{1}{3} - 1} = \frac{1}{3} . 2^{\frac{- 2}{3}} [Using \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Find the component statements of the following compound statements:

i. Number 7 is prime and odd.

ii. Chennai is in India and is the capital of Tamil Nadu.

iii. The number 100 is divisible by 3, 11, and 5.

iv. Chandigarh is the capital of Haryana and U.P.

v. $\sqrt$ 7 is a rational number or an irrational number.

vi. 0 is less than every positive integer and every negative integer.

vii. Plants use sunlight, water, and carbon dioxide for photosynthesis.

viii. Two lines in a plane either intersect at one point or they are parallel.

ix. A rectangle is a quadrilateral or a 5-sided polygon.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) p : N u m b e r 7 i s p r i m e . \\ q : N u m b e r 7 i s o d d . \\ (i i) p : C h e n n a i i s i n I n d i a . \\ q : C h e n n a i i s t h e c a p i t a l o f T a m i l N a d u . \\ (i i i) p : 100 i s d i v i s i b l e b y 3 . \\ q : 100 i s d i v i s i b l e b y 11. \\ r : 100 i s d i v i s i b l e b y 5. \\ (i v) p : C h a n d i g a r h i s t h e c a p i t a l o f H a r y a n a . \\ q : C h a n d i g a r h i s t h e c a p i t a l o f U . P . \\ (v) p : \sqrt[]{7} i s a r a t i o n a l n u m b e r . \\ q : \sqrt[]{7} i s a n i r r a t i o n a l n u m b e r . \\ (v i) p : 0 i s l e s s t h a n e v e r y p o s i t i v e i n t e g e r . \\ q : 0 i s l e s s t h a n e v e r y n e g a t i v e i n t e g e r . \\ (v i i) p : P l a n t s u s e s u n l i g h t f o r p h o t o s y n t h e s i s . \\ q : P l a n t s u s e w a t e r f o r p h o t o s y n t h e s i s . \\ r : P l a n t s u s e c a r b o n - d i o x i d e f o r p h o t o s y n t h e s i s . \\ (v i i i) p : T w o l i n e s i n a p l a n e i n t e r s e c t a t o n e p o i n t . \\ q : T w o l i n e s i n a p l a n e a r e p a r a l l e l . \\ (i x) p : A r e c t a n g l e i s a q u a d r i l a t e r a l . \\ q : A r e c t a n g l e i s a 5 - s i d e s p o l y g o n . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{h \to 0} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} \\ = \underset{h \to 0}{l i m} \frac{\sqrt[]{x + h} - \sqrt[]{x}}{h} * \frac{\sqrt[]{x + h} + \sqrt[]{x}}{\sqrt[]{x + h} + \sqrt[]{x}} [R a t i o n a l i z i n g t h e denominator] \\ = \underset{h \to 0}{l i m} \frac{x + h - x}{h [\sqrt[]{x + h} + \sqrt[]{x}]} = \underset{h \to 0}{l i m} \frac{h}{h [\sqrt[]{x + h} + \sqrt[]{x}]} \\ = \underset{h \to 0}{l i m} \frac{1}{\sqrt[]{x + h} + \sqrt[]{x}} \\ T a k i n g limit, w e h a v e \\ \frac{1}{\sqrt[]{x} + \sqrt[]{x}} = \frac{1}{2 \sqrt[]{x}} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fourteen

Which of the following sentences are statements? Justify.

i. A triangle has three sides.

ii. 0 is a complex number.

iii. Sky is red.

iv. Every set is an infinite set.

v. 15 + 8 > 23.

vi. y + 9 = 7.

vii. Where is your bag?

viii. Every square is a rectangle.

ix. Sum of opposite angles of a cyclic quadrilateral is 180°.

x. sin²x + cos² x = 0.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} A s w e k n o w, a s t a t e m e n t i s a s e n t e n c e w h i c h i s e i t h e r t r u e o r f a l s e b u t n o t b o t h simultaneously . \\ (i) I t i s t r u e s t a t e m e n t . A s t r i a n g l e i s a c l o s e d p o l y g o n o f t h r e e s i d e s . \\ (i i) I t i s t r u e s t a t e m e n t . A s 0 + 0 i \\ (i i i) I t i s f a l s e s t a t e m e n t . A s s k y i s b l u e . \\ (i v) I t i s f a l s e s t a t e m e n t . N o, e v e r y s e t i s n o t a n infinite s e t . \\ (v) I t i s f a l s e s t a t e m e n t . 15 + 8 = 23 > 23 \\ (v i) I t i s n o t a s t a t e m e n t . B e c a u s e y + 9 = 7, t h e v a l u e o f y i s n o t g i v e n . \\ (v i i) I t i s n o t a s t a t e m e n t . I t i s a q u e s t i o n . \\ (v i i i) I t i s t r u e s t a t e m e n t . A s s q u a r e i s a rectangle w h o s e a l l s i d e s a r e e q u a l . \\ (i x) I t i s t r u e s t a t e m e n t . \\ (x) I t i s f a l s e s t a t e m e n t . A s s i n^{2} x + c o s^{2} x = 1 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{1}{2}}{l i m} \frac{4 x^{2} - 1}{2 x - 1} = \underset{x \to \frac{1}{2}}{l i m} \frac{{(2 x)}^{2} - {(1)}^{2}}{2 x - 1} = \underset{x \to \frac{1}{2}}{l i m} \frac{(2 x - 1) (2 x + 1)}{2 x - 1} \\ = \underset{x \to \frac{1}{2}}{l i m} 2 x + 1 \\ T a k i n g, w e h a v e \\ 2 * \frac{1}{2} + 1 = 1 + 1 = 2 \end{array}$

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