Class 11th

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

The compound statement with 'And' is as follows 25 is a multiple of 5 and 8.

This is false statement because 25 is not a multiple of 8.

The compound statement with 'Or' is as follows 25 is a multiple of 5 or 8.

This is true statement because 25 is not a multiple of 8 but it is a multiple of 5.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) You watch television if and only if your mind is free.

(ii) You get an A grade if and only if you do all the homework regularly.

(iii) A quadrilateral is equiangular if and only if it is a rectangle.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) The statement p in the form 'if then' is as follows.

If you log on to the server, then you have password.

(ii) The statement q in the form 'if then' is as follows.

If it rains, then there is a traffic jam.

(iii) The statement r in the form 'if them' is as follows.

If you can access the website, then you pay a subscription fee.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) The negation of statement p is as follows.

There exists a positive real number x, such that x – 1 is not positive.

(ii) The negation of statement q is

There exists a CAT that does not scratch.

(iii) The negation of statement r is

There exist a real number x, such that neither x>1 nor x<1.

(iv) The negation of statement s is as follow.

There does not exist a number x, such that 0

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(i) False: As per the definition of a chord, it should intersect the circle at two distinct points.

(ii) False: The centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) True: The equation of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

If we put a = b = 1, then

x² + y² = 1, which is an equation of a circle

(iv) True: Given x > y. Multiplying by –1, –x < y.

(v) False: Since | | is a prime number, therefore √11$$ is irrational.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

(ii) q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.

(i) Let us consider an equilateral triangle ABC where $\angle A=\angle B=\angle C=60°,$ then triangle ABC is not obtuse angled since all its angles are equal

$\therefore$ The statement p is not true.

(ii) 1 lies between 0 and 2. Also, 1 satisfies the equation x² – 1 = 0

$\left[?1^2-1=1-1=0\right]$

$\therefore$ The statement q is not true.

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Let p: If x is an integer and x² is even, then x is also even.

Let q: x be an integer and x² be even

Let r: x is even

We have to prove, using the method of contra positive whether $q\Rightarrow r$ is true.

[i.e., its contra positive $\sim r\Rightarrow \sim q$ is true]

Let $\sim r$ be true i.e., r be false

Let us assume that x is not even (integer) i.e., x is odd.

$\therefore x=2m+1$ where m is an integer

$\therefore x^2{\left(2m+1\right)}^2=4m+1+4m$

$=4m^2+4m+1$

$=2\left(2m^2+2m+1\right)$ $=2t+1t=2m^2+2m$ is an integer.

$\Rightarrow$ x² is also odd $\Rightarrow$ x is not even

$\Rightarrow q\text{\hspace{0.17em}isfalse}\left[\text{\hspace{0.17em}Bydefof}q\right]\Rightarrow \sim q\text{\hspace{0.17em}istrue}$

$\therefore \sim r\Rightarrow \sim q$ is true

Thus, by the method of contrapositive $q\Rightarrow r$ is true

This is a  Mathematical Reasoning Solutions Type Questions as classified in NCERT Exemplar

Let p: If x is an integer and x² is even, then x is also even.

Let q: x be an integer and x² be even

Let r: x is even

We have to prove, using the method of contra positive whether $q\Rightarrow r$ is true.

[i.e., its contra positive $\sim r\Rightarrow \sim q$ is true]

Let $\sim r$ be true i.e., r be false

Let us assume that x is not even (integer) i.e., x is odd.

$\therefore x=2m+1$ where m is an integer

$\therefore x^2{\left(2m+1\right)}^2=4m+1+4m$

$=4m^2+4m+1$

$=2\left(2m^2+2m+1\right)$ $=2t+1t=2m^2+2m$ is an integer.

$\Rightarrow$ x² is also odd $\Rightarrow$ x is not even

$\Rightarrow q\text{\hspace{0.17em}isfalse}\left[\text{\hspace{0.17em}Bydefof}q\right]\Rightarrow \sim q\text{\hspace{0.17em}istrue}$

$\therefore \sim r\Rightarrow \sim q$ is true

Thus, by the method of contrapositive $q\Rightarrow r$ is true