RD Sharma Solutions Class 12 Mathematics Chapter 15 FBQ
RD Sharma Solutions Class 12 Mathematics Chapter 15 FBQ
Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:59 PM IST
The CBSE schools recommend the student use the RD Sharma solution books to prepare for their tests and exams. The students might face numerous doubts while finding the answers for the FBQs. The concept of Tangents and Normals make it even harder to find accurate answers. As the FBQ section is where students tend to lose marks, extra practice is required to avoid it. This is where the RD Sharma Class 12th FBQ comes to help.
Answer: Equation of normal, Hint: Use equation of normal formula, Given: Here given the curve , To find: We have to find the equation of normal to the curve at Solution: Here given, Differentiating both side with respect to , we get Slope of normal at Now, equation of normal to the curve at is Hence, this is the required equation of normal.
Answer: Hint: Since the given curve touches axis. i.e. Given: … (i) To find: We have to find the value of for which the given curve touches axis. Solution: We know that, Since the given curve, touches the axis Differentiating equation (i) with respect to , we get Putting the value of in equation (i), we get Since Hence, the required value of are
Answer: Hint: Gradient is zero. i.e. Given: Given curve, To find: We have to find the points on the given curve at which the gradient is zero. Solution: Here … (i) Given that gradient is zero i.e. Differentiating equation (i), we get Since Putting the value in equation (i), we get Hence the points are and
Given: Given curve, and the line To find: We have to find the co-ordinates of a point on the given curve at which the normal is parallel to the line Solution: We have, …(i) Differentiating both side with respect to , we get The Slope of normal The line, Comparing this equation with the formula , we can get the slope . Here the slope of line is As the slope of parallel lines are equal, Therefore, Putting the value of in equation (i) Therefore the coordinate of the point is
Answer: Hint: First we find slope of curve then equate with given slope Given: Given curve, and tangent has slope To find: We have to find the coordinates of the point on the given curve where tangent has slope . Solution: We have, … (i) Differentiating equation (i) on both side with respect to , we get And tangent has slope Squaring on both sides, Substituting the value of in equation (i), we get Hence the required coordinate is
Answer: Slope of tangent = Hint: First put in equation , then find and . Given: Given curve, and To find: We have to find the slope of tangent to the given curve at Solution: Given, … (i) … (ii) Put in equation (i), we get Differentiating equation (i) and (ii) with respect to , we get Thus we get the slope of tangent
Answer: Hint: First we need to find their point of intersection. So solve these given equations. Given: Given curves, and To find: We have to find the angle of intersection of the given curves at Solution: Here … (i) … (ii) From equation (i) and (ii), we get Which gives Therefore points of intersection of the curves are and On differentiating equation (i) and (ii), we get Similarly, To find the angle of intersection at So, slope of tangent to curve at So tangent is parallel to axis. = slope of tangent to curve at which is not defined. So tangent is parallel to axis. Now, one tangent is parallel to axis and other is parallel to axis. Hence angle between tangents is right angle. i.e.
Answer: Hint: Use is the point where tangent to the curve crosses axis, then proceed next to find slope of tangent. Given: Given curve, To find: We have to find the slope of tangent where the curves crosses axis. Solution: Given equation of curve is, … (i) Let be the point where tangent to curve crosses axis. So the point is Differentiating equation (i), we get Hence slope of tangent at is .
Given: Given equation of curve, To find: We have to find the point where curve at cut axis. Solution: The given equation of curve, … (i) Differentiating equation (i), we get Slope of tangent to the curve =2 So equation of tangent is Above tangent line cuts the axis, where Hence the required point is
Answer: Slope of normal Hint: Slope of normal Given: Given equation of curve, To find: We have to find the slope of normal to the given curve at the point Solution: Given curve, … (i) On differentiating with respect to , we get
Answer: Hint: First find the equation of normal then comparing with Given: The given equation of curve, To find: We have to find the value of Solution: We have, … (i) Differentiating equation (i) with respect to , we get Then the equation of normal at point is … (ii) As the normal is of the form … (iii) Comparing equation (ii) and (iii), we get and Hence
Answer: Should be negative. i.e. Hint: First find the slope of normal to the given curve then compare with slope of Given: Given the equation of curve, To find: We have to find the set of if the line is normal to the given curve. Solution: Given, Differentiating with respect to , we get Thus slope of normal Which is always positive and it is given is normal Slope = So and are of opposite sign. Hence => should be negative. i.e
Answer: Hint: First find the slope of given curve at point , then compare with Given: Given curve, To find: We have to find , if the normal to the given curve at makes an angle with positive x-axis. Solution: Here we have, On differentiating with respect to , we get Slope of tangent at Therefore, slope of normal But [Given]
Answer: Equation of tangent is Hint: Tangent is parallel to X-axis so slope becomes 0 Given: Given curve To find: We have to find the equation of the tangent to the given curve that is parallel to X-axis Solution: Given tangent is parallel to X-axis so slope is 0 … (i) Here … (ii) On differentiate both side with respect to we get Substituting the value of in equation (i), we get From (ii) we get Hence equation of tangent is [ equation of tangent ]
First we will find the slope of line then evaluate with the slope of curve
If tangent is perpendicular to the line then
Given: Given curve where the tangent is perpendicular to the line To find: We have to find the co-ordinate of the point on the given curve. Solution: We have, … (i) Comparing equation (i) with equation Again, if the line is perpendicular to the tangent then Given curve, … (ii) On differentiating both side with respect to , we get Here slope of tangent is Substituting the value of in equation (ii), we get Hence the point is
Answer: Points are and Hint: If the slope of tangent is equal to ordinate of point that means Given: The slope of tangent to curve at a point is equal to ordinate of point. To find: We have to find the point Solution: We have, … (i) On differentiating with respect to , we get Now, we know that the given slope of tangent to the given curve at a point is equal to ordinate of point … (ii) Putting value of in equation (i), we get and [From equation (i)] Thus the two points are and
Answer: Equation of normal, Hint: Use equation of normal, Given: Given curve, To find: We have to find the slope of normal to the given curve at point Solution: Given equation is On differentiating both side with respect to , we get [Taking common from each term] Now, equation of normal is Substituting these value, we get i.e. Hence, is required equation of normal.
Answer: Hint: First find the slope of curve , then compare with slope with x-axis. i.e. slope Given: Given curve, , the tangent at which makes an angle of with x-axis. To find: We have to find the point on the given curve. Solution: We have … (i) Differentiate both side with respect to , we get … (ii) Also tangent makes an angle of with x-axis … (iii) From equation (ii) and (iii), we get Put value of in equation Hence required point is
Answer: Hint: When both curves touch then slope of both curves should be same. Given: Given curves, and touch each other. To find: We have to find the point where the given curves touch each. Solution: The curves, and For first curve say For second curve When both curves touch the slope both curves should be same Solving the quadratic equation, we get Now consider, For first curve, For second curve, …(here there should be 10 inplace of 13) Thus at both curves touch Now consider, For first curve, For second curve, Thus at both curves do not meet. But their tangent are parallel Hence the only point where both curves touch is
Answer: is the required equation of normal Hint: Equation of normal, Given: Given curve, To find: We have to find the equation of normal to the given curve at the origin. Solution: We have, On differentiating both side with respect to , we get At origin i.e. Here the equation of normal, Hence the required equation of normal is .
Chapter 15, Tangents and Normals, has three exercises, ex 15.1, ex15.2, and ex 15.3. RD Sharma Class 12 Solutions Tangents and Normals FBQ include the equation of the normal to the curve, coordinates of a point, slope of a tangent, angle of intersection curves, and slope of normal to the curve. There are around 20 FBQ questions asked in the mathematics textbook. Students can utilize the RD Sharma Class 12 Chapter 15 FBQ book to find and verify their answers.
Experts provide accurate answers for these FBQs in the RD Sharma solution books with a broad knowledge of their respective domains. The NCERT pattern is strictly followed for the CBSE board students to adapt it quickly. Apart from the solutions for the textbook, the RD Sharma Class 12th FBQ also contains additional practice questions for the students to work out. This makes them increase their speed in finding answers during the examinations.
You will not find FBQs challenging to solve once you commence your practice with the Class 12 RD Sharma Chapter 15 FBQ resource material. The questions are solved in simple methods using shortcuts to find the answers quickly. This saves time during examinations. Moreover, once the students achieve full marks in FBQs, they can easily cross their benchmark scores.
The added benefit of using the RD Sharma Class 12 Solutions FBQ is that it is available for free of cost on the Career 360 website. Hence, you need not worry about its affordability. You can also download this resource material onto your device to refer to whenever necessary. As no kind of monetary charge is required, every student can use it easily.
Class 12 RD Sharma Chapter 15 FBQ Solution are widely used, and hence questions for the public exams are likely to be taken from this book. Preparing with the RD Sharma Class 12th FBQ book will prepare the students for their public exams. Many students of the previous set have been benefitted from using the RD Sharma Class 12 Solutions Chapter 15 FBQ book. Download your own set of RD Sharma books and start preparing for your exams from today.
1.Which is the prescribed book for the students to clarify their doubts regarding the mathematics chapter 15 FBQ?
The students can utilize the RD Sharma Class 12th FBQ book to clarify their doubts regarding the portions in the mathematics chapter 15.
2.Is it possible to find RD Sharma Class 12 Solutions Chapter 15 FBQ for free?
It is a boon that the students can now download the RD Sharma solution from the Career 360 website for free of cost. Hence, the students need not pay money to utilize this best resource material.
3.How can I get the RD Sharma solution book from the Career 360 website?
Visit the Career 360 website and type for the name of the solutions that you require. For example, if you search for RD Sharma Class 12th FBQ, you can download the solutions for the FBQs.
4.Are the solutions given in the RD Sharma books verified?
Experts in the teaching field prepare all the solutions provided in the RD Sharma books. Therefore, the answers are accurate and verified for the welfare of the students.
5.Do the RD Sharma books contain solutions only for the questions given in the exercises?
The RD Sharma books provide solutions for the exercise questions, MCQs, FBQs, and VSA portions.