Question 15 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by Teachoo

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The slope of tangent to the curve x = t
^{
2
}
+ 3t – 8, y = 2t
^{
2
}
– 2t – 5 at the point (2, –1) is:

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Question 15
The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is:
(A) 22/7 (B) 6/7
(C) − 6/7 (D) −6
Finding Slope of tangent 𝒅𝒚/𝒅𝒙 .
𝒅𝒚/𝒅𝒙= (𝒅𝒚/𝒅𝒕)/(𝒅𝒙/𝒅𝒕)
𝒙 = t2 + 3t – 8
Differentiating w.r.t t
𝒅𝒙/𝒅𝒕= (𝑑(𝑡^2 + 3𝑡 −8))/𝑑𝑡
𝑑𝑥/𝑑𝑡 = 2t + 3
𝒚 = 2t2 − 2t − 5
Differentiating w.r.t t
𝒅𝒚/𝒅𝒕= (𝑑 (2𝑡2 − 2𝑡 − 5))/𝑑𝑡
𝑑𝑦/𝑑𝑡= 4t − 2
Now,
𝑑𝑦/𝑑𝑥= (𝑑𝑦∕𝑑𝑡)/(𝑑𝑥∕𝑑𝑡)
𝒅𝒚/𝒅𝒙= (𝟒𝒕 − 𝟐)/(𝟐𝒕 + 𝟑)
∴ Slope of Tangent = (4𝑡 − 2)/(2𝑡 + 3)
Now, we need to find value of Slope at (2, –1)
But we need to find value of t first
To find value of t,
We put 𝒙 = 2 & 𝒚 = –1 in the curve
For x
x = t2 + 3t – 8
2 = t2 + 3t − 8
t2 + 3t – 8 – 2 =0
t2 + 3t − 10 = 0
t2 + 5t – 2t − 10 = 0
t (t + 5) – 2 (t − 5) = 0
(t − 2) (t + 5) = 0
So, t = 2 & t = −5
For y
y = 2t2 – 2t – 5
–1 = 2t2 – 2t – 5
2t2 – 2t – 5 + 1 = 0
2t2 – 2t – 4 = 0
2(t2 – t – 2 ) = 0
t2 – t – 2 = 0
t2 – 2t + t – 2 = 0
t (t − 2) + 1(t − 2) = 0
(t + 1) (t – 2) = 0
So, t = −1 & t = 2
Since t = 2 is common in both parts
So, we will calculate Slope of Tangent at t = 2
Finding Slope of Tangent
𝒅𝒚/𝒅𝒙= (4𝑡 −2)/(2𝑡 + 3)
= (4 (2) − 2)/(2 (2) + 3)
= (8 − 2)/(4 +3)
= 𝟔/𝟕
Hence, the correct answer is (B)

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